Simple Harmonic Motion - NetBadi.com

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Description of SHM’s SHM spring pendulum, simple pendulum, physical pendulum SHM Other systems Syllabus for IIT-JEE Simple Harmonic Motion: Linear and Angular Simple Harmonic Motions.

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SIMPLE HARMONIC MOTION NetBadi.com - Eclusive

Motion of earth around the sun (B)

Motion of a piston in a cylinder (C)

Motion of a ball in a bowl (D)

Motion of liquid in a U-tube (E)

Some Periodic Motions: Fig. 1

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Motion of hands of a clock (A)

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Introduction Periodic Motion A motion which repeats itself after equal intervals of time is called a periodic motion. In a periodic motion, force is always directed towards a fixed point which may or may not be on the path of motion. Examples of periodic motion: • Circular motion is a periodic motion. • Rotation of earth about her own axis is periodic. • Motion of a pendulum is periodic motion. Any motion which repeats itself after regular interval of time is called periodic or harmonic motion and the time interval after which the motion is repeated is called time period. A periodic motion can be either rectilinear or closed or open curvilinear (Fig. 1)

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Oscillatory Motion: A particle is said to be in oscillatory motion or vibratory if it moves periodically back and forth about a stable equilibrium position. Motion of pendulum is oscillatory. Oscillatory motion is a periodic motion between two fixed limits. Every oscillatory motion is a periodic motion but every periodic motion is not always oscillatory. Oscillatory or vibratory motion is a constrained periodic motion between certain precisely fixed limits. The world is full of oscillatory motions. The motion of the balance wheel of a watch, the motion of a piston in an automobile engine, the motion of the needle of sewing machine or the motion of a ball in a bowl are all oscillatory motions. The motion of molecules in a solid around fixed lattice sites, the perpetual motion of atoms within molecules, vibrations of the prongs of a tuning fork and the motion of electric and magnetic fields in an electromagnetic wave are few other examples of vibratory motion. Note: Every oscillatory motion is periodic but every periodic motion is not oscillatory, e.g., the uniform circular motion of a conical pendulum or the motion of electrons around the nucleus in an atom are periodic but not oscillatory while the motion of a liquid in a U- tube or that of the free end of a clamped strip are oscillatory and hence periodic.

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Some oscillatory motions: Fig. 2

T

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T

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Simple Harmonic Motion: If in case of oscillatory motion a particle moves back and forth (or up and down) about a fixed point (called equilibrium position or mean position) through a force or torque (called restoring force or torque) which is directly proportional to the displacement but opposite in direction, the motion is called simple harmonic and abbreviated as SHM. Further if the displacement in case of SHM is linear, the motion is called ‘Linear SHM’ and if the displacement is angular it is known as angular SHM. The up and down motion of a partially submerged floating body or of a mass attached to a spring are examples of linear SHM while the motion of a simple pendulum or physical pendulum or torsional pendulum (twisting and untwisting motion of a body suspended from a wire) are examples of angular SHM.

T

2S

L cos T g

T

2S

m k

T

L g

2S

T

2S

I mgL

T

2S

I C

T

2S

1 c

Different types of pendulum Fig. 3 12.1 SIMPLE HARMONIC MOTION: When a body repeats its motion after regular time intervals we say that it is in harmonic motion or periodic motion. The time interval after which the motion is repeated is called the time period. If a body moves to and fro on the same path, it is said to perform oscillations. Simple harmonic motion

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(SHM) is a special type of oscillation in which the particle oscillates on a straight line, the acceleration of the particle is always directed towards a fixed point on the line and its magnitude is proportional to the displacement of the particle from this point. This fixed point is called the centre of oscillation. Taking this point as the origin and the line of motion as the X-axis, we can write the defining equation of a simple harmonic motion as Z 2 x

… (1)

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D

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Where Z 2 is a positive constant. If x is positive, a is negative and if x is negative, a is positive. This means that the acceleration is always directed towards the center of oscillation. If we are looking at the motion from an inertial frame a = F/m. The defining equation (1) may thus be written as F/m = - Z 2 x F = - mZ2 x F = - kx.

… (2)

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Or, Or,

We can use equation (2) as the definition of SHM. A particle moving on a straight line executes simple harmonic motion if the resultant force acting on it is directed towards a fixed point on the line and is

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proportional to the displacement of the particle from this fixed point. The constant k mZ 2 is called the force constant or spring constant. The resultant force on the particle is zero when it is at the centre of oscillation. The centre of oscillation is therefore, the equilibrium position. A force which takes the particle back towards the equilibrium position is called a restoring force. Equation (2) represents a restoring force which is linear. Figure (4) shows the linear restoring force graphically. Example 12.1 The resultant force acting on a particle executing simple harmonic motion is 4 N when it is 5 cm away from the centre of oscillation. Find the spring constant. Solution: The simple harmonic motion is defined as F = - k x. The spring constant is k =

4N F =  5cm x

4N 5 u102 m

80 N / m. .

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12.2 QUALITATIVE NATURE OF SIMPLE HARMONIC MOTION: Let us consider a small block of mass m in placed on a smooth horizontal surface and attached to a fixed wall through a spring as shown in figure (12.2). Let the spring constant of the spring be k. k m 0 0 Q3 Fig. 12.2

0

P

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The block is at a position O when the spring is at its natural length. Suppose the block is taken to a point P stretching the spring by the distance OP = A and is released from there. At any point on its path the displacement x of the particle is equal to the extension of the spring from its natural length. The resultant force on the particle is given by F = - kx and hence by definition the motion of the block is simple harmonic. When the block is released from P, the force acts towards the center O. The block is accelerated in that direction. The force continues to act towards O until the block reaches O. The speed thus increases all the time from P to O. When the block reaches O, its speed is maximum and it is going towards left. As it moves towards left from O, the spring becomes compressed. The spring pushes the block towards right and hence its speed decreases . The block moves to a point Q when its speed becomes zero. The 1 k OP 2

2

and when the

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potential energy of the system (block + spring), when the block is at P, is

1 2 k OQ . Since the block is at rest at P as well as at Q, the kinetic energy is zero 2 at both these positions. As we have assumed frictionless surface, principle of conservation of energy

block is at Q, it is

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1 1 2 2 k OP k OQ or OP = OQ. 2 2 The spring is now compressed and hence it pushes the block towards right. The block starts moving towards right, its speed increases up to O and then decreases to zero when it reaches P. Thus, the particle oscillates between P and Q. As OP = OQ, it moves through equal distances on both sides of the centre of oscillation. The maximum displacement on either side from the center of oscillation is called the amplitude. Example 12.2 A particle of mass 0.50 kg executes a simple harmonic motion under a force F = –(50 N/m)x. If it crosses the center of oscillation with a speed of 10 m/s, find the amplitude of the motion. Solution: The kinetic energy of the particle when it is at the center of oscillation is

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gives

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1 1 2 mv 2 0.05 kg 10 m / s = 25 J. 2 2 The potential energy is zero here. At the maximum displacement x = A, the speed is zero and hence the E

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kinetic energy is zero. The potential energy here is

1 k A2 . As there is no loss of energy,, 2

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1 2 kA 25 J 2 The force on the particle is given by F = - (50 N/m)x. Thus, the spring constant is k = 50 N/m.

… (i)

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1 50 N / m A2 25 J . or A = 1 m. 2 12.3 EQUATON OF MOTION OF A SIMPLE HARMONIC MOTION: Consider a particle of mass m moving along the X – axis. Suppose, a force F = – kx acts on the particle where k is a positive constant and x is the displacement of the particle from the assumed origin. The particle then executes a simple harmonic motion with the center of oscillation at the origin. We shall calculate the displacement x and the velocity v as a function of time. Equation (i) gives

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F m

The acceleration of the particle at any instant is D Where

Z

Thus,

dv dt

Z 2 x

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Z 2 x

… (12.3)

dv dx dx dt

Z 2 x

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dv dx

v

k x m

k/m .

Or Or,



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Suppose the position of the particle at t = 0 is x0 and its velocity is v0. Thus, At t = 0, x = x0 and v = v0.

Z 2 x

vdv Z 2 x dx. The velocity of the particle is v0 when the particle is at x = x0. It becomes v when the displacement becomes x. We can integrate the above equation and write v

³

x

v dv

v0

³

Z 2 x dx.

x0

  Z 2 x 2  x02 .

Z

2

Z x

x02

2

2

0

x

ª x2 º Z 2 « » ¬« 2 ¼» x

0

v02  Z 2 x02  Z 2 x 2

or,

v2

or,

§ v2 v Z ¨ 02 ¨ ©Z

· 2 x ¸¸ ¹

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v

v02

v

ª v2 º « » ¬« 2 ¼» v

or,

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2 2 or, v  v0

or,

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Or,

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§ v 20 2· 2 ¨ 2  x0 ¸  x ©Z ¹

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Writing

2

§ v0 · 2 ¨ Z ¸  x0 © ¹

A2

… (12.4)

The above equation becomes v Z A2  x 2

… (12.5)

dx dx Z dt. Z A2  x 2 or, 2 A  x2 dt At time t = 0 the displacement is x = x0 and at time t the displacement becomes x. The above equation We can write this equation as

x

can be integrated as

³ x0

dx A2  x 2

t

³ Z dt 0

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or,

1 Writing sin

Zt

x0 A

t

sin 1

or,

0

G , this becomes sin 1

x A

x x  sin 1 0 A A

Zt  G

x = A sin Zt  G

or,

Zt

… (12.6)

dx A Z cos Z t  G . … (12.7) dt 12.4 TERMS ASSOCIATED WITH SIMPLE HARMONIC MOTION: (a) Amplitude:

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The velocity at time t is v

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x

ª 1 x º «sin A » ¬ ¼ x0

Equation (12.6) gives the displacement of a particle in simple harmonic motion. As sin Z t  G can

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take values between – 1 and + 1, the displacement x can take values between – A and + A. This gives the physical significance of the constant A. It is the maximum displacement of the particle from the center of oscillation, i.e., the amplitude of oscillation. (b) Time Period: A particle in simple harmonic motion repeats its motion after a regular time interval. Suppose the particle is at a position x and its velocity is v at a certain time t. After some time the position of the particle will again be x and its velocity will again be v in the same direction. This part of the motion is called one complete oscillation and the time taken in one complete oscillation is called the time period. T. Thus, in figure (12.4) Q to P and then back to Q is a complete oscillation, R to P to Q to R is a complete oscillation, O to P to Q to O is a complete oscillation etc. Both the position and the velocity (magnitude as well as direction) repeat after each complete oscillation.

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We have, x = A sin Z t  G .

If T be the time period, x should have same value at t and t + T. Thus, sin Z t  G

sin[Z t  T  G ].

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Now the velocity is (equation 12.7) v = AZ cos Zt  G As the velocity also repeats its value after a time period, cos Zt  G

cos[Z (t  T )  G ] .

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Both sin Zt  G and cos Zt  G will repeat their values if the angle Zt  G increases by 2S or its multiple. As T is the smallest time for repetition.

Z t T G

or,

ZT

or,

T

Zt  G  2S

2S 2S

Z

.

Remembering that Z 2S

k / m , we can write for the time period.

m ... (12.8) k Where k is the force constant and m is the mass of the particle.

T

Z

2S

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Example 12.3 A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by a spring of spring constant 80 N/m. Find the time period. Solution m k

2S

200 u103 kg 80 N / m

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2S

The time period is T

1 2S

Z 2S

... (12.9)

k m

... (12.10).

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1 T

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2S u 0.05 s 0.31 s. (c) Frequency and Angular Frequency: The reciprocal of time period is called the frequency. Physically, the frequency represents the number of oscillations per unit time. It is measured in cycles per second also known as hertz and written in symbols as Hz. Equation (12.8) shows that the frequency is

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The constant Z is called the angular frequency.. (d) Phase:

The quantity I

Z t  G is called the phase. It determines the status of the particle in simple harmonic

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motion. If the phase zero at a certain instant, x = A sin Z t  G = 0 and v = A Z cos Z t  G = A I Z .

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This means that the particle is crossing the mean position and is going towards the positive direction. If the phase is S / 2 , we get x = A, v = 0 so that the particle is at the positive extreme position. Figure (12.5) shows the status of the particle at different phases. We see that as time increases the phase increases. An increases of 2S brings the particle to the same status in the motion. Thus, a phase Zt  G is equivalent to a phase Zt  G  2S . Similarly a phase change of 4S , 6S , 8S ………… etc, are equivalent to no phase change. Figure (12.6) shows graphically the variation of position and velocity as a function of the phase.

S

S I

S

I S

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(e) Phase constant: The constant G appearing in equation (12.6) is called the phase constant. This constant depends on the choice of the instant t = 0. To describe the motion quantitatively, a particular instant should be called t = 0 and measurement of time should be made from this instant. This instant may be chosen according to the convenience of the problem. Suppose we choose t = 0 at an instant when the particle is passing through its mean position and is going towards the positive direction. The phase Zt  G should then be zero. As t = 0 this means G will be zero. The equation for displacement can then be written as x = – A sin Zt . If we choose t = 0 at an instant when the particle is at its positive extreme position, the phase is S / 2 at

x = A sin Z t  G

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this instant. Thus Zt  G S / 2 and hence G S / 2 . The equation for the displacement is x = A sin Zt  S / 2 . or, x = A cos Zt . Any instant can be chosen as t = 0 and hence the phase constant can be chosen arbitrarily. Quite often we shall choose G 0 and write the equation for displacement as x = A sin Zt . Sometimes we may have to consider two or more simple harmonic motions together. The phase constant of any one can be chosen as G 0 . The phase constants of the rest will be determined by the actual situation. The general equation for displacement may be written as

S § · A sin ¨ Zt   G ' ¸ = A cos Zt  G c . 2 © ¹

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Where G ' is another constant. The sine form and the cosine form are, therefore, equivalent. The value of phase constant, however, depends on the form chosen. Example 12.4: A particle execute simple harmonic motion of amplitude Along the X –axis. At t = 0, the position of the particle is x = A/2 and it moves along the positive x – direction. Find the phase constant G if the equation is written as x Solution:

A sin Zt  G . At t = 0, x = A/2. Thus, A/2 = A sin G or, sin G =1/2 or,, G = S / 6 or 5.

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We have x

A sin Zt  G .

The velocity is v

S

6

A Z cos Zt  G . At t = 0, v = A Z cos G . Now

3 5S and cos 2 6

w cos

dx dt



3 2

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As v is positive at t = 0, G must be equal to S / 6 12.5 SIMPLE HARMONIC MOTION AS A PROJECTION OF CIRCULAR MOTION: Consider a particle P moving on a circle of radius A with a constant angular speed Z (figure 12.7). Let us take the center of the circle as the origin and two perpendicular diameters as the X and Y – axes. Suppose the particle P is on the X – axis at t = 0. The radius OP will make an angle T Zt with the X – axis at time t. Drop perpendicular PQ on X – axis and PR on Y – axis. The x and y – coordinates of the particle at time t are x = OQ = OP cos Zt or, x = A cos Zt … (12.11) And y = OR = OP sin Zt or, y = A sin Zt . ... (12.12)

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Equation (12.11) shows that the foot of perpendicular Q executes a simple harmonic motion on the X – axis. The amplitude is A and the angular frequency is Z . Similarly, equation (12.12) shows that the foot of perpendicular R executes a simple harmonic motion on the Y – axis. The amplitude is A and the angular frequency is Z . The phases of the two simple harmonic motions differ by S / 2 (remember

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cos Z t sin(Zt  S / 2)). Thus, the projection of a uniform circular motion on a diameter of the circle is a simple harmonic motion. 12.6 ENERGY CONSERVATION IN SIMPLE HARMONIC MOTION: Simple harmonic motion is defined by the equation F = - kx. The work done by the force F during a displacement from x to x + dx is dW = F dx = – kx dx the work done in a displacement from x = 0 to x is

1  kx 2 . 2

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x

W

³

kx dx

0

Let U (x) be the potential energy of the system when the displacement is x. As the change in potential energy corresponding to a force is negative of the work done by this force 1 2 kx . 2 Let us choose the potential energy to be zero when the particle is at the center of oscillation x = 0. W

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U x U 0

1 2 kx . 2 This expression for potential energy is same as that for a spring and has been used so far in this chapter. 0 and U x

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Then U 0

k ,k m

Z

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As

m Z2

1 m Z 2 x2 ... (12.13) 2 The displacement and the velocity of a particle executing a simple harmonic motion are given by

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We can write U x

x

A sin Zt  G and v

A Z cos Zt  G . 1 m Z 2 x2 2

The potential energy at time t is, therefore, U 1 m Z 2 A2 sin 2 Zt  G . 2

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1 1 m Z 2 A2 cos2 Z t  G . m v2 2 2 The total mechanical energy at time t is E = U + K And the kinetic energy at time t is K

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1 m Z 2 A2 ªsin 2 Z t  G  cos 2 Zt  G º ¬ ¼ 2

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1 m Z 2 A2 . … (12.14) 2 We see that the total mechanical energy at time t is independent of t. Thus, the mechanical energy remains constant as expected. As an example, consider a small block of mass m placed n a smooth horizontal surface and attached to a fixed wall through a spring constant k (figure 12.8).

When displaced from the mean position (where the spring has its natural length), the block executes a simple harmonic motion. The spring is the agency exerting a force F = - kx on the block. The potential energy of the system is the elastic potential energy stored in the spring. 1 1 m v02 m Z 2 A2 . 2 2 All the mechanical energy is in the form of kinetic energy here. As the particle is displaced away from the mean position, the kinetic energy decreases and the potential energy increases. At the extreme positions x = r A, the speed v is zero and the kinetic energy decreases to zero. The potential energy is

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At the mean position x = 0, the potential energy is zero. The kinetic energy is

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increased to its maximum value

1 k A2 2

1 m Z 2 A2 . All the mechanical energy is in the form of 2

potential energy here.

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Example: 12.5 A particle of mass 40 g executes a simple harmonic motion of amplitude 2.0 cm. If the time period is 0.20 s, find the total mechanical energy of the system. Solution: 2

The total mechanical energy of the system is E 2 S 2 40 u10 2 kg =

0.20 s

2.0 u 102 m

2

2

7.9 u102 J .

10

1 § 2S · 2 1 A m Z 2 A2 = m ¨ 2 © T ¹¸ 2

2 S m A2 T2

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12.7 ANGULAR SIMPLE HARMONIC MOTION: A body free to rotate about a given axis can make angular oscillations. For example, a hanging umbrella makes angular oscillations when it is slightly pushed aside and released. The angular oscillations are called angular simple harmonic motion if. (a) There is a position of the body where the resultant torque on the body is zero, this position is the mean position T 0 . (b) When the body is displaced through an angle from the mean position, a resultant torque acts which is proportional to the angle displaced, and (c) This torque has a sense (clockwise or anticlockwise) so as to bring the body towards the mean position. If the angular displacement of the body at an instant is T , the resultant torque acting on the body in angular simple harmonic motion should be T = - k T . If the moment of inertia is I, the angular acceleration is D d2T

or,

Z 2 T



k I

T.

… (12.15)

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dt 2

T I

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Where, k/I Z Equation (12.15) is indentical to equation (12.3) except for the symbols. The linear displacement x in (12.3) is replaced here by the angular displacement T . Thus, equation (12.15) may be integrated in the similar manner and we shall get an equation similar to (12.6), i.e.,

T

T0 sin Z t  G

… (12.6)

Where T 0 is the maximum angular displacement on either side. The angular velocity at time t is given by, :

dT dt

T 0 Z cos Z t  G .

… (12.17)

T

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The time period of oscillation is

… (12.18)

1 T

… (12.19)

2S

Z

I k

2S

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And the frequency of oscillation is 1 2S

k I

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v

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The quantity Z k / I is the angular frequency.. Example: 12.6 A body makes angular simple harmonic motion of amplitude S / 10 rad and time period 0.05 s. If the body is at a displacement T S / 10 rad at t = 0, write the equation giving the angular displacement as a function of time. Solution: Let the required equation be T Here T 0 = amplitude

S 10

T0 sin Zt  G .

rad

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2S T

Z

2S  40 S s 1 0.05 s §S · 1 ª º ¨ 10 rad ¸ sin ¬ 40 S s t  G ¼ . © ¹

At t = 0, T

S /10 rad. Putting in (i)

§S · ¨ 10 ¸ sin G © ¹

10

or, sin G 1 Thus by (i),

or, G

S / 2.

