Solutions to Concepts of Physics by HC Verma Chapter 12

12 1.

Simple Harmonic Motion A particle executes simple harmonic motion with an amplitude of 10 cm and time period of 6 s. At time t = 0 it is at position 𝓍 = 5 𝑐𝑚 going towards positive 𝓍 direction. Write the equation for the displacement 𝓍 at time t. Find the magnitude of the acceleration of the particle at time 𝑡 = 4 𝑠. i)

Given that amplitude A = 10 cm and time period T = 6 sec. 2π 2π π ⇒ ω = T = 6 =3 and 𝑥 = 5 cm when 𝑡 = 0 ∴ 5 𝑐𝑚 = 10 𝑐𝑚 𝑠𝑖𝑛 (𝜔𝑡 + 𝜙) ⇒ 𝑠𝑖𝑛 𝜑 = ½ ⇒ 𝜑 = 𝜋/6 ∴ equation of displacement is πt π 𝓍 = 10cm 𝑠𝑖𝑛 3 + 6

ii)

When t = 4 𝓍 = 10 cm 𝑠𝑖𝑛 = (10 𝑐𝑚) 𝑠𝑖𝑛 ⇒ 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎 = −𝜔2 𝑥 𝜋 2

=− = 2.

3

10𝜋 2 9

9𝜋 6

4π 3

π

+6

= (10 𝑐𝑚) 𝑠𝑖𝑛

3𝜋 2

= −10 𝑐𝑚

−10

= 10.96 𝑐𝑚 ≈ 11 𝑐𝑚/𝑠 2

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 ms-1 and 10 ms-2at a certain instant. Find the amplitude and the time period of the motion. We have 𝓍 = 𝐴 𝑠𝑖𝑛 𝜔𝑡 = 2 𝑐𝑚 = 0.02 𝑚 𝑣 = 𝜔𝐴 𝑐𝑜𝑠 (𝜔𝑡) = 1 𝑚𝑠 −1 𝑎 = – 𝜔2𝐴 𝑠𝑖𝑛 (𝜔𝑡) = – 10 𝑚𝑠 −2 .. 1 ..

12

3.

Simple Harmonic Motion 𝒶

10

⇒

ω2 = 𝓍 = 0.02 = 500

⇒

ω =

500 = 10 5

⇒

𝐴=

𝓍2 + ω2

And

= 0.0004 + 0.002 = 0.049 m = 4.9 cm 2π 2π T = ω = 10 5 = 0.628 s = 0.28 cm

𝓋2

A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal? Given that amplitude A = 10 cm Let 𝓍 be the distance from the mean position where kinetic and potential energies are equal. Let K and U be the kinetic and potential energies at this displacement. But 𝐾 = ½ 𝑚𝑣 2 = ½ 𝑚𝜔2 𝐴2 cos 2 𝜔𝑡 and 𝑈 = ½ 𝑘𝓍 2 = ½ 𝑚𝜔2 𝐴2 sin2 𝜔𝑡 ∴ 𝐾 = 𝑈 ⇒ ½ 𝑚𝜔2 𝐴2 cos 2 𝜔𝑡 = ½ 𝑚𝜔2 𝐴2 sin2 𝜔𝑡 ⇒ cos 2 𝜔𝑡 = sin2 𝜔𝑡 ⇒ 𝜔𝑡 = ± 𝜋/4 ⇒ 𝓍 = A sin ωt π = 10 cm 𝑠𝑖𝑛 ± 4 = ±

4.

10

2

= ± 5 2cm

The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm s-1 and 50 cm s-2. Find the position(s) of the particle when the speed is 8 cm s -1. Let 𝑣𝑚 and 𝑎𝑚 be the maximum speed and acceleration of the particle and A its amplitude. Then 𝑣𝑚 = ωA = 10 cm s-1 𝑎𝑚 = 𝜔2 𝐴 = 50 𝑐𝑚 𝑠 −2 .. 2 ..

12

Simple Harmonic Motion 𝑎𝑚

= 5 𝑠 −1

⇒

𝜔=

And Again ⇒ ⇒

𝐴 = 𝜔 = 10/5 = 2 𝑐𝑚 𝑣 = 𝜔𝐴 𝑐𝑜𝑠 𝜔𝑡 = (10 𝑐𝑚 𝑠 −1 ) 𝑐𝑜𝑠 𝜔𝑡 = 8 𝑐𝑚 𝑠 −1 𝑐𝑜𝑠 𝜔𝑡 = 8/10 = 4/5 𝓍 = 𝐴 𝑠𝑖𝑛 𝜔𝑡 = ± 𝐴 (1 – cos 2 𝜔𝑡)

𝑣𝑚 𝑣𝑚

16

= ± A 1 − 25 = ± 5.

3A 5

6

= ± 5 = ± 1.2 cm

A particle having mass 10 g oscillates according to the equation 𝓍 = 2.0 𝑐𝑚 𝑠𝑖𝑛 {(100 𝑠 −1 )𝑡 + 𝜋/6}. Find (a) the amplitude, the time period and the spring constant (b) the position, the velocity and the acceleration at t = 0. (a)

Clearly (from the equation) the amplitude A = 2.0 cm Angular frequency 𝜔 = 100 𝑟𝑎𝑑 𝑠 −1 2𝜋 ∴ 𝑇 = 𝜔 = 2𝜋/100 = 0.0628 𝑠 Also

𝑘 = 𝑚𝜔2 =

10

1000

1002 = 100 𝑁𝑚−1

(b) At time t = 0 we have 𝜋 2.0 𝑐𝑚 𝓍 = 2.0 𝑐𝑚 𝑠𝑖𝑛 6 = 2 = 1.0 𝑐𝑚 𝜋

𝑣 = 100 𝑟𝑎𝑑 𝑠 −1 2.0 𝑐𝑚 𝑐𝑜𝑠 6 = 100 3/2 = 173 𝑐𝑚 = 1.73 𝑚𝑠 −1 𝒶 = (100 rad s-1)2 2.0 cm sin π/6 = (100 rad s-1)2 (2/100 m) /2 = 100 ms-2 6.

The equation of motion of a particle started at 𝑡 = 0 is given 𝜋 by 𝓍 = 5 𝑠𝑖𝑛 20𝑡 + 3 , where 𝓍 is in centimeter and t in second. When does the particle (a) first come to rest (b)first have zero acceleration (c) first have maximum speed? (a)

Putting 𝑣 = 0 gives

.. 3 ..

12

Simple Harmonic Motion 𝜋

𝑐𝑜𝑠 20𝑡 + 3 i.e. i.e.

𝜋

20𝑡 + 3 = 𝜋

= 0

𝜋

2

𝑡 = 120 𝑠

(b) Putting 𝒶 = 0 gives 𝜋 𝑠𝑖𝑛 20𝑡 + 3 = 0 𝜋

i.e.

20𝑡 + 3 = 𝜋

i.e.

𝑡 = 30 𝑠

𝜋

𝜋

(c)

𝑣 = 100 𝑐𝑜𝑠 20𝑡 + 3 ⇒

𝜋

speed is maximum when 𝑐𝑜𝑠 20𝑡 + 3 = 1 𝜋

i.e. when 20𝑡 + = 𝜋 ⇒ 7.

𝜋

3

𝑡 = 30 𝑠

Consider a particle moving in a simple harmonic motion according to the equation 𝓍 = 2.0cm cos 50π t + tan–1 0.75) where 𝓍 is in centimeter and t in second. The motion is started at t = 0. (a) when does the particle come to rest for the first time? (b) when does the acceleration have its maximum magnitude for the first time? (c) when does the particle come to rest for the second time? (a)

It will come rest when 𝓋 = 0 ∴ 𝑣 = – 100𝜋 𝑠𝑖𝑛 (50𝜋 𝑡 + 𝑡𝑎𝑛– 1 0.75) = 0 ⇒ 50𝜋 𝑡 + tan–1 0.75 = 𝜋 ⇒

𝑡 =

𝜋 –tan –1 0.75 50𝜋

= 1.6 𝑥 10–2 𝑠

(b) We know acceleration 𝑎 = – 𝜔2𝐴𝓍 where 𝜔 is angular frequency, A the amplitude and 𝓍 the displacement. Thus 𝑎 is maximum when 𝓍 is maximum i.e. when 𝑐𝑜𝑠 (50𝜋 𝑡 + tan–1 0.75 ) = 1 .. 4 ..