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S

… (i)

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So that T

Sº §S · §S · ª 1 1 ¨ 10 rad ¸ sin «(40 S s ) t  2 » = ¨ 10 rad ¸ cos ª¬ (40 S s ) t º¼ . © ¹ © ¹ ¬ ¼ Energy:The potential energy is U

1 2 kT 2

1 I Z 2 T 2 and the kinetic energy is K 2

1 I :2 . 2

1 1 I Z 2 T 2  I :2 . 2 2

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The total energy is E = U + K =

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T

T

Using

E

T0 sin Z t  G

1 1 I Z 2 T 02 sin 2 Z t  G  I T 02 Z 2 cos2 Z t  G 2 2

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1 I Z 2 T 02 . … (12.20). 2 Composition of two ‘parallel’ SHM’s We have already seen that, for a particle performing SHM, the restoring force is given by, F = - kx. Now, if the particle is acted upon by two restoring forces along same line instead of one, then the resultant motion can be studied using superposition principle. Hence, if F1 and F2 are the two forces acting on the particle, then using superposition principle, we first assume that only the force F 1 is acting on the particle and find the corresponding position of the particle as, x1 = A1 sin Z t  I1

... (73)

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Then we assume that only the force F2 acts on the particle and find the corresponding position of the particle as, x2 = A2 sin Z t  I2

... (74)

Here, we assume the frequency of the oscillation to be same. The resultant position of the particle is then given by, x = x1 + x2 ... (75) ?

x

A1 sin Z t  I1  A2 sin Z t  I2

?

x

A1 {sin Z t.cos I1  cos Z t.sin I1}  A2 {sin Z t.cos I2  cos Z t.sin I2 }

sin Z t [ A1 cos I1  A2 cos I2 ]  cos Z t [ A1 sin I1  A2 sin I2 ]

Let, A1 cos I1  A2 cos I2 = X

… (76)

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And A1 sin I1  A2 sin I2 = Y

… (77)

? x X sin Zt  Y cos Zt Let, X2 + Y2 = A2 We can write,

… (78)

A cos T & Y

X

tan T

Ÿ

A sin T ½ ° ¾ ° ¿

Y X

… (79)

?

x

A sin Z t cos T  A cos Z t sin T

?

x

A sin Z t  T

... (80)

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Equation (80) shows that the resultant of two SHM’s along the same path is itself a SHM. The amplitude of this resultant SHM is A and it’s relation with A1 and A2 is given by, A12  A22  2 A1 A2 cos I1  I2

A

... (81)

A1 sin I1  A2 sin I2 A1 cos I1  A2 cos I2

Y X

tan T

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This relation can be easily derived using equations (76) to (78). Also, the initial phase T , is given by,, ... (82)

et B

The difference I1  I2 in equation (81) is called the phase difference between the two motions. If I1

I2 , then the two simple harmonic motions are in phase and we get, A = A1 + A2 &

tan T

tan I1

If I1  I2

tan I2 . In this case, the amplitude is the maximum possible amplitude.

S then the two SHM’s are out of phase and we get, A

A1  A2 and tan T

tan I2

If A1 = A2, then A = 0 and the particle does not oscillate.

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For any value of I1  I2 other than 0 and S the resultant amplitude is between A1  A2 and A1 + A2. Vector Combination of Two Simple Harmonic Motion: Suppose the two individual SHM s are represented by A1 sin Z t

A2 sin Z t  G .

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x1 x2

w

)& The two SHM s can be represented by vectors with magnitude A1 and A2 respectively. The resultant A of these two vectors represent the resultant SHM.

A12  A22  2 A1 A2 cos G

and

tan 

w

)& | A|

Case: If the two SHM’s are perpendicular, then

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A2 sin G A1  A2 cos G

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x

A1 sin Z t

... (1)

y

A2 sin Z t  G

... (2)

y2





A22

2

ª x x2 º A2 « cos G  sin G 1  2 » «¬ A1 A1 »¼

§ y2 2 x y x2 x2 · 2 cos G  2 cos 2 G ¨¨1  2 ¸¸ sin G Ÿ 2  A2 A1 A2 A1 A1 ¹ ©

2 x y cos G A1 A2

sin 2 G 

x2

sin 2 G

A12

sin 2 G

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A12

A12

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§ y · y Ÿ¨  cos G ¸ © A2 A1 ¹ x2

x2

A2 sin Z t cos G  cos Z t sin G

By equation (2), y

Ÿ

1

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x By equation (1), sin, A , cos Z t 1

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The is equation of an ellipse and hence the particle moves in ellipse. Here note that the above cases are valid only when the frequencies of the two SHMs are equal. 15.3 A Few Graphs Concerning SHM (a) Graph of Displacement Versus Time Assuming the displacement equation to be x = a sin Z t , clearly the graph will be a sine curve. (Fig 15.3)

Putting x = 0, we have sin ( Z t ) = 0 n S , where n  I Ÿ tn

w

Ÿ Zt

nS

Z

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Putting n = 0, 1, 2, 3 etc, we get t0

0, t1

S , t2 Z

2S

Z

,

t3

3S

Z

etc, and putting x = r a (i.e., maximum displacement) we get sin

Z t r1 . Ÿ Z tn

S

where n  I . 2 Putting n = 0, 1, 2 etc, we get.

t0

2n  1

S , t1 2Z

3S , t2 2Z

5S 2 Z , etc

14

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(b) Graph of Velocity versus Time. Assuming x = a sin Z t , we get on differentiating a Z cos Z t .

v

Putting v = 0; we get cos Z t = 0

Ÿ

Z tn

2n  1

S 2

; where n  I

om

dx dt

Putting n = 0, 1, 2 etc. 3S , t3 2Z

5S 2 Z etc.

Putting v = r a Z (velocity amplitude), we get cos Z t

Z tn

Ÿ

tn

n S where n  I

nS

Z

, putting n = 0, 1, 2 etc, we get t0 x

S ;t Z 2

2 /

/

3 /2

/2

et B

O

0 : t1

time (t)

The graph is a cosine curve (Fig. 15.4) (c) Graph of Acceleration Versus Time:

Assuming x = a sin Z t , we have f = - Z 2 x Z 2 a sin Z t

.N

Putting f = 0, the instants at which f = 0, can be found; sin Z t = 0. Ÿ Z tn n S Putting n = 0,1, 2,……….. t0

0, t1

w

?

S , t Z 2

2S

Z

etc.

Putting f = # Z 2 a we have sin Z t

Z tn

w

Ÿ

r 2n  1

r1 (acceleration amplitude).

S 2

w

Putting n = 0, 1, 2, etc. t0

S 2Z

, t1

3S , t2 2Z

2S

ad

Ÿ

r1 .

i.c

S , t1 2Z

t0

5S 2 Z , etc.

Clearly the graph is a sine curve (figure 15.5)

15

Z

etc. .

om

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(d) Graph of Acceleration Versus Displacement

ad

i.c

Since acceleration (f) = - Z 2 (x). The variation is linear (figure. 15.6)

et B

(e) Graph of Velocity Versus Displacement we have seen earlier that v Ÿ

v

2

Z

2

2

a x

2

Z a 2  x2

Ÿ

v2

Z2

a2  x2 Ÿ

x2 a2



v2

Z 2 a2

1.

w

.N

The graph is an ellipse. (Figure. 15.7).

w

(f) Graph of Variation between Displacement (x) and Phase Z t  I .

w

From the equation x = a sin Z t  I x = 0 at Z t  I = n S where n  I . And x

? The graph is a sine curve (Fig. 15.8)

16

r a at Z t  I

2n  1

S 2

, where n  I .

om

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(g) Graph of Velocity Versus Phase.

v = 0 at Z t  I

S

r a Z at Z t  I

where n  I . v 2 The graph is shown in figure 15.9.

n S where n  I

et B

ad

2n  1

i.c

From the equation v = a Z cos Z t  I .

.N

Example:1 A particle executes SHM between two points separated by 10 cm. If the force experienced by it be 2N at a displacement of 1 cm from the mean position, then find the maximum force experienced by it. Solution: 10 5 cm. 2 For a particle executing SHM the force experienced is given by F = - kx ... (1). And Fmax = - k (xmax) ... (2).

w

Evidently, the maximum displacement

w

Dividing (2) by (1)

Fmax F

xmax F Ÿ max x 2

5 1

Ÿ Fmax

10 N .

w

Example 2 A particle executing SHM oscillates between two fixed points separated by 20 cm. If its maximum velocity be 30 cm/s, find its velocity when its displacement is 5 cm from its mean position. Solution: Evidently, the max, displacement =

20 2

10 cm

For a particle executing SHM the speed at a displacement ‘x’ is given by.

17

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v Z a2  x2

a a x 2

30 u

2 3 3

... (2) v vmax

Ÿ

2

10 2

Ÿ

2

10  5

10

2 3 3

5 3

om

v vmax v

Za

vmax

and

... (1)

20 3 cm/s.

Example 3: A particle executes SHM with an angular frequency Z

i.c

initially, then find the instants, when it is at a distance

4 S rad/sec. If it is at its extreme position

3 times, its amplitude from the mean position. 2

x

a sin 4 S t  I

Z

ad

Solution: Let a be the amplitude and I the phase constant for the oscillation. The displacement equation can be written as 4 S rad / sec

Since at t = 0; x = + a.

S

S

ŸI

.

et B

4S 0  I

?

2

2

Now, let tn be the instant, when the particle’s displacement is 3a Ÿ a cos 4 S ta 2 Taking only the positive values for ‘t’. t1

.N

Evidently r

1 s, t 2 24

5 s, t3 24

7 s , t4 24

11 s, t5 24

4 S tn

nS r

S 6

3a , nth time (where n  I ). 2

Ÿ tn

n2 1 r . 4 24

13 s, etc. 24

w

w

w

Example 4 A particle executes SHM with amplitude 8 cm and a frequency 10 sec-1. Assuming the particle to be at a displacement 4 cm, initially, in the positive direction, determine its displacement equation and the maximum velocity and acceleration. Solution Given a = 8 cm and Z = 2 S n = 20 S rad/sec. Let the phase constant be I . The displacement equation can be written as x = 8 sin (20 S t + I ) Given, at t = 0; x = 4 cm ? Ÿ

4 8sin 20 S 0  I sin I

1 2

ŸI

S 6

18

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S· § ? The displacement equation becomes x 8 sin ¨ 20 S t  6 ¸ . © ¹ Differentiating the above equation w.r.t time ‘t’

r 160 S cm / s [ when cos 20 S t  S / 6)

? vmax

om

S· § v 160 S cos ¨ 20 S t  ¸ 6¹ ©

dx dt

r1 ] .

Differentiating once again w. r. t time t. d t2

f

3200 S 2 sin 20 S t  S / 6

i.c

d2 x

Given | vmax | a Z

20 cm / s

And | f max | a Z 2

100 S cm / s 2

… (1). … (2).

5 S rad/sec.

et B

Dividing equation (2) by (1) we get Z ? Time period T

ad

? f max # 3200 S 2 cm / s [when sin (20 S t  S / 6) r1 ]. Example 5 If the maximum speed and acceleration of a particle executing SHM be 20 cm/s and 100 S cm/s2, find the time period of oscillation. Solution

2S

Z

2S 5S

0.4 sec.

.N

Example 6 If a particle executing SHM possesses a speed of 10 cm/s, when at mean position, find its speed when at a displacement of half the amplitude. Solution Let ‘a’ be the amplitude of oscillation. From velocity v

Z a2  x2

Z 2 a 2  x2

w

2 Squaring, we get v

Given at x = 0; v = 10

w

?

100 Z 2 a 2

w

And let at x = Ÿ

v c2

a ;v 2

… (1). v'

§ 3a 2 · ¸¸ © 4 ¹

Z 2 ¨¨

… (2).

§ v' · Dividing equation (2) by (1), we get ¨ ¸ © 10 ¹ Ÿ

v'

3 u10 2

2

3 4

5 3 cm / s .

19

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Example 7 Two particles execute SHM with same frequency and amplitude along the same straight line. They cross each other, at a point midway between the man and the extreme position. Find the phase difference between them. Solution

i.c

om

Let ‘a’ be the amplitude and I1 and I2 their respective phases at any instant, when they cross each other.

Then their displacement equations can be written as x = a sin (phase) for the two particles we have a sin I1 and

a a sin I2 2

S

Taking the least value, I1

Ÿ

6

ad

a 2

. And taking the next possible value  I1

I2

5S 6

S 6

a Z cos I2

, then I2

Ÿ

cos I2

& v2

cos S r I1

5S 7S or 6 6

.N

? If I1

et B

& Moreover, from v a Z cos (phase) & & & If v1 and v 2 be the respective velocities of the two particles, then v1

i.e., a Z cos I1

I2 .

? Required phase difference

I2  I2 Ÿ

2S rad. 3

w

Note: The phase difference can be in general

2S  2 n S ; n I  3

w

w

Example: 8 A particle executes SHM with time period of 2 sec; find the time taken by it to move from one extreme position to the nearest mid point between mean position and extreme position. Solution Let ‘a’ be the amplitude of oscillation then evidently Z The displacement equation can be written as x

2S T

S rad/sec.

a sin S t  I

Let, for simplicity, the time be reckoned from the instant when the particle was as its extreme position. Thus, at t = 0; x = a a

a sin I

Ÿ I

S 2

20

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Thus , x

S· § a sin ¨ S t  ¸ 2¹ ©

Let at t

t1  ve x

S· § ¨ S t1  2 ¸ © ¹

Ÿ

5S 6

S

Ÿ S t1

Ÿ

3

1 s. 3

t1

S· § y1 10 sin ¨ 4S t  ¸ and y2 2¹ © Solution y1

5 [sin 2 S t  8 cos 2 S t ] compare their amplitudes.

ad

For the equation

i.c

Example 9 If two SHMs are represented by

S· § 10 sin ¨ 4S t  ¸ the amplitude a = 10 units and for the 1 2¹ ©

5 [sin 2 S t  8  cos 2 S t ] . Multiplying and dividing by

et B

equation y2

ª 1 15 «sin 2 S t u  cos 2 S t 3 «¬

y2

om

S· § a sin ¨ S t1  ¸ 2¹ ©

a 2

?

a 2

8º » 3 »¼

15sin 2 S t  I where tan I

1 8

3

8

Whose amplitude a2 is obviously 15 units. a1 a2

.N

? The required ratio

10 15

2 3.

Example 10 A particle is executing SHM with amplitude ‘a’ and maximum velocity v0. Find the displacement when

w

its velocity is

3v , 2

w

w

Solution From the equation for speed v2

Z 2 a 2  x2

v02

Z2 a2

3 v02 4

And

Z 2 a 2  x2

Dividing (1) by (2), we get Ÿ

x2 a

2

1 4

... (1)

Ÿ

... (2) a2

4 3

a2  x2

x

r

or

3 4

a . 2

21

1

x2 a2

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maximum speed vmax

om

Example 11 A small particle moves along a horizontal circle of radius 5 cm with a uniform angular speed of 6 S rad/sec. The circle stands in front of a vertical wall, whereon a normal beam of light casts a sharp shadow. What is the maximum speed and maximum acceleration of the shadow? Also find the range between whose limits the shadow moves. Solution: The situation is similar to the one discussed above. The shadow executes SHM upon the wall. The

Z a 6 S u 5 30 S cm / s and the maximum acceleration f max 6S

2

u 5 180 S 2 cm / s 2

Z2 a

i.c

The shadow moves along a straight line over the projection of circle upon the wall and its range is the diameter i.e., 10 cm. Example 12 Find the position (s) of a particle executing SHM, when its K.E. is Solution 1 m Z 2 a 2  x2 2

ad

From equation (8) K = E or 4K = E 4

K

1 th its T.E. 4

and from equation (11) E =

1 m a 2 Z 2 from the problem 2

et B

ª1 º 1 3a 4 « mZ 2 (a 2  x 2 ) » mZ 2 a 2 Ÿ x r . 4 a 2  4 x 2 a 2 Ÿ 4 x 2 3a 2 Ÿ ¬2 ¼ 2 2 Example 13 Find the ratio of K.E. to P.E. for a particle executing SHM, when it is midway between the mean position and extreme position. Solution a 2

r

Now

K U

w

w

.N

Given x

K U

w

?

1 m Z 2 a 2  x2 2 1 m Z 2 x2 2

§ 2 a2 ¨¨ a  4 © 2 a 4

· ¸¸ ¹

a2  x2 x2

3 1.

Example 14 What fraction of total energy is carried by a particle executing SHM, in the form of kinetic energy when the displacement of the particle is

1 th the maximum displacement, from one of its extreme 4

position? Solution

22

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Displacement of the particle from its mean position is a 

a 4

3a 4

x (say)

Now, fraction of T.E. in the form of K.E.

Ÿ

1 m Z 2 a2  x2 2 1 m Z 2 a2 2

§ x· 1u  ¨ ¸ ©a¹

2

K E

Ÿ

§3· 1 ¨ ¸ ©4¹

2

7 . 16

om

K E

Example 15 A particle of mass 200 g is executing SHM with an amplitude 20 cm. If its kinetic energy be 32×10–3

i.c

joule initially, determine its displacement equation, assuming the phase constant to be

S

4

.

Solution Let Z be the angular frequency, for the SHM, then displacement equation can be written as

§S · 20sin ¨ ¸ ©4¹

20

Ÿ

x2

2 400 2

[Putting t = 0]

et B

Now, initially x Ÿ

ad

x = 20 sin Z t  S / 4 where t is in sec. and x is in cm.

200

Now, kinetic energy at a displacement x, will be K

Z2

32 u 2

200 u 200 u10

2

Z

Ÿ

w

Ÿ

1 § 200 · u¨ u Z 2 400  200 u102 ¸ 2 © 1000 ¹

.N

32 u 103 Ÿ

1 m Z 2 a2  x2 2

Thus, the displacement equation will be x

8 2

4 rad / sec

S· § 20 sin ¨ 4t  ¸ . 4¹ ©

w

w

Example 16 A particle executes SHM with force constant 2 × 106 N/m and amplitude 1 cm, and has a total mechanical energy of 200 J. Find the following (a) maximum K.E., (b) Maximum P.E., (c) minimum K.E., (d) minimum P.E. Solution Given k

mZ2

2 u106 ; a 1u 10 2 m

(a) Maximum K .E.

1 m Z 2 a2 2

1 u 2 u106 u 1u104 2

» vmax

100 J .

23

Z a ¼º

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(b) From E = K + U E U max [when K = 0; at the extreme position] ? Maximum potential energy Umax = 200 J.



1 2 m vmin = 0 [at the extreme portion] 2 (d) Minimum Potential energy Umin = E - Kmax = 200 – 100 = 100 J. Note: The particle’s potential energy is not zero at mean position

i.c

om

(c) Minimum kinetic energy

W

 mg

et B

ad

Example 31 A uniform rod of length l and mass m, is suspended by one end and is displaced slightly, to undergo angular SHM. Find its time period. Solution Consider the rod, in a position making an angle 'T ' with the vertical at any instant. (Fig. 15.43) Net torque acting on the rod about the point of suspension.  m g lT l

l sin T 2

W

ml 2 d 2 T 3 d t2

w

?

d2 T

W

3W

2

I

ml 2

dt

.N

Now, angular acceleration

[ for small angle sin T | T ]

 m g lT 2

w w

Z

? The period T

or, d2T

Comparing with the standard form

We get,

d2 T

d t2

dt

2

 Z2 T



3g T 2l

0.

0

3g . 2l 2S

Z

2S

2l 3g .

Example 32 a

A square lamina of edge ‘a’ is suspended through a hole punched on a diagonal at a distance of

2 2

from its center. Find the period of oscillations. Solution: Consider, the oscillating lamina (of mass m say) at any position making an angle 'T ' with the vertical. Net torque acting on the lamina about the point of suspension O, will be (Figure. 15.44)

24

W

m g

a 2 2

sin T | mg

a 2 2

om

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T

ad

7 ma 2 24

I d2T

W

Since,

2

dt 2

7 ma 2 24

§ d 2T ¨¨ 2 © dt

· ¸¸ ¹

d 2T dt 2

 g (12)T



7 2

m g aT

2 2

et B

I

§ a2 · ma 2  m¨ ¨ 2 2 ¸¸ 6 © ¹

i.c

Now, the moment of inertia of the lamina about an axis through O, and normal to its plane will be.

d 2T

.N

Comparing with the standard form 12 g

Z2

We get

w

? The period T

7 2

2S

2S

Z

Z 2T

dt 2

. 7 2 12 g

S

7 2 3g

w

w

15.11 Average Values of a Few Parameters: Average velocity and acceleration of a particle executing SHM. The average of any time dependent function (Z = f(t)) say) can be obtained, between the instants t = t1 to t = t2.

As

Z!

t2

t2

t1

t1

³ Zdt ³ f t dt . ³ dt ³ dt t2

t2

t1

t1

(a) Average Velocity: Since the motion in a SHM, repeats again and again over a period of one time period, therefore the average velocity over one complete time period is zero. However, the average velocity between one extreme position to the other will not be zero. This can be calculated as follows.

25

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V

t

Assuming the equation for velocity in terms of time as v

? v!

³

a Z cos Z t d t

T /4

³

3T / 4

ª sin Z t º aZ « » ¬ Z ¼T / A

dt

t

· § S ·½ ¸  sin ¨ 2 ¸ ¾ ¹ © ¹¿

[ Z T

2a {(1) (1)} T

·º ¸» ¹¼

2S ]

et B

2 a ­ § 3S ®sin T ¯ ¨© 2

ª § 3Z T · §Z T a «sin ¨  sin ¨ ¸ © 4 ¬ © 4 ¹ § 3T T · ¨ 4 4¸ © ¹

ad

T /4

3T / 4 T /4

a Z cos Z t

i.c

3T / 4

3T / 4

om

O

2 4a 2 a Z times maximum velocity ... (24). S T S Alternatively: Average velocity over one half oscillation = Average speed from one extreme position to another. [ there is no variation in direction]

?

v!

.N

total dis tan ce cov ered total time taken

2a T /2

4a T .

w

(b) Average Acceleration: Arguing in the same way as for velocity, the average acceleration over one complete oscillation will be zero. However if we are not interested in one complete oscillation, it may not turn out to be zero. Let us find out the average acceleration between the instants when the particle crosses its mean position two consequtive times. Z 2 a sin Z t

w

w

Assuming the equation for acceleration as f

26

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³ ³

T /2

0 T /2

sin Z t d t

T /2

dt

0

2Z 2 a ªcos S  0 º¼ ZT ¬ | f !|

or ,

2Z 2 a ª § Z T cos Z T «¬ ¨© 2

§ 2 · ª  cos Z t º Z 2 a ¨ ¸ « » ©T ¹ ¬ Z ¼0 2 Z2 a



S

2 Z2 a

2

S

S

º · ¸  cos 0 » ¹ ¼

om

 f !

Z 2 a

.

time maximum acceleration.

… (25)

i.c

Average Kinetic Energy, Potential Energy and Total energy. (a) Average kinetic energy over one complete oscillation.

The instantaneous K.E. of a particle executing SHM is given by K 1 m Z 2 a 2 cos 2 (Z t ) dt 2

0

K!

³

T

dt

0 r

1 m Z 2 a2 2

³

¨¨ ©

0

t

T 0

et B

1 m Z 2 a2 4 T

2Z t · dt ¸¸ 2 ¹

T § 1  cos

ad

?

³

T

1 m Z 2 a 2 cos 2 Z t 2

ª sin 2 Z t º «t  » 2 Z »¼ ¬« 0

1 m Z 2 a2 T 0 4 T

1 m Z 2 a2 4

... (26)

(b) Average potential energy over one complete oscillation.

U !

³

T

0

1 m Z 2 a 2 sin 2 Z t d t 2

³

w

?

.N

The instantaneous P.E. of a particle executing SHM is given by U

1 m Z 2 a2 2

T

dt

³

T

0

§ 1  cos 2 Z t ¨¨ 2 ©

t

0 T

1 m Z 2 a 2 ª sin 2 Z t º «t  » 4 T 2 »¼ 0 ¬«

· ¸¸ d t ¹

T 0

1 m Z 2 a2 T 4 T

w

1 m Z 2 a 2 sin 2 Z t . 2

0

1 m Z 2 a 2 ... (27) 4

w

(b) Average total energy over one complete cycle. Since the total energy is independent of time ?

E!

1 mZ2 a2 . 2

Example 37 What is the ratio between the maximum to average kinetic energy of a particle executing SHM. Solution

27

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We known that the maximum K.E., is given by K 1 mZ 2 a2 4

average

K

?

K

om

K

1 m Z 2 a 2 and average K.E., is given by 2

max

2 1.

max average

§ S2 4 sin 1 ¨ ¨ 4 S2 4 © 4

is

i.c

Example 38 The average speed over the period of complete oscillation of a particle executing SHM is 2 cm/sec. The particle attains this speed when it is at P, distant 4 cm from the mean position O. Find (i) the amplitude (ii) the time period. Also show that the least time taken by the particle to travel from O to P · ¸ ¸ . [Express your answers in terms of S ]. ¹

ad

Solution Let a cm and Z rad/sec be the amplitude and angular frequency of oscillation.

?

2

2 aZ

S

2 aZ

.

S

et B

Then average speed is given by =

Ÿ aZ S . Also, the speed (v) at a displacement (x) is given by

Z 2 a2  x2

Z 2 a 2  Z 2 x2

?

From 1 , a

... (2) §S2 4· ¸¸ Ÿ Z © 16 ¹

Z 2 ¨¨

Ÿ

4 S 2  Z 2 16

w

Ÿ

v2

.N

v2 Ÿ

... (1)

S Z

4S

Ÿ

S 4 2

S2 4

cm and From T

4 2S

Z

Ÿ

8S

S2 4

sec .

w

w

Let the displacement equation for the particle be x = a sin Z t. [Here for the sake of simplicity phase constant I is taken zero].

S2 4 t

4S

S2 4

sin

4

.

Let at t = t1 the particle be at 0 ?

0

4S

S2 4

S 2  4 t1 sin

4

Ÿ

t1

28

0

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Now, let the particle be at P, at t = t2 for the first time

S 2  4 t2

Then 4

sin

S2 4

4

S 2  4 t2

Ÿ

sin

S2 4 S

4

Ÿ

t2

§ S2 4 sin 1 ¨ ¨ 4 S2 4 © 4

§ S2 4 sin 1 ¨ ¨ 4 S2 4 ©

4

· ¸ ¸ ¹

i.c

? The required time interval = t2 – t1

· ¸ ¸ ¹

om

4S

Example 39 A particle executes SHM whose displacement changes with time as x = a sin Z t . Find the average

ad

speed between the instants when it passes its mean position and its displacement is time. Solution

a for the first 2

et B

Since the displacement equation is given as x = a sin Z t when x = 0; t = 0 and when x =

a t ; 2

S 6Z .

Also, instantaneous velocity v = a Z cos Z t .

³

S /6 Z

a Z cos Z t d t

0

.N

? Required value =

³

S /6 Z

dt

§ sin Z t a Z ¨¨ © Z t

0

S /6 Z 0

S /6 Z

· ¸¸ ¹0

ª º §S · a «sin ¨ ¸  sin 0 » ©6¹ ¬ ¼ § S ·  0¸ ¨ © 6Z ¹

w

w

w

§ 1 · 6Z 3 a Z u¨ ¸u S . ©2¹ S General Method to show that a given body (or a particle) executes SHM and to find its time period. Step 1: Establish the equilibrium condition (if any) for the body at its mean position. Step 2: Imagine the body to be displaced (linear or angular as the case may be) from its mean position slightly. Find the net restoring force (in case of linear SHM) or torque (in case of angular SHM) acting on the body in terms of displacement, ending up finally in the form of F = – kx (for linear SHM) (for angular SHM). or W kT Where k should be a constant. Step 3 : Dividing F by M or ( W by I, as the case may be) find the net acceleration as : d2x

Z 2 x

2

or

d 2T 2

Z 2 T

dt dt Knowing the value of Z , time period can be calculated as

T

2S

Z

.

29

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Alternatively: One can also find the total energy of the system, in step 2 and differentiate the expression with respect to time t and equate it to zero. [ The T.E. in SHM remains constant]. One should then eventually end up in equation of the form as obtained in step 3. [See example 25, alternative]. A few examples are solved stepwise to illustrate the idea.

450

S / 4 rad .

i.c

Initial phase G

om

EXAMPLES Example 1 Write down the equation of a simple harmonic motion with an amplitude of 5 cm if 150 oscillations are performed in one minute and the initial phase is 450. Solution Given: Amplitude (A) = 5 cm Frequency of oscillation = n = 150 per minute 150 per second 60 5 per second 2

5 2 S ( ) 5 S rad s 1 2 The displacement of the particle at any instant t is given by Angular frequency = Z

?

2S n

et B

?

S· § 5sin ¨ 5 S t  ¸ cm . 4¹ ©

A sin Z t  G

x

ad

=

Example 2 In what value of time t (i.e., after t = 0) will a particle, oscillating according to the equation x = 7 sin

.N

0.5 S t , move from the position of equilibrium to the maximum displacement? Solution

The displacement of the particle is given by x = 7 sin 0.5 S t Given: xmax = amplitude = 7

w

7 sin 0.5 S t

7

?

1 sin 0.5 S t

w

?

0.5 S t

w

?

?

t

S 2

1 sec ond

Alternate Method: 2S T

0.5 S

ŸT

4 sec

At t = 0, x = 7 sin (0) = 0 Time to reach from x = 0 i.e., equilibrium) to x = 7 (Amplitude) that is a quarter Oscillation t = 1 sec.

Example 3

(i) Period of oscillation (iii) Maximum acceleration and

S

S

2 sin ( t  ) cm . Find 2 4 (ii) Maximum velocity (iv) Initial displacement

The equation of an oscillating particle is given by x

30

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Solution Given that the displacement of the particle is x

S

S

2 sin ( t  ) cm 2 4

Compare it with x = A sin Z t  G

S

rad s 1 .

2

2S

(i) Period of oscillation

T

(ii) Maximum velocity

vmax

2S S /2

Z

ZA =

u 2 S cm s 1 2

Z2 A = S / 2 u 2 S

2 sin ( ) 4

= x0

(iv) Initial displacement

2

S2 2

cm s 2

ad

(iii) Maximum acceleration = a max

S

4s

i.c

Angular frequency Z

om

? Amplitude (A) = 2 cm

Example 4

2 cm .

(i.e., at t = 0).

Put x

et B

The equation of a simple harmonic motion is given as x 6 sin10 S t  8 cos 10 S t Where x is in cm and t is in seconds. Find the amplitude, period and initial phase. Solution 6sin10 S t  8 cos 10 S t

Expanding, we get, x

?

A cos D sin Z t  A sin D cos Z t

10 S rad s 1

.N

? We get , Z

A sin Z t  D

The Period

2S

T

2S 10 S

Z

0.2 s .

w

Also A cos D = 6 and A sin D = 8 Squaring and adding, we get A2 = 62 + 82 = 100 Amplitude = A = 10 cm. ?

w

tan D

A sin D A cos D

8 6

1.333

w

? Initial phase = D = 530 8’. Example 5 A particle performing S. H. M has velocities of 3 cm s-1 and 4 cm s-1 respectively when the displacements are 4 cm and 3 cm. Find the amplitude, period and maximum velocity of the particle. Solution The velocity v of the particle performing SHM when its displacement x is given by v Z

?

v2

A2  x 2

Z 2 A2  x 2

We have, v = 3 cm s-1 when x = 4 cm.

31

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9 Z 2 A2  16

… (I)

Also,

v = 4 cm s-1 when x = 3 cm.

?

16 Z 2 ( A2  9)

Dividing (I) by (II),

9 16

… (II)

A2  16

om

?

A2  9

Solving, A = 5 cm Substituting this in (I), we get Z = 1 rad s–1 ? Period T = 2 S / Z 2 S s Velocity is maximum, when x = 0

Then maximum velocity

... (I)

Z 2 A 50 cm s 2

... (II)

et B

max

Dividing (II) by (I), we get,

Z2 A Z

? Period of oscillation = T =

ZA Z

40 1.25

.N

Amplitude = A =

Z A 40 cm s 1

vmax a

Maximum acceleration

ad

i.c

? vmax Z A 1.0 u 5 cm s 1 Example 6 A particle executing linear SHM had a maximum, velocity of 40 cm s–1 and a maximum acceleration of 50 cm s-2. Find its amplitude and the period of oscillation. Solution Let A be the amplitude and Z be its angular velocity..

50 40

2S

2S 1.25

Z

i.e., Z

1.25 rad s 1

5.03 s

32 cm .

w

Example 7 A particle of mass 50 g executes SHM along a path of length 20 cm at the rate of 600 oscillations per minute. Find its total energy. Also find the potential and kinetic energies when the displacement x = A/2 where A is the amplitude. Solution: mass (m) = 50 g = 5 × 10-2 kg

Z

w

Angular frequency

w

Amplitude (A) ? Total energy =

When, x =

20 2

2S u 600 60

20 S rad s 1

10 cm 0.1 m

1 m Z 2 A2 2

1 u 5 u 102 u 400 S 2 u 102 2

A 1 , Potential energy = m Z 2 x 2 2 2

? Kinetic energy =

1 m Z 2 A2  x 2 2

A2 1 m Z2 2 4

1 A2 m Z 2 ( A2  ) 2 4

32

0.987 J . 0.247 J

0.740 J .

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om

Example 8 A body is supported by a light spiral spring and causes a stretch of 1.5 cm in the string. If the mass is now set in vertical oscillation of small amplitude, what is the periodic time of the oscillation? Solution Let m be the mass of the body in kg. Then, since 1.5 cm = 0.015 m, mg = k × 0.015, at equilibrium position ... (I) Where k is the spring constant in N m–1 mg N m 1 0.015 If the mass is displaced by x below its equilibrium position, the net restoring force F = K (0.015 + x) – mg = Kx i.e. the net force (restoring) is opposite to the displacement )& & i.e., F  K x i.e., the motion is Simple Harmonic

1.

period

2S

m 0.015

m = 2S K

2S

0.015 = 0.25 s 9.81

ad

? T

i.c

K=

mg

A body of mass 1 kg is executing harmonic motion which is given by x = 6.0 cos 100 t  S / 4 cm.

et B

What is the (i) amplitude of displacement, (ii) frequency, (iii) initial phase, (iv) velocity, (v) acceleration, (vi) maximum kinetic energy? Solution: The given equation of SHM is x = 6.0 cos 100 t  S / 4 cm Comparing it with the standard equation of SHM, x = A cos Z t  I , we have (i) (ii)

Amplitude A = 6.0 m Frequency Z = 100/sec.

(iii)

Initial phase I

Velocity v Z

A2  x 2

Acceleration Z 2 x

w

(v)

.N

(iv)

S /4

2

x

104 x

Kinetic energy

w w

 100

36  x 2

1 1 m v2 m Z 2 A2  x 2 . 2 2 When x = 0, the kinetic energy is maximum i.e.,

(vi)

2.

100

K .E

max

1 m Z 2 A2 2

1 § 36 · u 1u 10 4 u ¨ meter ¸ 18 joules. 2 © 100 ¹

A particle is executing SHM with amplitudeA and has maximum velocity V0. Its sped at displacement A/2 will be : a.

3 V0 / 2

b. V0 / 2

c. V0

d. V0/4

Solution : (A) v

Z A2  x 2

33

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For x = A/2 v

2

3 V0 [V0 Z A] 2 The velocity time graph of a harmonic oscillator is shown in the following figure. The frequency of oscillation is :

Z A 3/ 4

a.25 Hz Solution: (A)

b. 50 Hz

c. 12.25 Hz

v

et B

a. 2 S sec

b.

S

2

sec

c. S sec

.N

Solution: (C)

... (i)

& 8 Z A2  5

... (ii)

w

10 Z A2  4 2

w

(i) Ÿ

w

3S sec 2

2

A2

§ 10 · 2 ¨Z ¸ 4 © ¹

A2

§8· 2 ¨Z ¸ 5 Ÿ © ¹

2

(ii) Ÿ ?

d.

Z A2  x 2

Using v

36

Z2

9Ÿ

Z

100

Z2

 42

64

Z2

 52

6 3

2S 2S S 6/3 v A linear harmonic oscillator has a total mechanical energy of 200 J. Potential energy of it at mean position is 50 J. Find. ?

5.

d. 33.3 Hz.

1 1 25 Hz. T 0.04 A body is executing SHM. When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocity of the body is 10 cm per sec. and 8 cm per sec. respectively. Then the time period of the body is Ÿ

T = 0.04 s

4.

ad

i.c

om

v

3.

Z A2  A / 2

T

34

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?

U max

om

a. The minimum kinetic energy b. The minimum potential energy c. The potential energy at extreme positions. Solution: At mean position, potential energy is minimum and kinetic energy is maximum. Hence Umin = 50 J (at mean position) and Kmax = E – Umin = 200 – 50 = 150 J (at mean position). At extreme positions, kinetic energy is zero and potential eneryg is maximum. E = 200 J (at extreme position).

i.c

Single Answer Correct:

A particle is executing SHM. Which of the following is maximum when the bob is at the mean position? a. Time period b. Amplitude c. velocity d. acceleration. Solution: (C) At the mean position, velocity is maximum and acceleration is minimum. 2. The displacement of a particle in simple harmonic motion in one time period is a.A b. 2 A c. 4 A d. zero. Solution: (D) At the end of one complete vibration, the particle returns to the initial position. 3. Two bodies A and B of equal masses are suspended from two separate mass less springs of spring constants k1 and k2 respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is: a.k1 / k2 Solution : (D)

4.

b.

or

a1 a2

Z1 Z2

k1 / k2

c. k2/k1

d.

k 2 / k1 .

k2 k1

.N

a1 Z1 a2 Z2

et B

ad

1.

The equation of SHM of a particle is a + 4 S 2 x = 0, where ‘a’ is instantaneous linear acceleration at displacement x. The frequency of motion is b. 4 S Hz

c.

w

a.1 Hz

1 Hz 4

d. 4 Hz

Solution: (A)

w

w

Comparing with a  Z 2 x 0, Z 2 4 S 2 ; Z 2 S v 2 S , v 1 H z. 5. A particle executes SHM of amplitude 5 cm and period 3 s. The velocity of the particle at a distance 4 cm from the mean position (take S = 3) is a.8 cm s-1 b. 12 cm s-1 c. 4 cm s-1 d. 6 cm s-1. Solution:

35

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2u 3 2S 25  16 6cms 1 a2  x2 3 T 6. Figure shows the displacement – time graph of a simple harmonic oscillator. The amplitude, time period and initial phase of the oscillator, are respectively. b. 2 cm, 4 s, S /4 c. 2 cm, 4 s, zero d. 2 cm, 2 s, zero. a.1 cm, 2 s, S /2 Solution: (C) Graph begins from the origin. So, initial phase is zero.

om

v Z a2  x2

(Acceleration)max = Z 2 a

5S 2 .

Velocity at y = 4 cm is v

Z a2  y 2

3S

Z 2 a 2  16 Z a 2

a 2  16 a

?

9S 2

5S 2

9 5

9 or 5 a = 5 cm.

9 r 41 10

a

u 5 or - 3.2 cm.

w

or, ? T = 2 sec. So, choice (B) and (C) are correct and choice (A) and (D) are wrong. When the particle is half way to its end point in a S. H. M (where E is the total energy) :

E 4 Solution : (A), (D)

w

a. PE

w

... (2)

Z 2 u 5 5S 2 . Z S.