12

Simple Harmonic Motion

(c)

⇒

50𝜋 𝑡 + tan–1 0.75 = 𝜋

⇒

𝑡 =

= 1.6 𝑥 10–2 𝑠

50𝜋

The particle will come rest for second time when 𝑣 = 0 and 50𝜋 𝑡 + tan–1 0.75 = 2𝜋 ⇒

8.

𝜋 –tan –1 0.75

𝑡 =

2𝜋 –tan –1 0.75 50𝜋

= 3.6 𝑥 10–2 𝑠

Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement to change value from half the amplitude to the amplitude. Let the 𝓍1 =A/2 and 𝓍2 =A be the values of displacement at time t1 and t2. Then A = A 𝑐𝑜𝑠 ωt1 2 and A = A 𝑐𝑜𝑠 ωt 2 ⇒ 𝑐𝑜𝑠 ωt1 = ½ and 𝑐𝑜𝑠 ωt 2 = 1 π π ⇒ ωt1 = and ωt 2 = 6

⇒ ⇒ 9.

π

π

ωt 2 – ωt1 = 2 – 6 = π

T

t 2 – t1 = 3ω = 6

2 π

3

[∵ T = 2π/ω]

The pendulum of a clock is replaced by a spring-mass system having spring constant equal to 0.1 Nm –1. What mass should be attached to the spring? 𝑘 = 0.1 𝑁𝑚 −1 𝑇 = 2𝜋

𝑚 𝑘

𝑠

But the time period of pendulum of a clock = 2 second 𝑚

⇒

2𝜋

= 2

⇒

𝑚 = 𝜋 2 = 0.1/𝜋 2 = 0.01 𝑘𝑔 = 10 𝑔

𝑘 𝑘

.. 5 ..

12

Simple Harmonic Motion

10. A block suspended from a vertical spring is in equilibrium.

Show that the extension of the spring equals the length of an equivalent simple pendulum i.e., a pendulum having frequency same as that of the block. Let m be the mass of the block, 𝓍 the extension of the spring, 𝑙 the length of the pendulum and ω be the angular frequency of the spring and the pendulum then 𝜔 =

𝑘 𝑚

=

𝑙 𝑔

At the equilibrium position 𝑚𝑎 + 𝑚𝑔 = 0 ⇒ 𝑚𝑔 = – 𝑚𝑎 = – 𝐹 = 𝑘𝓍 ⇒ 𝑘𝓍 = 𝑚𝑔 ⇒ 𝓍 = 𝑚𝑔/𝑘 = 𝑙

[Q.E.D.]

11. A block of mass 0.5 kg hanging from a vertical spring executes

simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find maximum force exerted by the spring on the block. Maximum force 𝐹 exerted by the spring on the block is 𝐹 = 𝑚𝑔 + 𝑚𝑎𝑚 = 𝑚𝑔 + 𝑚 𝜔2 𝐴 = 0.5(9.8) + 0.5 = 0.5(9.8) + 0.5

2𝜋 2 𝑇 2𝜋 0.314

(0.1) 2

(0.1) = 25 𝑁𝑚−1

12. A body of mass 2 kg suspended through a vertical spring

executes simple harmonic motion of period 4 s. If the oscillations are stopped and the body hangs in equilibrium, find the potential energy stored in the spring. Let the potential energy at mean position of the spring be zero. Also 𝑚 = 2 𝑘𝑔 and 𝑇 = 4 𝑠. .. 6 ..

12

Simple Harmonic Motion Let the extension of the spring in equilibrium position be 𝓍. Now T = 2π

m

= 4s

k

π2

2π 2

⇒ k = m 4 = 4 = 5 Nm−1 But at equilibrium position 𝑚𝑔 = 𝑘𝓍 ⇒ 𝓍 = 𝑚𝑔/𝑘 = 2𝑥10/5 = 4 𝑚 ∴ 𝑈 = ½ 𝑘𝓍 2 = 40 𝑁𝑚−1 13. A spring stores 5 J of energy when stretched by 25 cm. It is

kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block? The energy stored in the spring when stretched by 25 cm is the potential energy U where 1 𝑘 𝑈 = ½ 𝑘𝓍 2 = ½ 𝑘 . = = 5𝐽 16 32 ⇒ 𝑘 = 32(5) = 160 𝑁 𝑚−1 also ⇒

𝑇 = 2𝜋 𝑚 =

𝑘 100 𝜋 2

𝑚 𝑘

1

=5

=

160 100 𝜋 2

= 0.16 𝑘𝑔

14. A small block of mass m is kept on a bigger block

m

of mass M which is attached to a vertical spring of M spring constant k as shown in the figure. The system oscillates vertically. (a) Find the resultant k force on the smaller block when it is displaced through a distance 𝓍 above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude? (c) What can be the maximum amplitude with which the two blocks may oscillate together? .. 7 ..