From (1),

13.

3S

.N

or,

9S 2

Z a 2  42

et B

2 2 Squaring, Z a  16

So,

... (1)

ad

or, Z a 2  16

i.c

Multiple Answers: 12. A particle vibrates in SHM along a straight line. Its greatest acceleration is 5 S 2 cm s-2 and when its distance from the equilibrium position is 4 cm, the velocity of the particle is 3 S cm s-1. Then: a. The amplitude i 10 cm b. The period of oscillation 2 sec. c. The amplitude is 5 cm d. The period of oscillation 2 sec. Solution: (B), (C)

Given, y

b. PE

5E 4

c. PE = KE

d. KE = 3 PE.

a and total energy = E 2

We also know that total of energy a particle in SHM E Potential energy of particle at some point (PE) 1 mZ2 y2 2

1 §a· m Z 2 .¨ ¸ 2 ©2¹

2

1§1 · m Z 2 a2 ¸ 4 ¨© 2 ¹

36

E 4

1 m Z 2 a2 2

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1 mZ2 a2  y2 2

§ 1 a2 m Z 2 ¨ a2  ¨ 2 4 ©

· ¸¸ ¹

1 3 a2 mZ 2 2 4

3 E 4

om

KE

So, KE = 3 (PE) Choice (D) is correct and choices (B) and (C) are wrong. EXERCISE Single Answer Correct.

i.c

a.

27.

1 4

b.

3 4

3 and 30 m s-1

d.

3.5 . 4

b. 20

3 and 30 m s-1.

d. 15 m and 5 3 m s-1. 3 m and 30 m s-1 The equation of a SHM of amplitude ‘a’ and angular frequency ' Z ' in which all distances are measured from one extreme position and time is taken to be zero at the other extreme position is b. x = a cos Z t a. x = a sin Z t

.N

c. 15

c. x = a – a sin Z t d. x = a + a cos Z t . The amplitude and the periodic time of SHM are 5 cm and 6 s respectively. At a distance of 2.5 cm away from the mean position, the phase will be: b. S /4 c. 5 S /12 d. S /3. a. S /6 Two particles execute SHM of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions, each time their displacement is half of the their amplitude. The phase difference between them is: a. 900 b. 300 c. 1200 d. 600. The maximum displacement of a particle executing SHM is 1 cm and the maximum acceleration is (1.57)2 cm s-2. Then the time period is a. 0.25 s b. 4.00 s c. 1.57 s d. (1.57)2 s. Which of the following quantities are always positive in a simple harmonic motion? [Letters have usual meanings]. )& & & & & & )& )& b. v. r c. a. r d. F . rs . a. F . a

w

29.

c.

S· § The displacement of a particle executing simple harmonic motion is given by : y = 10 sin ¨ 6 t  ¸ . 3¹ © Here, y is in meter and t is in second. The initial displacement and velocity of the particle are respectively. a. 5

28.

2 4

ad

26.

If the displacement equation of a particle be represented by y = A sin pt + B cos pt, the particle executes. a. a uniform circular motion b. a uniform elliptical motion. c. a SHM d. a rectilinear motion. What fraction of the total energy is potential when the displacement is one - half of the amplitude?

et B

22.

w

30.

w

31.

32.

33.

The ratio of the KE and PE possessed by a body executing SHM when it is at a distance of 1/n of its amplitude from the mean position is a. n2

b.

1

c. n2 + 1

n2

37

d. n2 – 1.

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34.

The velocities of a body executing SHM are v1 and v2 when the displacements are x1 and x2 respectively. The period is a. 2 S

x22  x12 v12

 v22

1 b. 2 S

x22  x12 v12

 v22

c. 2 S

v12  v22 x22



x12

d. 2 S

x2  x1 v2  v1 .

om

Using the following paragraph, solve Q. 50 & 51 A particle is said to possess SHM if (a) acceleration is directly proportional to displacement, (b) acceleration is directed towards a fixed point. The average of such a function in a complete cycle is

complete cycle is given by, Ya v

T

0

0

³Y d t ³ d t .

If the average PE of an oscillating system is E, the average KE of a system with double amplitude and frequency will be:

ad

50.

T

i.c

§1 2 2· zero. The PE and KE of such SH function will be half the total energy ¨ m Z A ¸ , where A is the ©2 ¹ amplitude and Z is the angular frequency of the mass m. The average of a function of time in a

E c. 4 E d. 16 E. 4 51. The ratio of average of displacement’s A sin Z t and A sin 2 Z t over half of cycle is: a. 0 c. 1 : 1 d. 1 : 2. b. f Assertion Reason Type: In each of the following questions two statements are given as Assertion ‘A’ and Reason ‘R’. Examine the statements carefully and answer the questions according to the instructions given below. Mark (A) if both A and R are correct and R is the correct reason of A. Mark (B) if both A and R are correct and R is not the correct reason of A. Mark (C) if A is correct and R is wrong. Mark (D) if A is wrong and R is wrong. 52. Assertion (A): Every oscillatory motion cannot be treated as simple harmonic under small oscillation Reason (R): Even under larger oscillation restoring force should be proportional to displacement for being SHM.

b.

w

.N

et B

a. E

w

w

53. Assertion (A): Two SHM’ s along x and y axes with angular frequency ratio Z1 : Z2 = 1 : 2, with same amplitude results in a parabola path on super – position. Reason (R): The x and y displacements are related as y D x2.

22. (C) 27. (A) 32. (A) 37. (B) 42. (B), (C) 47. (D) 52. (B)

Simple Harmonic Motion 23. (B) 28. (D) 33. (D) 38. (B), (C) 43. (A), (D) 48. (D) 53. (A).

24. (B) 29. (A) 34. (A) 39. (A), (C) 44. (A), (B), (C) 49. (B)

38

25. (D) 30. (C) 35. (C) 40. (A), (B) 45. (A), (B), (C) 50. (D)

26. (A) 31. (B) 36. (B) 41. (A), (B), (C) 46. (A), (B), (D) 51. (B)

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15.7 Spring Pendulum (a) Horizontal oscillations of mass attached to an ideal spring

C

P x

om

F=kx

ad

i.c

Consider a block of mass m, attached to one end of a mass less spring, obeying Hooke’s law, with its other end attached to a rigid support. The block rests on a horizontal frictionless surface. (Fig. 15.29). Let the block be slightly pulled towards right and left free to oscillate. It executes SHM. Consider the position of the block, at any instant, when its displacement is ‘x’ from the mean position. (towards right). The net force acting upon the block is the restoring force due to spring. (towards left). )& & Thus, F k x [where ‘k’ is the force constant] )& )& F k & f  x . or, m m )& & Comparing with the standard from f Z 2 x . k m Now, time period of motion.

et B

2 We have Z

2S

T

Z

m . k

2S

Energy Conservation in Spring Oscillations: Let ‘x0’ be the maximum compression (or extension) in the spring. In such situation. 1 m 0 2

2

0;

U

1 2 k x0 2

.N

K

1 2 k x0 ... (a). 2 Now, if v0 be the maximum speed, of the block, when at mean position, then E

w

And

w

K

w

K

K U

1 m v02 ; 2

U

1 K 0 2

0

1 m v02 ; 2

U

1 K 0 2

0

1 m v02 2 But, as the block moves from its mean position, to either of its extreme position, it utilizes, its kinetic ?

E

K U

§1 2· energy ¨ m v0 ¸ to elongate or compress the spring, which gets stored in the spring in the form of 2 © ¹ §1 2· potential energy ¨ k x0 ¸ . Thus, from (a) and (b), it is evident at the total energy of the system (i.e., ©2 ¹ spring + block) remains constant.

39

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1 2 k a0 2

1 2 k a1  P m g 2

i.c

om

Note: Oscillations of a spring pendulum is independent of the value of ‘g’ at that place. Horizontal oscillations, of a block attached to a spring, on a rough surface: Consider a block of mass ‘m’, attached to a spring of force constant k, oscillating on a rough horizontal surface with coefficient of friction ' P ' (Fig. 15.30). Let a0 be the initial compression, of the spring and a1 be the maximum extension, for the first time. From energy considerations.

a0  a1

k P m g a0  a1 Ÿ a0  a1 a0  a1 2 This gives the decrease in amplitude, during each a swing. (b) Vertical oscillations of block attached to an ideal spring.

a0  a1

2Pm g k

.N

et B

ad

Ÿ

Consider a block of mass m attached to one end of an ideal spring, with its other end fixed to a rigid

w

w

w

support. Let ' yT ' be the initial extension in the spring. (Fig. 15.31). If ‘k’ be the force constant of the spring, then for equilibrium (vertical) of the block ky0 = mg (numerically). Now, if the block be slightly displaced from its mean position it starts executing SHM. Let ‘y’ be the displacement of the block, at any instant from its mean position (downward). The net force acting on the block upward will be F = k (y0 + y) – mg )& )& or, F k y )& )& F § k · )& ¨ ¸ y or, acceleration f m ©m¹ )& )& Comparing the above equation with the standard from f Z 2 y , it is evident that the block executes SHM, with an angular frequency Z

k . m

40

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2S

? Time period of oscillations T

Z

m . k

2S

om

Note: Even the vertical spring oscillations is also independent of the value of ‘g’ at that place. (c) Horizontal spring oscillations, taking into consideration, the mass of the spring. As the block oscillates, so does the spring. However the velocity at each point of the spring is not the same. In the elongated (or compressed) state, the displacements of each section can be assumed to be linearly proportional from their respective distances from the fixed end. Thus, if ‘y’ be the displacement of the block (i.e., the end of the spring connected to the block) at any instant, then the displacement of a section distant ‘x’ from the fixed end will be

yx where L is the L

i.c

dy d t (Fig. 15.32).

et B

ad

length of spring; similarly if v be the velocity of block then v

According the velocity v’ of the aforementioned section will be v '

x dy L dt

xv L .

Now, kinetic energy of the entire spring can be obtained as the sum (integral) of the kinetic energies of each sections.

.N

Thus, total K.E. of the spring =

­1 § M 'd x · 2 ½ v' ¾ L ¸¹ ¿

³ ®¯ 2 ¨©

[Here M’ is the mass of the spring and dx, the length of the elemental section].

³

L

v2 x2 d x

1 M 'v2 2 L3

2

L

0

w

1 M' 2 L

L

ª x3 º « » «¬ 3 »¼ 0

1 2

§ M ' v2 ¨¨ 3 ©

w

? Total kinetic energy of the spring + block system

1 2

· ¸¸ ¹ § M ' v2 ¨¨ © 3

· 1 2 ¸¸  M v [where M is the mass of 2 ¹

w

the block]. Now, the total mechanical energy of the system is given by E = K + U. 1 2

§ M ' v2 ¨¨ © 3

· 1 1 2 2 ¸¸  M v  k y 2 2 ¹

Since the total mechanical energy should remain conserved. ?

dE dt

0

M ' § 2vd v · M § 2vd v · k § d y· ¨ ¸ ¨ ¸ ¨2 y ¸ 6 © dt ¹ 2 © dt ¹ 2 © dt ¹

Ÿ

§M' · dv ¨ 3 M ¸ dt k y © ¹

0

ª dy « ¬ dt

º v» . ¼

41

0

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d2 y

?

d t2



d §d y· ¨ ¸ dt © dt ¹

k y §M' · y ¨ 3 M¸ © ¹

d2 y d t2

0

d2 y Comparing with the standard form

dt

2

 Z2 y

om

dv Again, since d t

0

i.c

Z

We find, that the complete system executes SHM, with an angular frequency

2S

Note:

ad

? Time period of oscillations T

§M' · ¨ 3 M ¸ © ¹ k

The above equation can be rewritten as T

2S

et B

(1) (2)

M eff k

where M eff

k

§M' · ¨ 3 M¸ © ¹

§M' · ¨ 3 M ¸. © ¹

w

.N

The oscillations of a block attached to a spring, remains SHM only for small elongation (or compression) when the restoring forces generated in the spring remains linear. For large deformations in spring, the restoring forces become non linear. (3) Unless, otherwise specified, a spring should always be treated as ideal and mass less. Composite spring oscillations. (a) When springs are connected in series: Consider three springs with stiffness constants k1, k2 and k3 respectively. If x1, x2 and x3 be the respective elongations of the three springs, then, the displacement of the block x is given by x = x1 + x2 + x3. ... (a) Now, if F be the restoring force acting along each of the spring. Then F = k1 x1 = k2 x2 = k3 x3 ... (b) Now, the net force acting on the block will also be F.

w

From (b), x1

F ; x2 k1

F k2

and x3

F F F   k1 k2 k3

w

Substituting them in equation (a). x

F k3 .

42

)& § 1 1 1 · or F ¨   ¸ © k1 k2 k3 ¹

& x .

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1 1 1 1 )& Replacing k  k  k with K we have f eff 1 2 3

Time period of oscillation = T

2S

Z

keff m

& x §1 1 1 ·  ¸ m¨  © k1 k2 k3 ¹

& keff x

m

which proves the fact that the oscillation

.

Ÿ 2S

m 1 . keff where keff

1 1 1   k1 k2 k3 .

ad

(b) Where springs are connected in parallel:

.

i.c

are SHM, with angular frequency Z



)& F m

om

)& If m be the mass of the block, then acceleration f

T

2S

et B

It can be shown that the oscillations of the block are SHM with a time period T given by m keff where kefff = k1 + k2 + k3.

.N

(c) Figures 15.35 below shows a special arrangement of two springs virtually appearing to be in series; however they are not.

w

w

w

& If x be the displacement of the block at any instant then, the net force acting on it & )& ))& ))& & & F F1  F2 k1 x  k2 x =  k1  k2 x

? The effective spring constant for the above set up is kefff = k1 + k2 ? Time period T

2S

m keff .

Example 26

43

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i.c

om

A small block of mass m is attached to three springs with respective stiffness constants k1, k2 and k3 (see fig 15.37). Find the period of oscillations, if the block is slightly pressed against the spring with stiffness constant k3. Will your answer be the same, if the block be pressed against the spring with stiffness constant k1?

et B

ad

Solution Figure 15.38 below shows, the respective elongations occurring in the springs of stiffness constant k 1 and k2 when the spring of stiffness constant k3 is compressed by ‘x’.

)& F

& 1 &º )& )& ª & 1 F 1 cos 450  F 2 cos 450  F3  « k1 x u  k2 x u  k3 x » 2 2 ¬ ¼

ªk k º & « 1  2 » x ¬2 2¼

w

)& F

.N

Net restoring forces acting on the block along the spring of stiffness constant k3 will be

w

Thus, the effective stiffness constant is given by keff

§ k1  k2 ¨ 2 ©

· ¸  k3 ¹

w

? Time period of oscillations

Ÿ

T

2S

m . keff

If the block be compressed against spring of spring constant k1, the net force will be (See Fig. 15.9)

44

)& F

2

F1  F3 cos 450

 F3 sin 450 1/ 2

2 ª§ k · § k ·º «¨ k1  3 ¸  ¨ 3 ¸ » 2 ¹ © 2 ¹» «¬© ¼

2

&  x

2

k3 x · § k3 x · § ¨ k1 x  ¸ ¨ ¸ 2 ¹ © 2 ¹ ©

2 2 ª§ k · §k · º «¨ k1  3 ¸  ¨ 3 ¸ » 2¹ © 2¹ » «¬© ¼

1/ 2

.

et B

? Effective stiffness constant keff

2

ad

)& F

i.c

om

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w

w

.N

Example 27 Figure 15.40 shows three springs A, B and C with force constants k1, k2, and k3 receptively, connected to a block of mass m, passing through a light pulley. Assuming all other things to be mass less, find the period of oscillations.

Solution

w

Evidently, springs A and B are in series, with effective spring constant k '

parallel with the spring C ? Effective spring constant keff

? Time period T

2S

m keff

§ k1 k2 · ¨ ¸  k2 © k1  k2 ¹ Ÿ

2S

m k1  k2 k1 K 2  k2 k3  k1 k3

45

k1 k2 k1  k2 which in turn is in

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i.c

om

Alternative: (from energy considerations). Let y0 be the initial compression in the spring C and, for the sake of simplicity, let the spring A and B in the relaxed position, when the block is at rest. (Fig. 15.41) For vertical equilibrium of the block mg = k3 y0. Let ‘y’ be the displacement of the block (below its mean position), which would also be the additional compression in the spring C.

et B

ad

Let y1 and y2 be the respective elongation of the springs, A and B respectively. Since total extension in spring A = Net compression in springs B and C ? y1 = y – y2 ... (1) Ÿ y = y1 + y2 Also, since the tension in the two ends of the string (passing through the pulley) should be same therefore, k1 y1 = k2 y2 ... (2) § k2 y · From (1) and (2) y1 ¨ k  k ¸ and y2 © 1 2¹

§ k2 y · ¨ ¸ © k1  k2 ¹

1 1 1 k1 y12  k2 y22  k3 y0  y 2 2 2 Since, total mechanical energy remains constant, during oscillation ?

.N

Total mechanical energy will be E

dE dt

w

2

w w

Ÿ

Ÿ

Or ,

k1 k22 y

k1  k2

2

k1 k22 y k1  k2

2

2

d2 y 2



1 m

2



k k2 y d y dy  2 1 2  k3 y0  y dt dt k1  k2 

k2 k12 y k1  k 2

2

 k3 y 

§ k1 k2 · m d2 y  k3 ¸ y  ¨¨ ¸ d t2 © k1  k2 ¹

dt



1 m v 2  mgy 2

0

k2 § k1 y · k3 d ­° k1 § k2 y · i.e., d t ® 2 ¨ k  k ¸  2 ¨ k  k ¸ 2 y  y © 1 2¹ ¯° © 1 2 ¹ Ÿ

2

§ k1 k2 ·  k3 ¸ y ¨¨ ¸ © k1  k2 ¹

md v dt

½ m v2 °  m g y¾ 2 ¿°

0

d y mv d v m g d y   dt dt dt

0

0

0

[

0

46

dy dt

v and m g

k3 y0 ]

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d2 y Comparing with the standard form

 Z2 y

0

2S

m k1  k2

2S

Z

k1 k2  k2 k3  k1 k3

om

1 § k1 k2  k2 k3  k1 k3 · ¨ ¸ m© k1  k2 ¹

2 we find that Z

Time period T

d t2

.

ad

i.c

Example 28 A horizontal spring block system (of force constant k) and mass M executes SHM with amplitude A. When the block is passing through its equilibrium position an object of mass m is gently placed on it and the two move together. Find the new amplitude. Solution Let the initial angular frequency be Z and the velocity of the block at its mean position be v..

et B

k M

Then Z =

k M Now, if ‘v’ be the combined speed, then, from law of conservation of linear momentum. (M + m) v’ = Mv.

A

Also v = max. vel. Of the block

Mv M m

v'

.N

Ÿ

w

And, the new angular velocity will be Z ' Again from vmax '

w

w

Ÿ

Ÿ

Ÿ

Mv M m

A 'Z ' A'

k M m

M k A M m M A'

A

k . M m

A'

[where A’ is the new amplitude] k M m

[From eqn. (1)]

M M m

Example 29 Two bodies A and B of equal mass are suspended from two separate mass less springs of force constant k1 and k2 respectively. If the bodies oscillate vertically such that their maximum velocities are equal, then find the ratio of the amplitudes of A and B.

47

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Solution If Z1 and Z2 be the respective angular frequencies of the oscillations related to A and B. k1 and Z2 m

Then Z1

k2 m

[where m is the mass of each body A and B]

a1 Z1

vmax

a1 a2

Ÿ

om

Now, if a1 and a2 be their respective amplitudes then: a2 Z 2

Z1 Z2

k2 k1 .

T

2S

ad

i.c

Example 30 If a body attached to a spring executes vertical oscillations with time period T, find the new time period, if the same body be attached to a new arrangement in which the spring is subdivided into four parts of equal length each and all of them connected in parallel. Solution Let k be the stiffness constant of the spring and m be the mass of the body. Then time period ‘T’ of the oscillations will be given by m k

... (1)

et B

Now, since the stiffness constants of a spring varies inversely as its length ? Stiffness constant of each part = 4 k Further, the effective stiffness constant, when all of them are connected in parallel will be 4 (4K) = 16 k. Now, the new time period T '

m 16 k

Ÿ

2S 4

m k

.N

T [From (1)] 4 A solid cylinder is attached to a horizontal mass less spring so that it can roll without slipping along a horizontal surface. The spring constant for the spring is k. If the system is released from rest, by slightly stretching it, show that the center of mass of the cylinder executes SHM and find its frequency of oscillation (Fig. 15.55). ?

T'

w

w

3.

2S

w

Solution: Step: 1 At equilibrium, no force acts horizontally, so, as such no equilibrium condition exists. Step: 2 Let the system be, at any instant, such that, the spring is stretched by an amount x. If v be the linear velocity of the center of mass of the cylinder then total energy of the system will be E = translational K.E. of cylinder + Rotational K.E. of cylinder + Elastic P.E. in the spring. 1 1 1 M v2  I Z 2  k x 2 2 2 2

1 1 § M R2 M v2  ¨ 2 2 ¨© 2

48

· v2 1 2 ¸¸ 2  k x 2 ¹R

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[ Moment of inertia I for the cylinder about the axis of rotation is

M R2 and for pure rolling 2

v ]. R Now, since no work is done against friction in pure rolling motion, so T.E. remains conserved. dE dt

0.

0

Ÿ

d ª3 1 º M v2  k x2 » « d t ¬4 2 ¼

?

3M v d v k x d x  2 dt dt

d2 x

an angular frequency Z

2k 3M . n

dt

2



2k x 3M

0

 Z 2 x, it is evident that, the cylinder executes SHM, with

1 2S

2k 3M .

A uniform rod of length L and mass M is pivoted at its center. It is held in position by a system of springs as shown in fig 15.56. Find the time period of oscillation, if the rod be slightly turned and released.

w

.N

4.

d2 x

et B

? Frequency of oscillation

Z 2S

d t2

Ÿ

ad

Comparing the above equation with

0

i.c

or,

om

motion Z

w

w

Solution: Step 1: Evidently, there is no requirement of any equilibrium conditions, for no torque exists or acts when the rod is in its equilibrium position. Step 2: Consider the rod, at any position, when displaced by an angle T about point O. (Fig. 15.57) The net torque acting upon the rod clockwise.

W

F1

L L  F2 2 2

49

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[where F1 and F2 are the restoring forces generated in the two springs respectively] L L  ky 2 2

1

d 2T d t2

M L2 d 2T 12 d t 2

?

M L2 d 2T 12 d t 2

 K L2 T 2

Ÿ d 2T

Comparing with the standard from

d t2

d 2T

dt

2

6 k T M

Z 2 T .

i.c

W

But,

om

 k L2 T 2

W

?

L L tan T | T . 2 2

y

Now,

kLy

ad

ky

? Time period of oscillation T

2S

Z

2S

M 6k

Two blocks of masses m1 and m2 are connected to the two ends of a mass less spring of spring constant k, and are placed on a smooth horizontal surface. The stwo blocks are pressed slightly against each other so as to compress the spring and released. Show that the oscillations are SHM and find the period of oscillations.

.N

12.

6k . M

et B

It is evident that the rod executes angular SHM with angular frequency Z

w

w

w

Solution: Let l be the original length of the spring. If x1 and x2 be the displacement of the two blocks at any instant from their respective mean positions, the net compression in the spring will be x1 + x2 = x (say) [Fig. 15.66].

The net force on block m1 = F = - k x Ÿ

m1

d 2 x1 d t2

k x

... (1)

50

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dt

kx m1 .



2

Similarly the net force on block m2 will be F = – kx d 2 x2

Ÿ

dt



2

Ÿ

m2

K x m2

d 2 x2 d t2

... (2)

Now from x1 + x2 = x

1 1 Now let m  m 1 1

dt

2



frequency Z

k

P k

P

x

d2x dt

2

Ÿ

... (3)

d t2

d2x

dt

2

§ 1 1 · k ¨  ¸ x © m1 m2 ¹

which proves the fact that the oscillations are SHM with an angular

.

2S

Z

2S

.N

? Time period T

(say)

P

d t2

d2x

et B

d2x Then

1

d 2 x2

ad

kx kx From (1),(2) and (3) we get  m  m 1 2

d t2



i.c

Differentiating twice w. r. t time t, we get

d 2 x1

k x

om

d 2 x1

Ÿ

P

K

2S

m1 m2 k m1  m2

Thus, the two blocks of masses m1 and m2 connectd to the two ends of the spring oscillate a a single block of mass. m1 m2 m1  m2

w

P

which is known as the reduced mass.

w

w

14. A solid uniform cylinder of mass m performs small oscillations due to the action of two springs each of stiffness factor k. Assuming pure rolling find the period of oscillations (Fig. 15.68)

Solution As the cylinder is slightly disturbed from its rest position, the spring force come into action, leading the cylinder to perform angular oscillations about the point of contact of the cylinder with the horizontal surface as the instantaneous center of rotation. If T be the angular displacement of the cylinder about the instantaneous center of rotation C. (Fig. 15.69).

51

om

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Then, the net torque acting on the cylinder about C will be W = F (2R) + F (2R) [where F is the restoring force on each cylinder] ID .

§ m R2 · d 2T  m R2 ¸ 2 ¨¨ ¸ © 2 ¹dt 3 d 2T m R2 2 d t2

8 k R 2T

Ÿ

d 2T dt

2



16 k T 3m

et B

?

3 d 2T m R2 2 d t2

ad

Now since W

Ÿ  8 k R 2T .

2R

i.c

2 k 2 RT

Ÿ

d 2T

Comparing the above equation with the standard form

d t2

performs angular SHM with an angular frequency Z

16 k 3m

2S

.N

? Period of oscillation T

Z

2S

3m 16 k

S

2

3

Z 2 T , it is evident that the cylinder

m k

w

OBJECTIVE QUESTIONS: 18. The period of oscillation of a spring pendulum is T. If the spring is cut into four equal parts, then the time period corresponding to each part will be : T T c. 2 T d. . 2 4 A vertical spring of mass 120 g is fixed to the ceiling at its upper end and a mass of 370 g is suspended at its lower end. If the spring constant be 810 Nm-1 then its frequency of oscillation is:

a. T

w

w

19.

30.

45 1 a. 2 S s

b.

b.

45 2S

10 1 s 37

1 1 c. 45 S s

d.

1 180 S

37 1 s .. 10

Three identical springs with spring constants k each are arranged in two different manner (A) and (B) as shown in figure 15.72. The ratio of frequencies of the mass, in the two cases will be:

52

b. 2 : 3

2 3

2 3

Answers: 18. (B)

b. 4 S 19. (A)

1 m , we have T D k K

2S

1 Now, k eff ? ?

30. (A)

m k

c. 2 S

2m k

d. f .

32. (C).

et B

From T

18.

m k

ad

a. 2 S

1 1 1 1 1    or k1 k2 k3 k 4 k

4 k'

[Here keff = k and k1 = k2 = k3 = k4 = k’]

1 2S

k M m  s the formula for where m = mass of block Ms = mass of spring. 3

w

19.

.N

k’ = 4k. Time period will be twice the original value. n

2S

m , keff

keff

2K k 2 k  k in either case.

w

T

For series connection effective spring constant keff is given by keff

w

32.

d.

Springs of spring constants k, 2k, 4k, 8k …………. f are connected in series. A mass m kg attached to the lower end of the last spring a allowed to vibrate. The period of oscillation is:

32.

30.

c. 3 : 2

i.c

a. 1 : 1

om

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Ÿ

keff

ª1 1 º 1 1    ..........f » «  k 2 k 4 k 8 k ¬ ¼

1

53

1 1 1 1   ..........  k1 k2 kn

1 ª 1 1 º k «1   2  3  ..........f » 2 2 2 ¬ ¼

1

ª º « 1 » » k « « § 1 ·» «1  ¨ 2 ¸ » ¬ © ¹¼

1

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[Summing up G.P. with a = 1; r =

T

2S

m keff

2S

a ] 1 r

m k /2

om

?

1 ; Sf 2

Only one Option is correct 17 (C) 35. (C)

Answers:

21. (B) 63. (A)

23. (C) 66. (A).

32. (B)

33. (B)

ad

i.c

More than one option is correct. 2. A simple pendulum of length 1 m with a bob of mass m swings with an angular amplitude 300. Then: (g = 9.8 m/s2) (a) Time period of pendulum is 2s (b) Tension in the string is greater than mg cos 150 at angular displacement 150. (c) Rate of change of speed at angular displacement 150 is g sin 150. (d) Tension in the string is mg cos 150 at angular displacement 150. Ans: 2, b, c.

9.. a q; b t; c p; d t.

.N

Answers:

et B

Match the column: 9. In case of second’s pendulum, match the following (consider shape of earth also) : Table – 1 Table – 2 (A) At pole (P) T > 2 s (B) On a satellite (Q) T < 2 s (C) At mountain (R) T = 2 s (D) At center of earth (S) T = 0 (T) T = f

w

Assertion – reason 22. Assertion: If a pendulum clock is taken to a mountain top, it gains time. Reason: Value of acceleration due to gravity is less at heights. (a) A (b) B (c) C (d) D (e) E Assertion: The frequency of a second pendulum in an elevator moving up with an acceleration half the acceleration due to gravity is 0.612 s-1. Reason: The frequency of a second pendulum does not depend upon acceleration due to gravity. (a) A (b) B (c) C (d) D (e) E Answers: 22. d 24 c. 7. On a smooth inclined plane a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has a force constant k, the period of oscillation of the body Is : (assuming the spring as mass less):

w

w

24.

54

8.

b. 2 S

2M k

c. 2 S

M sin T 2k

d. 2 S

2M sin T . k

i.c

M a. 2 S 2 k

om

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A block of mass m is suspended by different springs of force constant shown in figure.

Let time period of oscillation in these four positions be T1, T2, T3 and T4. Then: b. T1 = T2 and T3 = T 4 d. T1 = T3 and T2 = T 4.

An object suspended from a spring exhibits oscillations of period T. Now the spring is cut in two halves and the same object is suspended with two halves as shown in figure. The new time period of oscillations will become :

w

w

.N

9.

et B

ad

a. T 1 = T2 = T 4 c. T 1 = T2 = T 3

w

15.

a. 4 : 1 b. 1 : 4. A mass m = 8 kg is attached to a spring as shown in figure and held in position so that the spring remains Unstretched. The spring constant is 200 N/m. The mass m is then released and begins to undergo small oscillations. The maximum velocity of the mass will be : (g = 10 m/s2) a. 1 m/s b. 2 m/s c. 4 m/s d. 5 m/s.

55

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b. T

c. T 2

d.

T1 T2 T1  T2

1

T12  T22

T

1

2

T12



1 .

T22

A ball of mass 2 kg hanging from a spring oscillates with a time period 2 S seconds. Ball is removed when it is in equilibrium position, then spring shortens by : a. 10 m b. 5 m c. 20 m d. 2 S m. In the figure, the block of mass m, attached to the spring of stiffness k is in contact with the completely elastic wall, and the compression in the spring is e. The spring is compressed further by e by displacing the block towards left and is then released. If the collision between the block and the wall is completely elastic then the time period of oscillations of the block will be:

m k

a. 2 S m/s

m k

c.

S

3

m k

d.

S 6

m . k

b. S m/s

c.

1

S

1 d. 2 S m/s.

m/s

The two blocks of mass m1 and m2 are kept on a smooth horizontal table as shown in figure. Block of mass m1 but not m2 is fastened to the spring. If now both the blocks are pushed to the left so that the spring is compressed a distance d. The amplitude of oscillation of block of mass m1, after the system is released is:

w

w

34.

b. 2 S

Frequency of a particle executing SHM is 10 Hz. The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is Unstretched. Maximum speed of the particle is: (g = 10 m/s2)

w

31.

2S 3

.N

a.

et B

ad

24.

a. T = T1 + T 2

om

22.

Let T1 and T2 be the time periods of two springs A and B when a mass m is suspended from the separately. Now both the springs are connected in parallel and same mass m is suspended with them. Now let T be the time period in this position. Then:

i.c

18.

a. d

m1 m1  m2

b. d

m2 m1  m2

c. d

56

2 m2 m1  m2

d. d

2 m1 m1  m2 .

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41.

Two masses M and m are suspended together by a mass less spring of force constant k. When the masses are in equilibrium, M is removed without disturbing the system. The amplitude of oscillation is: M m g M m g mg Mg b. c. d. k k k k A ball of mass when dropped from certain height as shown in diagram, strikes a wedge kept on smooth horizontal surface and move horizontally just after impact. If the ball strikes ground at a distance d from its initial line off all, then the amplitude of oscillation of wedge after being hit by the ball will be

51.

md M

Mg 2k h

b.

d m

kM g 2h

et B

a.

ad

i.c

48.

om

a.

c. d

kg 2M h

d. d

kg 2mh

The angular frequency of a spring block system is Z0 . This system is suspended from the ceiling of an elevator moving downwards with a constant speed v0. The block is at rest relative to the elevator. Lift is suddenly stopped. Assuming the downward as a positive direction, choose the wrong statement:

.N

v0 a. The amplitude of the block is Z 0

b. The initial phase of the block is S

w

v0 c. The equation of motion for the block is Z sin Zt . 0

d. The maximum speed of the block is v0. Two blocks with masses m1 = 1 kg and m2 = 2 kg are connected by a spring of spring constant k = 24 N/m and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0 = 12 cm/s to the right. The amplitude of oscillation is : a. 2 cm b. 1 cm c. 3 cm d. 4 cm. Two bodies M and N of equal masses are suspended from two separate mass less springs of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is :

w

54.

w

56.

61.

a. k1/k2

b.

k2 / k1

c. k2/k1

d.

k1 / k2 .

A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of: a. 2/3 k b. 3/2 k c. 3 k d. 6 k.

Answers : 7. a 8. b 15. b 18 d, 22. a. 24 a. 31. d 34. a, 41. a 48. a 51. b 54. a 56 b, 61. b.

57

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M 2 k . Period (or kefff) is independent of T .

2S

? T

2k

keff

8.

Effective force constant in case (iii) and (iv) is keff = 2k + 2k = 4k. Therefore, T1

T2

2S

m k

T3

T4

2S

m 4k

And 15.

om

7.

m k .

S

Mean position will be at k, x = mg.

i.c

m g 8 u10 2 m 0.4 m . k 200 5 This is also the amplitude of oscillation i.e., A = 0.4 m.

or, x

200 8

4S 2 m

T1

2S

m or k1 k1

T2

2S

m k2

or k 2

T

2S

Now,

0.4

T12

2 m/ s .

et B

18

k m

A

ad

AZ

now, vmax

4S 2 m T22

4S 2 m

m or k k

T2

.

1 1  T12 T22

22. T

2S

w

w

1 T2

.N

In parallel k = k1 + k2 Substituting the values of k, k1 and k2 we get :

k

w

?

24.

m k

4S 2 m T2

4S 2 2S

2 2

2 N /m

Now, mg = k x0 ?

Z

x0

mg k

2u g = g meter = 10 m. 2

k m

General equation of motion, x = 2e cos Z t

58

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e = 2e cos Z t

S

Zt

3

?

31.

S

m k

3 T

2t

2S 3

m k

Mean position of the particle is amplitude of oscillation is A

m k

?

2S f 1

mg distance below the unstretched position of spring. Therefore, k

mg k

20 S

(f = 10 Hz)

et B

k m

Z

i.c

t

.

ad

?

3

om

S

k t m

400 S 2

.

Therefore, the maximum speed of particle will be : vmax

AZ

§ g ¨¨ 2 © 400 S

· ¸¸ 20 S ¹

1 m/s . 2S

w

.N

More than One Option is correct: 3. A constant force F is applied on a spring block system as shown in figure. The mass of the block is m and spring constant is k. The block is placed over a smooth surface. Initially the spring was Unstretched.

w

Choose the correct alternative (s) : a. The block will execute SHM

w

F b. Amplitude of oscillation is 2 k .

12.

c. Time period of oscillation is 2 S

m k

d. The maximum speed of block is

2 F x  k x2 . m

(Here, x = F/k)

A block of mass m is attached to a mass less spring of force constant k, the other end of which is fixed

59

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from the wall of a truck as shown in figure. The block is placed over a smooth surface and initially the spring is Unstretched. Suddenly the truck starts moving towards right with a constant acceleration a0. As seen from the truck:

om

k

m

m k

b. The time period of oscillation will be 2 S

c. The amplitude of oscillation will be

ad

m 2 a02 . k In a spring – block system force constant of spring is k = 16 N/m, mass of the block is 1 kg. Maximum kinetic energy of the block is 8J. Then : a. Amplitude of oscillations is 1 m. b. At half the amplitude potential energy stored in the spring is 2 J. c. At half the amplitude kinetic energy is 6 J. d. Angular frequency of oscillation is 16 rad/s.

et B

d. The energy of oscillation will be

16.

Answers: 3. a, c, d

12. a, b, c

16. a, b, c.

The situation is similar as if a block of mass m is suspended from a vertical spring and a constant force mg acts downwards. Therefore, in this case also block will execute SHM with time period:

.N

3.

m a0 . k

i.c

a. The particle will execute SHM

2S

m k

w

T

At compression x where F = k x ;

x

F . k

F k At mean position speed of the block will be maximum. Applying work energy theorem.

w

w

This is also the amplitude of oscillation. Hence A

12.

or,

F. x =

v

1 1 k x2  m v2 2 2 2 F x  k x2 . m

Potential energy at extreme position. Free body diagram of the truck from non inertial frame of reference will be as follows :

60

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This is similar to a situation when a block is suspended from a vertical spring. Therefore, the block will execute simple harmonically with time period m a0 m (ma0 = kx) . Amplitude will be given by A = x = k k

2S

k m

Z

16 = 4 rad/s. 1

1 2 m vmax 2

8.

?

1 u 1u Z A 2

?

1 u 1 u 16 u A 2 8 2

At half the amplitude x 1 K x2 2

1 §1· u 16 u ¨ ¸ 2 ©2¹

w

U

§ m a0 · ¨ k ¸ © ¹

2

m 2 a02 . 2k

8

or

A 2

.N

2

1 k 2

et B

16.

1 k A2 2

ad

Energy of oscillation will be E

i.c

om

T

A = 1m.

0.5 m, potential energy stored in the spring will be

2

2J

Only when spring block system is horizontal. At half the amplitude v 2

1  0.5

4

v

2 3m/s

w

v

Z A2  x 2

2

w

2 1 1 u 1u 2 3 u m u v2 6J 2 2 Match the Column 3. In spring – block system match the following: Table – 1 Table – 2 (A) If k (the spring constant) (P) Speed will become 16 times is made 4 – times (B) If m (the mass of block) is (Q) Potential energy will becomes four times

? KE

61

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made 4 – times (C) If k and m both are made 4 – times

(T) None. In the two block spring system, force constant of spring is K = 6N/m. Spring is stretched by 12 cm and then left. Match the following:

Table – 1 (A) Angular frequency of oscillation (B) Maximum kinetic energy of 1 kg (C) Maximum kinetic Energy of 2 kg

i.c

(Q) 3 SI unit

(R) 2.4 × 10-3 SI unit

ad

(S) None. In a spring – block system; match the following: Table – 1 Table – 2 (A) If mass of the block is (P) Energy of oscillation becomes 4 times Doubled (B) If amplitude of oscillations (Q) Speed of particle becomes 2 times Is doubled

et B

12.

Table – 2 (P) 4.8 × 10-3 SI unit.

om

8.