12

Simple Harmonic Motion (a)

Since the mass attached to the spring is m+M, at the equilibrium position the angular frequency 𝑘

𝜔 = 𝑚 +𝑀 ∴ acceleration of smaller block at position 𝓍 is 𝑘𝓍 – 𝜔2 𝓍 = 𝑚 +𝑀 ∴ The resultant force 𝐹 on block of mass m when it is displaced through a distance 𝓍 above equilibrium position is 𝑚𝑘 𝓍 𝐹 = – 𝑚𝜔2 𝓍 = 𝑚 +𝑀 (b) The normal force N on block of mass m exerted by block of mass M when it is displaced through a distance 𝓍 above its equilibrium position is 𝑚𝑘 𝓍 𝑁 = 𝑚𝑔 – 𝑘𝓍 = 𝑚𝑔 – 𝑚 +𝑀 Clearly this force will be smallest at the highest point. (c)

The two blocks will oscillate together as long as N is greater than zero i.e. till 𝑚𝑘 𝓍 𝑚𝑔 = 𝑚 +𝑀 ⇒

𝓍 = 𝑚+𝑀

𝑔 𝑘

15. The block of mass m1 shown in the figure is fastened to the

spring and the block of mass m2 is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a farther distance (2/k)(m 1+m2)gsin𝜃 against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?

.. 8 ..

12

Simple Harmonic Motion (a)

m2 m2g

m1

k

m1g

θ

Since the mass attached to the spring, as long as the blocks oscillate together, is m1+m2 the angular frequency of the SHM shall be 𝜔 =

𝑘 𝑚 1 +𝑚 2

Let 𝓍 be the compression in the equilibrium position. Then 𝑚1 + 𝑚2 𝑔 sin 𝜃 = 𝑘𝓍 𝑚 +𝑚 𝑔 sin 𝜃 ⇒ 𝓍 = 1 𝑘2 (b) After the farther displacement 2 𝑚1 + 𝑚2 𝑔 sin 𝜃 on 𝑘 the equilibrium position and the release, the spring blocks system will start SHM with ω =

𝑘 𝑚 1 +𝑚 2

Now the two blocks will separate when 𝑁 = 𝑚2 𝑔 sin 𝜃 – 𝑚2 𝜔2 𝑥2 the normal force on the upper block is zero ⇒ 𝑚2 𝑔 𝑠𝑖𝑛𝜃 = 𝑚2 𝜔2 𝓍2 𝑘 𝑔 𝑠𝑖𝑛𝜃 = 𝑚 +𝑚 𝓍2 1

2

𝑚 1+𝑚 2 𝑔 𝑠𝑖𝑛𝜃

𝓍2 = 𝑘 Thus the blocks will separate at the equilibrium position of the spring blocks system. (c)

At the time of separation the speed is given by 𝑣 = 𝜔𝐴 𝑐𝑜𝑠 𝜔𝑡 = ωA 1–sin2ωt = A

k m 1 +m 2

1–

𝓍2 2 A

.. 9 ..

12

Simple Harmonic Motion k

= = =

m 1 +m 2 2 k

1–

m 1 +m 2 gsin θ 2 k

m1 + m2 g sinθ 3

k m 1 +m 2

2

k 2 m 1 +m 2 gsin θ

k m 1 +m 2

1

1 –4

gsinθ

16. In the figure k =100 Nm-,1

k M=1 kg and F = 10N. (a) F M Find the compression of the spring in the equilibrium position. (b) A sharp blow by some external agent imparts a speed of 2 ms-1 to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant. (c) Find the time period of the resultant simple harmonic motion. (d) Find the amplitude. (e) Write the potential energy of the spring when the block is at the left extreme. (f) Write the potential energy of the spring when the block is at the right extreme.

(a)

Let 𝓍 be the compression at equilibrium position. Then 𝐹 – 𝑀𝑘𝓍 = 0 ⇒ 10𝑁 = 1𝑘𝑔 100𝑁𝑚 − 1 𝑥 10 ⇒ 𝓍 = 100 = 0.1 𝑚 = 10 𝑐𝑚

(b) Due to sharp blow the block gets speed of 2 ms-1. Let K and U be the kinetic and potential energies at this instant. Then 𝐾 + 𝑈 = ½ 𝑀𝑣 2 + ½ 𝑘𝓍 2 = ½ 1 2 2 + ½ 100 0.1 2 = 2.5 𝐽 (c)

k

ω = ⇒

T =

m 2π ω

= 2π

100 = 10 π

= 10 = 5 s

.. 10 ..

12

Simple Harmonic Motion (d) The amplitude 𝑣2

𝓍2 + ω2

A = =

4

0 + 100 =

0.04 m = 20cm

(e)

At the left extreme 𝓍 = A+10cm = 3cm ∴ 𝑈 = ½ 𝑘𝓍 2 = ½ (100𝑁𝑚−1 )(.09) = 4.5 𝐽

(f)

At the right extreme 𝓍 = A–10cm = 1cm ∴ 𝑈 = ½ 𝑘𝓍2 = ½ (100𝑁𝑚−1 )(.01) = 0.5 𝐽

17. Find the time period of the oscillation of mass m in figures a, b,

c. What is the equivalent spring constant of the pair of springs in each case? (a)

When the block is displaced by distance 𝓍 the force on the block will be –(k1+k2)𝓍 ∴ k = k1+k2 ⇒

T = 2π = 2π

k1

m k m

k 1 +k 2

(b) In this case when the block is displaced by distance 𝓍 the force on the block will be 𝐹 = − k1 – k 2 𝓍 ∴ 𝑘 = 𝑘1 – 𝑘2 ⇒

M

k2

T = 2π

m k

= 2π

k1

M

k2

m k 1 –k 2

.. 11 ..

12

Simple Harmonic Motion (c)

Let’s suppose that the springs are extended by k2 k1 M 𝓍1 and 𝓍2. Thus total elongation of the two springs is 𝓍 = 𝓍1 + 𝓍2. The force exerted on the mass is –k2𝓍2 because first spring doesn’t exert any force on the block. If k is the spring constant of the combined spring then. 𝑘(𝓍1 + 𝓍2 ) = 𝑘2 𝓍2 Also by Newton’s third law on the two springs 𝑘1 𝓍1 = 𝑘2 𝓍2 ⇒ 𝑘(𝓍1 + 𝓍2 ) = 𝑘1 𝓍1 = 𝑘2 𝓍2 1 k

and ⇒ and ⇒

1

= =

k 𝓍1 𝓍2 𝓍2 𝓍1

𝓍 1+ 2 𝓍1

k1 𝓍 1+ 1

= =

𝓍2

k2 k2

–1

k k1

–1

k

k2

1 =

k

–1 k 1 +k 2

1 = 1–

k

k1 k

+

k1k2

k =k

–1 k1k2 k2

1 +k 2

𝑇 = 2𝜋

𝑚 𝑘

= 2𝜋

𝑚 𝑘 1 +𝑘 2 𝑘1 𝑘2

18. The spring shown in the figure k is unstretched when a man F M starts pulling on the chord. The mass of the block is M. If the man exerts a constant force F, find (a) the amplitude and the time period of the motion of the block, (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at

.. 12 ..

12

Simple Harmonic Motion this position. (a)

The total force on the block is F–k𝓍 when the displacement is 𝓍 from the equilibrium position. At equilibrium position 𝐹 = 𝑘𝓍0 ∴ Force on the block = 𝐹 − 𝑘𝓍 = 𝑘(𝓍0 − 𝓍) Clearly the block will oscillate. The amplitude A is the value of 𝓍 where force on the block is zero i.e. 𝐹 𝐴 =𝑘 The time period T = 2π

M k

(b) The displacement at the mean position from the equilibrium position = 𝐹/𝑘. Thus the energy stored in the spring at equilibrium position is 1 kF 2

1

U = 2 k𝓍02 = 2 (c)

k2

=

F2 2k

At the mean position 𝓋 is maximum and the maximum value of 𝑣 is Aω. ∴ K = ½ M𝓋^2 = ½ MA2 ω2 = ½M

F2

k

k2

M

=

F2 2k

19. A particle of mass m is attached to the

springs A, B and C of equal force constants k as shown in the figure. If the particle is pushed slightly against the spring C and released, find the time period of the oscillation.