(R) Kinetic energy will remain unchanged

(C) If force constant is doubled (R) Time period becomes 2 times (D) If angular frequency is doubled (S) Potential energy becomes 2 times. 3. a q, b t, c. q

Answers:

8. a q, b s, c p

12. a r, b p, c s, d q.

w

w

w

.N

Assertion – Reason 17. Assertion: A block of mass m is attached to a spring inside a trolley of mass 2m as shown in figure. (i) all surfaces are smooth. Spring is stretched by A and released. Both trolley and block oscillate simple harmonically with same time and same amplitude. Reason: The problem is similar to a two – body problem as shown in figure. (ii).

18.

a. A b. B c. C d. D e. E. Assertion: If a man with a wrist watch on his hand falls from the top of a tower, its watch gives correct time during the free fall. Reason: The working of the wrist watch depends on spring action and it has nothing to do with gravity. a. A b. B c. C d. D e. E.

Answers: 17. d.

18 a.

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Comprehension Based questions: Concept: [6 – 8]: In perfectly inelastic collision between two bodies momentum remains constant and K and maximum speed m

the bodies stick together. Angular frequency of a spring block system is Z

om

of particle is SHM is Z A , where A is the amplitude. Question: Two identical blocks P and Q have mass m each. They are attached to two identical springs A and the right spring 2 (along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initially time period of both the block was T.

a. 7.

T

2T

2 The amplitude of combined mass is:

c. T

A A b. 4 2 What is energy of oscillation of the combined mass?

a. 8.

b.

ad

The time period of oscillation of combined mass is :

1 k A2 2 Answers: 6. c 7. a

1 k A2 4 8. d

b.

c.

d.

c.

1 k A2 8

2A 3

d.

T . 2

d.

3A . 4

1 k A2 . 16

w

.N

a.

et B

6.

i.c

initially unstretched. Now the left spring (along with P) is compressed by

1 kg is attached with two springs each of force constant k = 10 N/m 2 as shown. The block is oscillating vertically up and down with amplitude A = 25 cm. When the block is at its mean position one of the spring breaks without changing momentum of block. What is the new amplitude of oscillation. a. 10.6 cm b. 43.3 cm c. 25 cm d. 62.2 cm.

w

w

Question [9.10] : A block of mass m

9.

10.

If instead of breaking one spring amplitude? a. 21.6 cm

b. 26.2 cm

m mass would have fallen from the block, then what will be new 2 c. 15.8 cm

63

d. 16.9 cm.

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k . Maximum acceleration in SHM is m

Statement [11 – 12] : Angular frequency in SHM is given by Z

Z 2 A and maximum value of friction between two bodies in contact is P N, where N is the normal reaction between the bodies. In the figure shown, what can be the maximum amplitude of the system so that there is no slipping between any of the blocks? P 

2 a. 7 m

4 10 3 m. m c. m d. 9 3 4 Now the value of k, the force constant is increased then the maximum amplitude calculated in above question will : a. rem ain same b. increase c. decrease d. data is insufficient.

b.

ad

12.

i.c

P

om

11.

g 2

18.

.N

et B

Question [18 – 19] : There is a spring block system in a lift moving upwards with acceleration a

In mean position (of block’s oscillations) spring is :

w

mg a. compressed by 2 k mg c. elongated by 2 k

b. elongated by

mg k

3m g d. elongated by 2 k .

5m g If maximum extension in the spring is 2 k , then maximum upward acceleration (a1) and maximum.

w 19.

w

Downward acceleration (a2) of the block are:

g 3g g 3g b. a1 d. a2 c. a2 . 2 2 2 2 Statement [20 – 21]: In case of pure rolling a = R D , where a is the linear acceleration and D the angular acceleration. Question: A disc of mass m and radius R is attached with a spring of force constant k at its center as shown in figure. At x = 0, spring is unstretched. The disc is moved to x = A and then released. There is no slipping between disc and ground. Let f be the force of friction on the disc from the ground. a. a1

64

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f versus t (time) graph will be as :

21.

In the problem if k = 10 N/m, m = 2 kg, R = 1 m and A = 2 m. Find linear speed of the disc at mean position : a.

40 m/s 3

b.

20 m / s

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i.c

20.

c.

10 m/s 3

d.

50 m/s 3

w

.N

et B

Answers: 9. b, 10. a 11 c. 12. c. 18. d 19. b, c 20. c 21. a. 15.6 Simple Pendulum: A small heavy spherical bob connected, to one end of a light inextensible and flexible string, with its other end fixed to a rigid and unyielding support, constitutes a simple pendulum. Let P be the position of the bob at any instant, when the string makes an angle T , with the vertical. (Figure. 15.14). If ‘l’ be the length of the string then, the net unbalanced force on the bob is F = mg sin T . [where m is the mass of the bob].

w

l

w

Assuming T to be very small sin T | T Net unbalanced force F = mg ( T )

Now, if ‘x’ be the displacement of the bob from the mean position C, then T = ?

)& F

& § x · m g ¨¨ ¸¸ © l ¹

Acceleration f

F m

g x l

65

x . l

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? f D  x , which proves that the bob executes SHM for small amplitudes (less than 40).

Z 2 x,

g l

2 We get Z

g . l

Ÿ Z

om

Comparing with f

Alternatively: The torque acting about the point of suspension, is T = mg l sin T Now, moment of inertia I of the system about the point of suspension is I = ml2 From T = I D , we get m g l sin T

Comparing with g or T l

Z

g T

i.c

dt

2

l

d 2T dt

2

2S

g sin T . l

Z 2 T

ad

d 2T

Ÿ

ml 2 D Ÿ D

l g

w

(ii)

[Law of lengths] T D l [other things remaining constant] If a child sitting on a swing stands up, then the effective length l decreases, due to an uplift of the center of gravity of the system. This reduces the time period consequently. If a hollow sphere filled with a liquid, and having a hole at its base, be made the bob of a simple pendulum, the time period, initially increases, due to an increase in the effective length as the center of gravity gets lowered due to the outflow of liquid from the sphere. The time period once again, gets restored, when the whole liquid flows out, due to the fact that the center of gravity once again, comes to the center of the sphere. If the string be elastic with Young’s modulus Y, then the length of pendulum will be l + ' l [where l is the original length and ' l, increase in length]

.N

(i)

et B

Note: Here ‘l’ is the “effective length” of the simple pendulum, measured from the point of suspension to the center of gravity of the bob. A Few Important Facts Regarding Simple Pendulum (a) Change of time period T due to a change in effective length.

(iii)

w

From Y

w

Ÿ

'l l

Stress ŸY Strain

M gl A u 'l

[where M is the mass of the bob]

Mg AY

§ 'l · ? Net length = l ¨ 1  l ¸ © ¹ Now, Time period T

2S

§ Mg· l ¨1  ¸ AY ¹ © l § Mg· ¨1  ¸ g © AY ¹

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Now, since Mg < < A Y Mg  1 AY

?

§ Mg· ¨1  ¸ AY ¹ . [From Binomial expansion] © Obviously, the time period increases, in this case. If the effective length be comparable to the radius of earth (Fig. 15.15)

ad

i.c

(iv)

l g

2S

om

T

The net unbalanced force on the bob = mg sin T  I ?

F

m g T  I [Assuming T  I to be small]

?

)& F

& & §x x· m g ¨¨  ¸¸ © l R¹

et B

§1 1 · & m g ¨  ¸ x ©l R¹

§1 1 · & g ¨  ¸ x . ©l R¹ )& & Which is of the form f Z 2 x , indicating that the bob, still executes SHM and its time period is not

.N

)& or acc. f

w

2 infinity. Comparing the above equation with std. form Z

T

1 §1 1 · g¨  ¸ ©l R¹

w

? Time period

2S

§1 1 · g¨  ¸ ©l R¹

w

Putting l = R, we can determine, the corresponding time period as T

If l be too large i.e., l o f . 1 o0 l

Ÿ

?

T

2S

R g

84.6 min .

If l be too small compared to earth’s radius, then

67

1 1  . R l

2S

R | 1 hr 2g

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1 1 in comparison to . l R

? Neglecting

2S

l g as derived earlier..

om

? T

(b) Change of time period with a change in acceleration due to gravity. (Law of gravity) 1

TD

If a simple pendulum be taken to higher altitudes, its time period increases, due to a decrease in the value of g. Consider a place P at height ‘h’ above the earth’s surface. (Fig 15.16). The value of g at that place g’ (say) is given by :

ad

i.c

(i)

g [other things remaining constant]

Rh

et B

g R2

g'

2

Where g = acc. due to gravity on the earth’s surface.

2S

l Rh

gR

w

Ÿ

2S

2

w

Now, if h < < R or

w

Then , T

(ii)

l g' .

.N

Time period T

2S

2

Ÿ

T

§ Rh· l 2S ¨ ¸ © R ¹ g

2S

l g

h· § ¨1  R ¸ © ¹

h g. (Fig. 15.18).

w

For a simple pendulum oscillating in a vehicle accelerating with an acceleration ‘a’, the time period decreases due to an increase in the value of ‘g’.

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Let us first determine the equilibrium position of the string (see figure 15.20). If D be the angle made by string with vertical. Then : T cos D = mg and T sin D = ma

om

D

§a· a Ÿ D tan 1 ¨ ¸ ©g¹ g (Opposite to the direction of acceleration of vehicle).

Dividing tan D

? Effective acceleration g '

2S

l g'

2S

1/ 2

l

2

g  a2

1/ 2

.

If the bob of simple pendulum is acted upon by additional forces of electrical, magnetic, buoyant, gravitational origin, the time period will increase or decrease according as the net acceleration in the downward direction is decreased or increased (see problem No. 18, 19) (C) Law of Mass: The time period of a simple pendulum is independent of the mass of the bob, other things remaining constant. For example, if two pendulums, be constructed of same effective length, but whose one of the bobs is solid and the other hollow; both oscillate with the same time period. (d) Law of Isochronism: The time period of a simple pendulum is constant for every oscillation provided the amplitude remains small. Example 17

.N

(vi)

T

a2  g 2

et B

? Time period

ad

Effective external force = m a 2  g 2

i.c

D

w

Show that the maximum tension in the string of a simple pendulum, with angular amplitude T 0 is given

2 by mg 1  T 0 . Hence find the maximum mass of bob that can be used, if the breaking load of the

w

w

string be F. Solution Evidently, the tension is maximum, at the lowest position of the bob. (Fig. 15.21) m v 2 max … (1) l Where vmax is the maximum velocity of the bob (attained in the mean position). Tmax

mg 

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vmax



a Z

2 vmax

aZ Ÿ

2

2

Ÿ T0 l

2

§ g· ¨¨ ¸¸ © l ¹

2

T02 g l

[Where a = linear amplitude and Z = ang. Freq.] Substituting for mg 

Tmax

in eqn. (1)

m T02 g l l

om

2 vmax

m g 1  T 02

Ÿ

M

M g 1  T 02

i.c

If M be the maximum mass, that can be employed for the bob. Then F F g 1  T 02

ad

Example 18 A simple pendulum of length l, is made to oscillate with a solid sphere (density D), in a non viscous fluid of density U ; find the period of oscillation (D > U ) Solution Net force on the bob will be (D - U ) Vg (down). Net Acceleration downward (See Fig. 15.22)

DU V g DV

w

f

.N

et B

U

w

? Time period

T

U· § ¨1  D ¸ g © ¹ 2S

l U· § g ¨1  ¸ . © D¹

w

? The time period increases, due to a decrease in effective value of g. Example 19 A simple pendulum consists of a small sphere of mass m, suspended by a thread of length L. The sphere carries a negative charge ‘q’. The pendulum is placed in a uniform electric field of strength E directed (i) vertically upwards (ii) vertically downwards, and (iii) horizontal. Find the time period in each case. Solution (a) The net force downward F = mg + qE

Net acceleration downward

F m

f

§qE· g ¨ ¸ © m ¹

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L §qE · . g ¨ ¸ © m ¹

2S

T Time period

om

Note: The electric force experienced by a body carrying a negative charge is opposite to the direction of electric field. (b) The net force downward = mg – qE [Assuming mg > qE] [Fig. 15.24 (a)] § qE · g ¨ ¸ © m ¹

L §qE· g ¨ ¸ © m ¹

et B

2S

Time period T

ad

i.c

? Acceleration f

The net force upward = qE – mg

[Assuming mg < qE] [Fig. 15.24 (b)]

§ qE · ¨ m ¸g © ¹

Acceleration f

2S

L §qE · ¨ m ¸ g © ¹

.N

T

? Time period

w

(c) If 'T ' be the angle made by the string with the vertical (Fig. 15.25) Then T cos T = mg T sin T = qE

w

Ÿ

tan T

qE mg

ŸT

mg

2

 qE

2

.

w

Net force on the bob =

§ qE · tan 1 ¨ ¸ © mg ¹

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2S

? Time period T

§ qE · g2  ¨ ¸ © m ¹

2

L 1/ 2

­° 2 § q E · 2 ½° ®g  ¨ ¸ ¾ © m ¹ °¿ °¯

.

om

? Net acceleration f

Example 20 A simple pendulum is suspended from the ceiling of a carriage, moving along a horizontal curve of radius of curvature r, with a speed v. If ‘l’ be effective length of the pendulum find (i) the tension in the

et B

ad

i.c

string, while oscillating in the mean position and (ii) the period of oscillation (Assume T 0 as the angular amplitude). Solution Let us adopt the carriage as the frame of reference. Therefore, in addition to the tension and weight acting on the bob, a pseudo (centrifugal) force should also be applied on it. (Fig. 15.26)

.N

At the mean position, the pendulum is stable. ? For its equilibrium net vertical and horizontal forces should be zero. ? T cos D = mg ª ma « where a «¬

And T sin D

v2 º » r »¼ .

a g

w ?

tan D

w

or, g sin D - a cos D = 0. … (1) Now, let T be the displacement (from mean position). at any instant Then, the net unbalanced force tangential to the motion of the bob.

w

F = mg sin D  T - ma cos D  T . = m [ g sin D .cos T  cos D .sin T )  a (cos D .cos T  sin D .sin T )] Since T is too small x l Where ‘x’ is the displacement of the bob from the mean position

? cos T | 0 and sin T |

)& ? F

& & ª § § § x · §  x · · ·º « ¨ m g sin D  cos D . ¨¨ ¸  a ¨¨ cos D  sin D ¨¨ ¸¸ ¸¸ ¸ » . ¸ l ¹¸ l « ¨ © © ¹ 73 © ¹ ¹ ¼» ¬ ©

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m ª¬ g sin D  a cos D  x / l g cos D  a sin D º¼  g cos D  a sin D x

Z 2 x, the standard form of a simple harmonic motion, we get

§ g cos D  a sin D · ¨ ¸ ŸZ l © ¹

§ v 2 sin D ¨ g cos D  r ¨ l ¨ ¨ ©

1/ 2

· ¸ ¸ ¸ ¸ ¹

Now, maximum velocity (vel. at the mean position) vmax § gr cos D  v 2 sin D l 2T 02 ¨ ¨ rl ©

2 vmax

· ¸¸ . ¹

T0 l Z

ad

Or,

om

? Comparing with f

Z2

[By virtue of 1]

l

i.c

? Acceleration f

et B

Now, from the FBD of the pendulum in its mean position, it is evident that, for equilibrium along the string.

w

§ v 2 sin D m 1  T 02 ¨ g cos D  ¨ r ©

w

Now, from tan D

w

sin D

?

T

a 2

a g

2

ª v2 v 2 sin D º m « g cos D  sin D  T02 g cos D  T 02 » r r »¼ ¬«

2 m vmax l

.N

m g cos D  m a sin D 

T

· ¸¸ ¹

a g we have

and cos D

g 2

a  g2

ª g2 a2  m 1  T02 « « a2  g 2 a2  g 2 ¬

(ii) Time period =

2S

Z

2S

º » » ¼

1/ 2

m

l § 2 v4 ¨¨ g  2 r ©

1/ 2

· ¸¸ ¹

74

1  T02

ª 2 v4 º «g  2 » r »¼ «¬

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i.c

om

Example 21 A simple pendulum, of length l, installed between two inclined walls, each making an angle D with the vertical, oscillates with a maximum speed of u, colliding elastically with the walls. Find the time period of oscillation. For the above problem, how is D related to u? (Fig. 15.28)

Solution Had there been no walls, the linear amplitude of oscillation would be u

g l

But since, Z l g

u

et B

a

a Z]

[ From vmax

Z

ad

a

Now, let the equation of motion for the pendulum be T T 0 sin Z t [where T 0 = angular amplitude] The time taken by the pendulum to move from O (mean position) to A (extreme position), can be § g · sin ¨ t ¨ l ¸¸ gl © ¹

u

sin

g t l

D gl

Ÿ

t

u

§D gl · g sin 1 ¨ ¸ ¨ u ¸ l © ¹

§D gl · l sin 1 ¨ ¸ ¨ u ¸ g © ¹

w

Ÿ

.N

obtained as D

Ÿ

t

w

Evidently, the period of oscillation T = 4t

w

Ÿ

T

4

§D gl · l sin 1 ¨ ¸ ¨ u ¸ g © ¹

Now, since, the bob collides with walls, the angular amplitude should exceed the value D . u i.e.,

gl

! D or u ! D

gl

This is the required relation. Note: If the bob just misses the collision with wall. Then, in the limiting case, angular amplitude = D

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T

Then ,

4

l sin 1 (1) Ÿ T g

4

l §S · g ¨© 2 ¸¹

2S

l g .

2S

From T

l g

we have l

om

This proves the validity of above derivation. Second’s Pendulum: It is a pendulum with period of oscillation 2 sec or period of swing 1 sec. Length of a second’s pendulum at mean sea level, may be calculated as follows. T2 ug 4S 2

.

Putting T = 2 and taking S 2 | g

et B

ad

i.c

l | 1m . ? Note: A second’s pendulum always bears a time period of 2 sec. However, the length changes with a change in acceleration due to gravity. Loss or Gain of Time by Pendulum Clocks: Pendulum clocks gain time, if the time period of the pendulum decreases (due to a decrease in length in winter or due to increase in acc. due to gravity, as when taken from equator to poles). For example, let the time period of a second’s pendulum (originally) decrease to 1.99 sec. Now, each time. It oscillates, it counts 2 sec rather than 1.99 s. Thus, after a considerable no. of oscillations, the time counted or shown by the clock is more than the actual time, and consequently, is goes fast. Similarly, Pendulum clocks loose time and go slow, when their time period increases. Gain or loss of time due to a change in surrounding temperature: Let T be the time period of clock pendulum of effective length l, showing the correct time. T

Then,

2S

l g

… (1)

Now, let ' T be the change in temperature of the surrounding, the new length will be

.N

l 1  D ' T [where D is the coefficient of linear expansion of the material of the pendulum]

l'

? The changed time period

2S

l' g

w

T'

2S

l 1D 'T g

… (2)

Dividing equation (2) by (1).

w

T' T

w

Ÿ

Ÿ

1 D 'T

1  D 'T

1/2

§ 1 · ¨ 1  2 D ' T ¸ [Expanding by binomial expansion] © ¹ § 1 · T c T ¨1  D ' T ¸ © 2 ¹

§ T ' T · Also; ¨ ¸ © T ¹

1 D 'T 2

i.e., Relative change in time period

1 D 'T 2

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1 D 'T . 2 Note: Unless otherwise stated, adopt the time period of clock pendulum as 2 sec. Example 22 A clock with an iron pendulum keeps correct time at 200 C, how much time will it loose or gain, in a day, if the atmospheric temperature changes to 400 C. ( D for iron = 0.000012/0 C) Solution The relative change in time (i.e., time lost or gained per swing i.e., in 1 sec is given by) 'T T

om

or, time lost or gained each second is

1 1 D u ' T Ÿ u 12 u106 40  20 2 2

Ÿ 12 u 105 s

et B

ad

i.c

? Total time lost per day = 12 u105 u 86400 12 u 864 u 103 10.37 sec Example 23 The pendulum of a certain clock has a time period 2.02 sec. How fast or slow does the clock run in an interval of one week. Solution Since the correct time period for a clock pendulum is 2 sec. Therefore, for every 2.02 sec. (one oscillation period) of actual time elapsed, it would count only 2 sec. In other words, as successive oscillations are repeated it goes on loosing time, and would run slow. Loss of time per oscillation ( = 2 sec). = 2.02 – 2 = 0.02 sec. ? Loss of time per sec = 0.01 sec. ? Time lost in 1 week = 0.01 × 86400 × 7 = 6048 sec = 108 min. Example 24 The linear displacement ‘x’ of a simple pendulum’s bob from its mean position varies as x = a sin (0. 707 S t) where a is a constant and t is in sec. Find its length (take g = S 2 m/s). Solution Since the simple pendulum executes SHM. Therefore comparing the displacement equation with

.N

standard form x = a sin Z t , we get Z

w

Also from, T 2 We have, Z

2S

l g

T

2

rad/sec.

2S

Z .

g S2 Ÿ l 2

g l

Ÿ l

2m.

w

Example 25 A simple pendulum of 1 meter, suspended from the ceiling of a lift executes oscillation with a period

w

of 6.25 sec. If the lift is moving up with an acceleration ‘a’ find a (take g = S 2 m/s2). Solution If ‘a’ be the acceleration of the lift upwards then, net acceleration on the bob downwards, relative to the lift is g + a ?

Ÿ

T

2S

l ga

§ 1 · 6.25 4 S 2 ¨ 2 ¸Ÿ ©S a¹

S2 a

16 2 S Ÿa 25

77

9 2 S 25

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Thus, the lift is retarding with a retardation of

9 2 S m / s2 25

4.

The formula T

See text T

2S

w

5.

l g .

6.

w w

l T  40 . For large amplitudes g holds good only for small angular amplitude 0

R g

2S

64 u 105 s. 9.8

Let the amplitude A be given by A = k1 e- k2t . Then T.E.

7.

2S

.N

T z2S

et B

ad

i.c

om

Objective Questions: 1. The time period of a simple pendulum is independent of : Ans: (d). a. The length of the string b. The radius of the bob c. Amplitude of oscillation d. None of these. 2. The time period of a simple pendulum of infinite length suspended on earth’s surface is :Ans: (c) a. Inifinte b. 59.8 minutes c. 84.6 minutes d. 24. hrs. 3. The ampliutude of oscillation of a simple pendulum decays exponentially with time due to air resistance and friction at the point of suspension. The total energy of the pendulum. Ans: (A) a. Decays exponentially with time b. Decays linearly with time c. Decays at a steady rate d. Decays at a rate less than the amplitude. 4. A hollow sphere is filled with water. It is hung by a long thread. As the water flows out of a hole at the bottom of the sphere, the frequency of oscillation will : Ans: (D) a. Go on increasing b. Go on decreasing c. First increases and then decreases d. First decreases and then increases.

1 m Z 2 A2 2

1 m Z 2 e  k2 t 2

Freq. of a simple pendulum f

1 2S

2

or , T .E. D e2 k2 t

g 1 Evidenty f D at any place. l l

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w

w

w

.N

15.

The time period of a simple pendulum is T. In which of the following situations will the period increase? Ans : B a. The pendulum is suspendend in lift, going with an acceleration (< g) b. The bob is given a negative charge, while the point of suspension a positive charge. c. The pendulum is suspended in a cabin, moving along a circle with uniform speed” d. The pendulum is suspended in a cabin, moving along a straight path, with a uniform acceleration. The electrostatic force of attraction acting on the bob is up, which decreases the net force of gravity downwards.

et B

6.

ad

i.c

11.

om

5.

The effective length l is the distance of the point of suspension O’ form the center of gravity (C.G) of the bob. As, water flows out, of the hole at the bottom, the C.G descends from center towards the bottom, increasing the effective length and consequently, f decreases. However, when all the water has flown out, the C.G. of a hollow sphere is once again at its center and hence the effective length would decrease, thereby increasing the frequency. A simple pendulum oscillates slightly above a large horizontal metal plate. The bob is given a charge. The time period: Ans : B a. Has no effect, whatever be the nature of charge. b. Always decreases, whatever be the nature of charge. c. Always increases, whatever be the nature of charge. d. May increase or decrease depending upon the nature of charge When any charge is given to the bob of the pendulum, it induces opposite charge on the metal plate, and hence a net force of attraction acts on the bob. Thus, the effective value of g increases.

?

geff  g .

1 So, since T D g a decrease in g eff

increases the time period. In the rest of all the cases, the effective value of g increases thereby decreasing the value of T.

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35.

The bob of a simple pendulum of mass m executes oscillations with a total energy E. At any instant is at one of the extreme positions. Its linear momentum, after a phase change of

S 6

rad takes place, will

mE 3m E d. c. m E . 2 4 A simple pendulum of length l has a time period T for small oscillations. A fixed obstacle is placed directly below the point of suspension, so that only the lower quarter of the string continues oscillations (see figure 15.74). The pendulum is released from rest at a certain point O. The period of oscillation, assuming small angles will be: 2m E

b.

i.c

a. 36.

om

be:

ad

3 /4

O

et B

T 3T T 1 3 d. T. b. c. 4 4 2 Two simple pendulums of time periods 3s and 5s respectively start oscillating simultaneously from their mean positions. What is the minimum time elapsed, before which they oscillate with the same phase? a. 15 s b. 2 s c. 7.5 s d. 8 s. In the above problem no. 38. if the bobs of the pendulums start their oscillations, from two opposite extreme positions, then, the answer would be : a. 3.75 s b. 7.5 s c. 4 s d. 1.5 s. Figure 15.75 shows three physical pendulums consisting of identical spheres rigidly connected by a mass less stiff rod. The pendulums are vertical and are made to oscillate about the point O. If their respective frequencies be f1, f2 and f3 then which of the following is correct :

38.

39.

w

w

w

40.

.N

a.

a. f1 f 2  f 3 b. f1 f 2 ! f 3 c. f1  f 2  f3 d. f1 ! f 2 ! f 3 . Answers: 35. a 36. a, 38. c 39. a 40. a. Only one option is correct 17. A simple pendulum has time period T = 2 s in air. If the whole arrangement is placed in a nonviscous liquid whose density is

1 times the density of bob. The time period in the liquid will be: 2

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32.

om

23.

The maximum tension in the string of a pendulum is two times the minimum tension. Let T 0 be the

i.c

21.

2

s b. 4 s d. 4 2 s. c. 2 2 2 An accurate pendulum clock is mounted on the ground floor of a high building. How much time will it lose or gain in one day if it is transferred to top storey of a building which is h = 200 m higher than the ground floor. Radius of earth is 6.4 × 10 6 m. a. it will lose 6.2 s b. it will lose 2.7 s. c. it will gain 5.2 s d. it will gain 1.6 s. A simple pendulum 4 m long swings with an amplitude of 0.2 m. What is its acceleration at the ends of its path? (g = 10 m/s2) a. zero b. 10 m/s2 c. 0.5 m/s2 d. 2.5 m/s2. a.

angular amplitude. Then cos T 0 is :

3 2 1 3 b. c. d. . 5 3 2 4 A pendulum has time period T for small oscillations. An obstacle P is situated below the point of

33.

ad

a.

3l . The pendulum is released from rest. Throughout the motion the moving 4 string makes small angle with vertical. Time after which the pendulum returns back to its initial position is :

.N

et B

suspension O at a distance

4T . 5 Time period of a simple pendulum of length L is T1 and time period of a uniform rod of the same length L pivoted about one end and oscillating in a vertical plane is T 2. Amplitude of oscillations in both the cases is small. Then T1/T2 is:

b.

3T 4

c.

3T 5

d.

w

a. T

w

35.

1 3 d. . 3 2 The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which moves without friction, down an inclined plane of inclination D , is given by :

a.

w

63.

66.

4 3

a. 2 S

L g cos D

b. 1

c.

L b. 2 S g sin D

L c. 2 S g

d. 2 S

L g tan D

A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = Kt2, (K = 1 m/s2) where y is the vertical displacement. The time period now becomes T2. The ratio of

T12

T22

is: (g = 10 m/s2)

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6 5 4 b. c. 1 d. . 5 6 5 15.9 Physical Pendulum or Compound Pendulum: Any rigid body, suspended about a fixed point and capable of oscillating about a horizontal axis passing through the point, constitutes a physical or a compound pendulum. Figure 15.45 along side shows a rigid body (of any shape) suspended about a fixed point O (the center of suspension) capable of rotating about a horizontal axis through O. Let G be the center of gravity at a distance l from the point of suspension O. If ‘R’ be the vertical reaction upwards at the support O, then for vertical equilibrium, we have.R = mg

ad

l

i.c

om

a.

et B

When at rest, the line of action of R and mg is the same and hence, there is no net torque on the body. If the body be given a slight angular displacement about O, it starts oscillating, the oscillations being angular SHM. Consider the position of the body; at any instant (during oscillatory motion) when the line joining O and G makes an angle 'T ' with the vertical .

.N

The net torque on the body about the point O will be given by W If I be the moment of inertia of the body about O, then W

I

d 2T dt 2

m g l sin T [where

d 2T dt 2

= angular acceleration]

d 2T

 m g l sin T dt 2 [The – ve sign, here indicates that the direction (or sense] of T and W are opposite] [ for small angles sin T = T ] ?

w

I

w

m g lT 0. I dt Which reveals the fact that, the body performs angular SHM with angular frequency

w

?

d 2T 2



mgl I Now, if T be the time period for each oscillation, then

Z

T

2S

Z

2S

I mgl

... (20).

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If K be the radius of gyration of the body about a parallel axis through the center of gravity

then, I = mK2 + ml2 2S

T

K 2  l2 . gl

… (21)

2S

L g

ad

T

Then ,

i.c

om

If L be the length of an equivalent simple pendulum (i.e., a pendulum whose time period is same as that of the compound pendulum)

K2 l . l

et B

Comparing the above equation with equation (21) we get L

K2 more than the distance between the point of l suspension O and the center of gravity G. If we choose a point O’, on the line OG produced such that

Thus, the length of equivalent simple pendulum is

K2 , then O’ is known as the center of oscillation. l

.N

GO’ =

Interchangeability of Center of Suspension and Center of Oscillation:

w

Suppose we shift the point of suspension from O to O’. Then time period T '

2S

w

eq. 21) Where l’ is the distance of O’ (new point of suspension) from center of gravity G.

w

But, GO '

K2 . l

? l'

K2 l

?T'

§ K2 · K  ¨¨ ¸ l ¹¸ © 2S § K2 · g ¨¨ ¸¸ © l ¹

2

2

Ÿ

2S

§ K2 K 2  ¨¨1  2 l © 2 §K · g ¨¨ ¸¸ © l ¹

83

· ¸¸ ¹ .

K 2  l2 . (from gl '

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2S

g l2 K 2  l2 which is same as equation (19). gl

om

2S

l2  K 2 l

i.c

? T' T Thus, it is evident that, the time period remains the same for both the center of suspension and center of oscillation i.e., both are interchangeable. Minimum Period of Oscillation:

1 §§ K · ¨¨ 2S ¸  g ¨© l ¹ © 2

Which can be rewritten as T

1 g

l

2

· 2K 2K ¸ ¸ ¹

2 ª§ K º · «¨  l ¸ 2K» . «© l » ¹ ¬ ¼

et B 2S

2S

ad

From equation (21), the time period T for a compound pendulum is given by T

§ K2 · ¨¨ l  l ¸¸ © ¹ g

2

.N

§ K ·  l ¸ is also min. Since the min value of a  2K is constant, therefore T will be min, when ¨ © l ¹ square can be zero. ? K=l Thus, the time period of oscillations is minimum when the distance of the center of suspension from the center of gravity is same as the radius of gyration of the body about a parallel axis through C.G.

w

Tmin

Also,

2S

2k g

... (22)

w

w

Example 33 A uniform circular disc of radius R, is suspended about a point (not its center) such that it can oscillate about a horizontal axis through that point. What is the distance of this point from the center, for the time period to be minimum? What is the minimum time period? Solution The radius of gyration for a disc about its center of gravity (i.e., geometrical center) is given by R

K

2

ª «k «¬

I M

M R2 2M

º ». »¼

? This is the required distance

Also, Tmin

2S

2R (See equation 22).

g 2

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2S

R 2 . g

i.c

om

Example 34 Find the frequency of oscillations of a uniform thin ring of radius ‘R’ suspended about a point on its circumference.

From equation (20) T Here I

2 m R2 , l

I mgl .

R 1 2S

mgR

2mR

2

1 2S

g 2R .

et B

? Frequency n

2S

ad

Solution

w

.N

15.10 Torsional Pendulum: A rigid body suspended by means of a wire, with its one end fixed rigidly to the ceiling, and about which the body can perform horizontal angular oscillation, is known as a torsional pendulum. Figure 15.48 shows a body attached to a wire OO’ whose one end O, is fixed to the ceiling. When the body is at rest, there is no torque on the body. If the body be given a slight “twist”, and left free, it starts performing angular SHM. Let O’ P be a horizontal line chosen on the body at rest. Further let O ‘P’ be its position at any instant, during oscillations. Since the upper end O of the wire is fixed, and the lower end O’ is rotated by an angle 'T ' we say that the wire has a twist 'T ' . Due to this twist, a torque (restoring) is developed in the wire due to elastic forces, which tend to bring the wire and the body back to a position T = 0 (i.e., the mean position). This restoring torque is directly proportional to the twist, but opposite in sense. Then,

W D T

Or, W

 kT .

w

w

Where k is a positive constant of proportionality known as torsional constant (for the wire).

If I be moment of inertia of the body about the wire, then

85

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I d 2T

W

dt 2

I d 2T

d 2T

k  T 0. I dt dt Which proves the fact that, the body executes angular SHM, with an angular frequency. 2

Z

Ÿ

2

om

k T

k . I

? Time period T

2S

Z

I k

2S

... (23)

i.c

?

ad

Example 35 A circular disc of mass m = 500 gm and radius r = 5 cm is suspended by means of a wire of torsional constant k = 0.36 N – m/radian. Find the time period of oscillations. Solution m r2 . 2 ? From equation (23), time period of the torsional pendulum

2S

m r2 I Ÿ 2S 2k k

.N

T

et B

Moment of inertia of the disc about the wire

Substituting the values of m, r and k.

w We get T

w

?T

S

72

2

5 u102 1 2S u 2 2 u 36 u 102

2S

25 u102 4 u 36

2S u

5 20 u 36

sec.

w

Example 36 A torsional pendulum consists of a square lamina, suspended by means of a wire with torsional constant k = 0.25 N.m/rad. If the time period of oscillation be 2.5 sec, find the moment of inertia I of the square lamina, about the wire. Solution: The time period for oscillations of a torsional pendulum is given by T

or,

I

2

T 2k

2.5 u 0.25

4S 2

4 u 3.14

2

0.04 kg  m 2 .

86

2S

I . k

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(ii) equal strains in bob wires. 119.

ad

i.c

om

PRACTICE PROBLEMS: PROPERTIES OF MATTER: 116. In a Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young’s modulus of the material of the wire from these data. 117. A 5 m long cylindrical steel wire with radius 2 × 10-3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. (For the steel wire: Young’s Modulus = 2.1 × 1011 Pa; Density = 7860 kg/m3; Specific heat = 420 J/kg – K). 118. A light rod of length 200 cm is suspended 200 cm is suspended form the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross – section 0.1 sq cm. Find out the position along the rod at which a weight may be hung to produce (i) equal stress in both wires. Ybrass

1u 1012 dyne / cm2 , 2 u 1012 dyne / cm 2

.N

et B

A sphere of radius 10 cm and mas 25 kg is attached to the lower end of steel wire which is suspended from the ceiling of a room, the point of support is 521 cm above the floor. When the sphere is set swinging as a simple pendulum, its lowest point just grates the floor. Calculate the velocity of the ball at its lowest position. (Young’s modulus of steel = 20 × 1011 dyne/cm2. Unstretdched length of wire = 500 cm, Radius of steel wire = 0.05 cm) g = 980 cm/s2. A uniform cylindrical pulley of mass M and radius R can freely rotte about the horizontal axis O. The free end of a thread tightly wound on the pulley carries a mass m0. At a cetain angle D it counter balances a point mass m fixed to the rim of the pulley. Find the frequency of small oscillations of the arrangement. (Fig. 15.25) Solution: Step 1 : Under equilibrium of the system, the total

w

The net unbalanced torque on the cylinder about its center = r] m g R D  T  m0 g R Ÿ g R [m (sin D  cos T sin T )  m0 ]

= g R [m (sin D  cos T )  m0 ]

w

[ T is small cos

m R cos D T . Now, it I be the moment of inertia of the system about the horizontal axis of rotation then

w

W

T 1 and sin T  T ]

I

Now, from W

?

M R2  m R 2  m0 R 2 2 I d2 T dt 2

§ M R2 · d2T  m R 2  m0 R 2 ¸ ¨¨ 2 ¸ © 2 ¹ dt

 m g R cos D T

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Comparing with the standard form

Z with an angular frequency

Frequency n

?

Z 2S

d2 T dt

2

Note: From equation (1) m0 = m sin D n

1 S 2

Z 2 T it is evident that the system executes angular SHM,

m g cos D §m · R ¨  M0  m ¸ . 2 © ¹ 1 2S

om

 m g cos D T §m · R¨  M0  m¸ ©2 ¹

i.c

Step 3 : dt

2

2 m g cos D R M  2 m  m0

ad

d2T

2 m g cos D R [ M  2 m (1  sin D )]

?

g m0 R

et B

Clockwise moment of due to mass m0 should be equal to the anti – clockwise moment due to mass m about the center of the pulley.

g m R sin D

… (1)

w

w

.N

Step 2: Let the cylinder be given a slight angular displacement, the cylinder starts executing SHM. (which is to be proved). Consider, the position, in which the cylinder has a displacement T (in the clockwise direction). (Fig. 15.53)

S

v R [ M  2 m (1  sin D )] . 2 2. A point mass ‘m’ is suspended at the end of mass less wire of length l and corss – sectional area ‘A’. If ‘y’ be the Young’s modulus of elasticity for the wire, find the time period for small vertical oscillations. Solution: Step 1: Let ' l be the initial extension in the wire, then since strain = stress / Y

w

n

?

'l l

mg AY

... (1)

88

om

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l

u AY  m g

[ Restoring force = strain × Y × A]

ad

'l  y

i.c

Step 2: Let the mass be displaced slightly and left free to oscillate. Consider the position, at any instant, when the mas is at distance ‘y’ below the mean position. Net unbalanced force (upwards) acting on the block. F = Restoring force – weight f block.

AY y

[By virtue of eqn. (1)]

l

§ AY · )& ¨ ¸ y © l ¹ Which proves the fact that the mass executes SHM.

et B

)& F

Ÿ

Step 3: The net acceleration (up) can be obtained as d2 y d t2

§ AY · ¨ ¸y © ml ¹

2S

2S

Z

AY ml

ml AY

An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving pistorn of mass M. The piston and the cylinder have equal cross sectional area A. Atmosphere pressure is P 0 and when the piston is in equiblibrium the volume of the gas is V0. Assuming the system to be completely isolated form its surroundings, show that the piston executes SHM, when the piston is displaced slightly form its equilibrium position, and find the time period (ratio of sp. heat capacity for the gass = J ). Solution Step 1: When the piston was in equilibrium net force (acting vertically) on it must be zero.

w

w

5

F m

w

? The period T

dt

2

Z 2 y , the angular frequency for the oscillations is Z

.N

Comparing with

d2y

Mg P1 ... (1). A Step 2 : Let the piston, be at any instant, displaced (up) by a distance ‘y’ from its mean position. If P2 be its new pressure, then for an adiabatic process. P0 

or,

Ÿ

P2 [(l  y) A]J

P1 [(l A)]J

P2 [V0  A y ]J

P1 [V0 ]J

[where lA = V0]

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J

ª Ayº ª Ayº Ÿ P2 V0 «1  P1 V0J Ÿ P2 «1  J » » P1 V0 ¼ ¬ V0 ¼ ¬ [Expanding by binomial theorem and neglecting the terms containing higher powers of y] Ÿ

P2

P1 ª A yº «1  » ¬ V0 ¼

... (2).

Now, the net unbalanced force acting on the piston (downwards)

P0 A  M g  P2 A .

i.c

ª§ º Mg·  P2 » . A «¨ P0  ¸ A ¹ ¬© ¼

om

J

ªJ Aº A P1 « »y ¬ V0 ¼

J Ay

in comparison with 1]

et B

[Neglecting

ad

ª§ ·º «¨ ¸» P1 «¨ ¸» A «¨ P1  § J A y · ¸ » [From equations (1) & (2)] «¨ ¨1  ¸ ¸¸ » «¨© V 0 © ¹ ¹ »¼ ¬

V0

A2 J § Mg· y P0  [From equation (1)] ¨ V0 © A ¸¹ ? Now comparing with the standard form of equation for SHM: F = – K (y), it is evident that the piston executes SHM.

.N

F

d2y

Step 3 : The net acceleration on the piston d2y

w

Compring with

d t2

dt

2

w w

2S

J A2 §

Mg· P0  y . M V0 ¨© A ¸¹

Z 2 y .

The angular frequency Z can be obtained as Z

? The period T

F M

J A2 §

Mg· P0  ¨ M V0 © A ¸¹

M V0 §

Mg·

J A2 ¨ P0  A ¹¸ ©

6.

A particle of mass m is attached to the mid point of a wire of length L and stretched between two fixed points. If T be the tension in the wrie, find the frequency of lateral oscillations. Solution: Let the mass m be displaced by an amount ‘x’ from its mean position (while oscillating) laterally. The net unbalanced force (towards the mean position) will be :

90

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2 T sin T



& 4T x

ad

)& or F

i.c

om

F

§ x· ¨ ¸ 2T ¨ L ¸ ¨¨ 2 ¸¸ © ¹

L

d2x i.e.,

dt

2

et B

)& F Which proves the fact that the mass executes SHM. The net acceleration of the mass m is m

4 T ( x ) mL .

d2x

.N

Comparing the above equation with

d t2

w

It is evident that the angular frequency Z

? Frequency of oscillations n

w

1 2S

4T mL . 4T mL

An elastic string of natural length 2a can support a certain weight extending to a total length of 3a. One end of the string is now attached to a point on a smooth horizontal table and the same weight is attached to the other end and can move on the table. Show that, if the weight is pulled to any distance

w

7.

Z 2S

Z 2 x .

S §a·

and let go, the string slacks after a time 2

¨ ¸. ©g¹

Solution If ‘m’ be the mass of the weight, and assuming the string to obey Hooke’s law.

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mg ( x) a The acceleration of the weight 

d2x dt

2



[By virtue of eqn. (1)]

g ( x) a

………….. (2).

i.c

Ÿ

Which shows that the weight executes SHM. [where I is the phase constant] Now, initially i.e., at t = 0; x = a0 a0

Ÿ

I

a0 sin Z (0)  I sin 1 (1)

S

et B

2 ? Displacement equation becomes

a0 sin Z t  I .

ad

Let the displacement equation for the weight be given as x

?

om

i.e., F = Kx ... (1). We have mg = k ( 3a – 2a) = ka ... (1) [where K is the force constant for the wire] Let the wire be stretched by an amount ‘a0’. The net restoring force F0 = K (a0) acting on the weight. The weight starts moving under the action of this restoring force and becomes slack when the extension in it vanishes. If ‘x’ be the extension in the wire, then restoring forces acting on it will be F = - Kx.

S· § a0 sin ¨ Z t  ¸ a0 cos Z t 2¹ © If t1 be the time taken by the weight to reach the point where extension becomes 0. x

Ÿ

a0 cos (Z t1 )

Z t1 t1

cos 1 (0)

S

S

2Z

Required time

S 2Z

S 2

a g .

[ from eqn (2) Z

w

?

[least value]

2

w

Or

0

.N

Then

A small body is dropped in a tunnel dug along the diameter of earth from a height h (< < R) above the earth’s surface. Find the time period of motion. Is the motion simple harmonic? Solution Let AB be the tunnel dug along the diameter of earth with O as its center. Let P be the pt. (at a height h above the surface of earth) from which the body (of mass m, say) is dropped.

w

8.

g ]. a

Its velocity at A will be v A

2 g h . [ h < < R variation of ‘g’ with h is neglected].

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Let the body be at a distance of ‘y’ from the earth’s center at any instant, (with in the earth). The net force (gravitational) exerted on ‘m’ due to earth will be F

G M 'm 10 y 2 . diwb

Where M c is mass of the spherical portion of earth, with center at O and radius y.. [where U is the average density of earth]

)& 4  SGUm y 3

d2 y The net acceleration of the body will be

dt

2

)& F m

4  S G U ( y) 3

i.c

)& F

4 S y3 U . 3

om

Now, M '

d2 y d t2

Z 2 y, it is evident that, the body

ad

Comparing the above equation, with the standard from

ocillates simple barmonically within the earth, with an angular frequency Z Now, since g

GM

4 SGU . 3

Ÿ

g

G §4 3 · ¨ S R U¸ R2 © 3 ¹

4 S GU 3

g R

4 SGU R. 3

.N

?

et B

. R2 [where M is the total mass of earth and g the acc. due to gravity on the earth’s surface].

g . ... (1) R Now, let us consider the body at a distance y from the center (during its downward motion) i.e., at C. Kinetic energy possessed by the body there will be the sum of the kinetic energy possessed by it at A and the work done by the gravitational force (restoring force) on it. i.e., if v be the velocity of the body at C, then

Z

w

?

w

w

1 m v2 2

v2

Ÿ

?

?

v

2gh g r

1 1 m v A2  m Z 2 2 2

R2  y2

g 2 R  y2 R

(2 h  R ) R  y z

d y (2h  R ) R  y

z

g d t [ v R



dy ] dt

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?

If ‘t’, be the time taken by the body to travel from point A to point O, then 0

d y

R

(2h  R ) R  y

Ÿ

ª sin 1 « « ¬

Ÿ

t1

³

2

t1

g dt . R

0

R

º » 2h  R R »¼ 0 y

g t1 R

om

³

1/ 2

.

i.c

R ª R º sin 1 « » g ¬ 2h  R ¼

If ‘t2’ be the time taken by the body to fall from point P to earth’s surface (i.e, point A), then 2h g

t2

[ From, s

ut 

1 2 at ] . 2

ad

Now, we know that the total time period of oscillation of a particle executing SHM is equal to four times the min, time taken by it to travel from its mean position to extreme position or vice versa. Moreover, the time taken by the body to rise to the maximum height from the earth’s surface, or vice versa, at both points A, as well as B is the same. So, period of oscillation T = 4 (t1 + t2).

et B

· ½° 4 °­ R 1 §  2 h ¸¾ . ® R sin ¨¨ ¸ g ¯° © R  2h ¹ °¿

Note: If the body be dropped into the tunnel at point A (or B) with an initial velocity u, then in the above problem t2

u g

[From v = u + at]

.N

Alternatively: t1 can be evaluated as follows: Force acting on the body at C. mg ( y ) . R So, work done by the gravitational force, on the body, during the time it comes from point A to C, during its downward motion. W

w



³

R

R

mg y dy R

ª «¬ W

³ F d y »º¼

mg 2 R  y2 2R

w

w

F

Now, total K.E of the body at C will be equal to the sum of K.E. at A and work done by gravitational force on the body. 1 2 i.e., 2 m v

1 mg 2 m v A2  R  y2 2 2R

v2

Ÿ

vA2  g R 

Now, the velocity at point O can be obtained by putting y = 0 i.e, Also, y

v0

R g

v 2A  g R

… (3)

v A2  g R  v 2 .

94

g y2 . R

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dv Now, from equation (1) d t dv v2A

 g R v

g dt . [ The negative sign has been dropped, since v increases with t] R

2

At : t = 0; v = vA and at t = t1 ; v = v0.

vA

dv

³

0

v A2  g R  v 2

Ÿ

ª sin 1 « « ¬

Ÿ

t1

t1

g dt R

º ª »  sin 1 « « v A  g R »¼ ¬

º » v A  g R »¼

v02

R g

Ÿ

v 2A

1/ 2 ½ ­° S ° 1 ª 2 h º ®  sin « ¾. » 2 ¬ 2h  R ¼ ¿° ¯°

sin

1

ª « « ¬

v

º0 » v A  g R »¼ vA v0

g t1 R

t1 0

i.c

v0

g t1 R .

ad

³

?

om

i.e.,

g y . R

[from eqn. (3)]

Notice that the two expressions for ‘t1’ deduced by the two methods are the same ª R º sin 1 « » ¬« R  2 h ¼»

ª R º cos 1 « » 2 ¬« R  2 h ¼»

S

et B

Ÿ

[sin 1 T  cos1 T

S / 2] .

ª 2h º sin 1 « ». 2 «¬ R  h »¼

S 9.

w

.N

A particle of mass m is located in a unidimensional potential field where the potential energy of the particle depends on the co – ordinate x as U (x) = U0 (1 – cos ax) where U0 and a are positive constants. Find the period of small oscillations,w hich the particle performs about the equilibrium position. Solution Evidently, the equilibrium position of the particle will be the position corresponding to the least potential energy. ? U will be minimum when cos ax is maximum

w

i.e., ax 0 Ÿ x 0 [ (cos ax) max 1] . Thus, the particle oscillates about the point x = 0. Now, net force acting on it will be given by

w

F



dU dx



d [U 0 (1  cos a x)] = – U a sin ax. 0 dx

 Oscillations are small

? sin ax | ax

?F d 2x

or acceleration

d t2

U 0 a 2 ( x) . F m



U0 a2 ( x) m d2x

Comparing the above equation with

d t2

Z 2 .x .

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It is evident that, the particle executes SHM. a2 U 0 . m

2 with an agular frequency Z

2S

m

2S

Z

2

a U0

.

om

? Time period T

A uniform rod of mass m suspended by two identical threads each of length 1 is turned through a small angle about the vertical axis passing through the middle point C (See fig 15.62). Then the rod is released to perform small oscillations. Find the time period. Solution The threads, here are ssumed to be mass less, and inextensible. As, the threads are inextensible, the rod of mass m, slightly gets elevated, when it is turned through a small angle ( T , say).

et B

ad

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10.

Let the threads deviate in this process by an angle \ . Total mechanical energy of the system (i.e., rod) will be E = rotational K.E. + potential energy. Rotational K .E.

1 I Z2 2

1 2

§ M L2 ¨¨ © 12

2

· § dT · ¸¸ ¨ ¸ . ¹ © dt ¹

E

1 § M L2 ¨ 2 ¨© 12

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?

.N

Where L is the length of the rod. And potential energy U = mgl (1 - cos \ ) [Taking the equilibrium postion of the rod as datum level] 2

· § dT · ¸¸ ¨ ¸  m g l 1  cos\ . ¹ © dt ¹

Since mechanical energy remains conserved in SHM. dE dt

Ÿ

d dt

Ÿ

§ m L2 ¨¨ © 12

w

w

?

0

­° 1 § M L2 ® ¨¨ °¯ 2 © 12

2

· § dT ·  mgl 1  cos \ ¸¸ ¨ d t ¸ ¹ ¹ ©

· § d T · § d 2T ¸¸ ¨ ¸ ¨¨ ¹ © dt ¹ © dt

· § d\ · ¸¸  mgl sin \ ¨ ¸ © dt ¹ ¹

Now, linear displacement of either end of rod. Ÿ

l sin \

½° ¾ °¿

L sin T 2

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L T 2

l\ ld\

Also,

dt

[ \ & T are small ]

... (2)

L § dT · ¨ ¸ 2 © dt ¹

... (3).

Making use of equations (2) and (3), equation (1) can be rewritten as L2 § dT · § d 2T ¨ ¸¨ 12 © d t ¹ ¨© d t

· § L · dT g¨ ¸ \ ¸¸ © 2¹ dt ¹

L 6

0 Ÿ

§ d 2T ¨¨ d t 2 ©

· ª LT º g« » ¸¸ ¬ 2l ¼ ¹

d 2T 2

 Z2 T

0 Ÿ

dt

l 3g .

3T 2l

0

3g . l

ad

Z

2S



A ring of radius R carries a uniformly distributed charge + Q. A point charge – q of mass m placed at the center of the ring and is given a small displacement a (< < R), along the axis and released. Does the particle execute SHM? If so what is the time period? (Fig. 15.63)

w

.N

et B

11.

2S

dt

2

0.

It is evident that, the rod executes angular SHM with angular frequency Z

? Time period T

d 2T

i.c

Comparing the above equation with the standard from

om

Ÿ

w

w

Solution When the charge – q is placed at the center of the ring, it is attracted with equal force (or intensity) in all directions by the positive charge uniformly distributed along the ring, and hence remains at rest.

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om

However if the charge – q be slightly displaced from the center (along the axis) it starts executing oscillatory motion due to the electrostatic force of attraction always directed towards the center. Let us find out whether the motion is SHM or not. Consider the instant, when the charge (- q) is at a distance of x from the center (x d•a). Consider two elemental lengths on the ring (A and B say). (Fig. 15.64). They exert forces of attraction F1 and F2, on the charge (- q) at P, making angles T , with the axis (See Fig 15.64). Their components normal to the axis F1 sin T and F2 sin T being equal and opposite cancel each other. However, their components along the axis F1 cos T and F2 cos T add up being in the same direction. Thus, we find that, the component of force normal to the axis due to any elemental length, is neutralized (or cancelled) by the corresponding component of force, due to an elemental length diametrically opposite to the former (of the type A and B). Further, their component along the axis add up. F

6

[Where (dq) is the charge on the elemental length] § Q · ( q ) cos T dl ¸ ¨ 2 2 2 R S © ¹ 4S H0 R  x q Q 4S H0 R  x 2

2

u

q Q

4S H0 R  x 2

x R 2  x2

1/ 2

4 S H 0 R 2  x2 .

§ cos T · u¨ ¸ l © 2S R ¹

2S R

ad

0

.

2

et B

³

2S

(dq ) ( q ) cos T

i.c

? The net force on charge (- q) along the axis is

0

Now, since x < < R ? x2 is negligible small compared to R2 q Q

? F

( x) .

.N

4 S H 0 R3

The linear acceleration of the point charge (-q) will be

d2x dt 2

F m



qQ 4 S H 0 m R3

( x)

w

Which porves the fact that the particle executes SHM, with an angular frequency Z

w

? Time period of oscillation T

2S

Z

2S

qQ 4 S H 0 m R3

.

4 S H 0 m R3 . qQ

A solid cylinder of radius r rolls without sliding along the inside surface of a cylinder of radius R 1 performing small oscillations. Find the period of oscillation. Solution Consider any position of the solid cylinder of radius r, when the line joining the centres of the two cylinders make an angle T with the vertical. (Fig. 15.67) Let v be the velocity of the center of mass of the solid cylinder and Z its angular velocity.. The total mechanical energy of the solid cylinder will be given by E = translatory K. E. + rotational K.E. + potential energy.

w

13.

?

E

1 1 m v 2  I Z 2  m g R  r 1  cos T 2 2

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dv dZ dT  IZ  mg R  r sin T . dt dt dt dT Rr dt

Zr

Now, v

and

dv dt

0

r

… (1)

dZ dt

Rr

d 2T dt 2

.

ad

mv

i.c

[Here I is the moment of inertia of the solid cylinder about the axis through its center]. Differentiating the expression for E w. r. t time t and equating it to zero. [Since M.E. remains conserved]

dv dZ Making use of these in equation (1) we get m Z r d t  I Z d t  m g sin T Z r mr

Ÿ

R r d 2T 2

d 2T

1 d 2T ( R r )  u  d t2 2 d t2



dt

w

 ( g ) (T )

d 2T

Comparing with the standard form with angular frequency Z

[ sin |T ]

dt

2

Z 2 (T ) it is evident that the solid cylinder performs SHM

2g T 3( R  r ) and time period

2S

3( R  r ) . 2g

A uniform rod of mass m length l performs small oscillations about the horizontal axis passing through its upper end. Find the mean kinetic energy of the rod averaged over one oscillation period if, at the

w

15.

0

2g (T ) 3( R  r )

.N

Ÿ

d v m r2 d Z   m g sin T r 2 dt dt

et B

Ÿ

0.

w

initial moment, it was deflected from the vertical by an angle T 0 and then imparted an angular velocity

Z0 .

Solution Consider the rod, at any instant making an angle T with the vertical. (Fig. 15.70)

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The net torque acting on the rod, about O will be W

?

ml 2 3

l m g l (T ) sin T | 2 2

m l 2 d 2T 3 d t2

ID § d 2T ¨¨ 2 ©dt

· ¸¸ ¹

m g lT 2



Ÿ

d 2T dt

2

om

But W

mg

§ 3g · ¨ ¸T © 2l ¹

i.c

Which shows that the rod makes angular oscillations with angular frequency Z

3g . 2l

Now if a0 be the angular amplitude, the equation for angular displacement for the rod can be written as a0 sin Zt  I .

Since at t = 0; T ? T0

T0

a0 sin (I )

T0

sin I

Ÿ

[where I is the phase constant]

... (1)

a0

Z a0 cos Z t  I

Again at t

0;

dT dt

et B

dT Also, d t

? Z0

ad

T

Z0

Z a0 cos I

Z0 Z a0

.N

Ÿ cos I

... (2)

Squaring and adding equations (1) & (2) we get

T02



Z02 Z 2 a02

w

Ÿ

a0

1

Ÿ a02

T 02 

Z02 Z2

… (3)

w

w

Now, the instantaneous K.E. (rotational) of the rod will be K

Ÿ

1 § ml 2 ¨ 2 ©¨ 3

· 2 ¸¸ [Z a0 cos (Z t  I )] ¹

1 § dT · 1¨ ¸ 2 © dt ¹

ml 2 Z 2 a02 cos 2 (Z t  I ) 6

? Average K.E. over one complete time period will be

K!

ml 2 Z 2 a02 { cos 2 (Z t  I )} 6

Ÿ

ml 2 Z 2 a02 12

[? Average of cos 2 (Z t  I ) over one complete oscillation is

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1 ] 2

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ml 2 § 3 g · § 2 2l 2 · Z0 ¸ ¨ T0  12 ¨© 2l ¹¸ © 3g ¹

If a particle moving with SHM has velocities u, v and w at the distances a, b ad c respectively from an arbitrary point on the straight line, prove that its period T is given by the equation: u 2 v 2 w2 T

2

a b c

4S

1 1 1

2

1

1 1

a

b c

2

a b2 c2

Z 2 A2  x 2 we have

2 Then from v

Z 2 A2  (a  d ) 2

v2

Z 2 A2  (b  d ) 2

w2

Z 2 A2  (c  d ) 2

... (1)

et B

and

u2

ad

i.c

Solution Let A and Z be the amplitude and angular frequency of the SHM.

om

16.

mgl T 02 ml 2 2 Z0  g 12

.N

Now the three equations can be rewritten as u2

Ÿ

v2

Z2

 a 2  2ad  d 2  A2

 b 2  2bd  d 2  A2

w

Similarly,

Z2

w

And

w2

Z2

 c 2  2cd  d 2  A2

u2

Z2

... (2) ... (3)

 A2  a 2  d 2  2ab

0

... (4)

0

... (5)

0

... (6)

w

Eliminating d and A from equations (4), (5) and (6) we get § u2 2· ¨¨ 2  a ¸¸ a 1 ©Z ¹ §v · 2 ¨¨ 2  b ¸¸ b 1 ©Z ¹

u2

2

0

Ÿ

1

Z2

§ w2 2· ¨¨ 2  c ¸¸ c 1 ©Z ¹

101

v

2

w

2

a 1

a2

a 1

2

b 1

2

c 1

b 1 b c 1

c

0

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[Resolving the original determinant into two determinats] T2

Ÿ

a 1

v2 b 1

w2

a2

b2

u2

c2

c

a

b

c

1

1

1

1

2

or

T a

1

v2 b

1

w2 c

1

4S a

1

2

a

2

1

1

b

c

b

2

c2

w

w

w

.N

et B

ad

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om

4S 2

u2

102