B

90°

A

45° C

Let the displacement along the direction of spring C i.e. 225 0 with the positive x-axis direction be 𝓍. The particle is pushed against spring C and therefore it exerts a force k𝓍 at 450 with .. 13 ..

12

Simple Harmonic Motion x-axis. The net force FAB of the springs A and B has the following components viz., Fx = k𝓍 cos 450 Fy = k𝓍 sin 450 FAB = k𝓍 cos2 450 + sin2 450) = k𝓍 Therefore the total net force that force of all three springs is 2k𝓍 T = 2π

m 2k

20. Repeat the previous exercise if angle between each pair of

springs is 1200. Let length of each spring be 𝑥 and the displacement along the direction of spring C be ∆𝓍. The spring C therefore compresses by ∆𝓍. Let spring B stretch by ∆y. Then from the figure cos 120° = ∆𝑥 2 −∆𝑦 2 −2𝑥∆𝑦

𝑥 2 +∆𝑥 2 − 𝑥+∆𝑦 2 2𝑥 ∆𝑥

=

B 120°

120°

A

C

2𝑥∆𝑥

x+∆y

x

But ∆𝑥, ∆y are both small and therefore 120° their squares are negligible quantities. ∆x ⇒ ∆𝑦 = −∆𝑥 cos 120° = ∆𝑥 cos 60° Thus the springs A and B each stretch by ∆𝓍 cos 60° . Therefore the force contributions of the springs A, B and C on the block along the direction of spring C are 𝑘 ∆𝓍 𝐹𝐴 = – 𝑘 ∆𝓍 cos 60° cos 60° = – 𝐹𝐵 = – 𝑘 ∆𝓍 cos 60° cos 60° = –

4 𝑘∆𝓍 4

𝐹𝐶 = – 𝑘∆𝓍 Therefore the net force that is the force of all three springs is 1 1 3 𝐹 = – 𝑘∆𝓍 1 + 4 + 4 = − 2 𝑘∆𝓍 T = 2π

2m 3k

.. 14 ..

12

Simple Harmonic Motion

21. The springs shown in the

k1

figure are all unstretched F k2 k3 M in the beginning when a man starts pulling at the block. The man exerts a constant force F on the block. Find the amplitude and the frequency of the motion of the block. The spring constant of the springs (of spring constant) k 2 and k3 is given by k2k3/(k2 + k3) (see question 17(c) for proof). Therefore the spring constant of the three springs taken together is k k k k +k k +k k k = k1 + k 2+ k3 = 1 2 k 2+ k3 3 1 2

1

2

3

M k2+ k3

T = 2π f = 2π

3

k 1k 2+ k 2 k 3 + k 3k 1 k 1k 2 + k 2k 3 + k 3k 1

The amplitude F A=k=k

M k2 + k3 F k2 + k3 1k2

+ k 2 k 3 + k 3k 1

22. Find the elastic potential energy stored in each spring shown

in the figure, when the block is in equilibrium. Also find the time period of the vertical oscillation of the block.

.. 15 ..

12

Simple Harmonic Motion The forces acting on the block are the spring force k3𝓍3 and the gravitational force Mg. ⇒ 𝑀𝑔 = 𝑘3 𝓍3 Also by Newton’s third law, the forces of the contiguous springs on each other are equal i.e. 𝑘1 𝓍1 = 𝑘2 𝓍2 = 𝑘3 𝓍3 = 𝑀𝑔 [∵ 𝑘3 𝓍3 = 𝑀𝑔] 𝑀𝑔 ⇒ 𝓍1 = 𝑘

k1

k2

k3

M

1

𝓍2 = 𝑀𝑔/𝑘2 𝓍3 = 𝑀𝑔/𝑘3 If k be the effective spring constant of the three springs then 𝑘 𝓍1 + 𝓍2 + 𝓍3 = 𝑀𝑔 Substituting for 𝓍1, 𝓍2 and 𝓍3 we get 1 1 1 1 = 𝑘 +𝑘 +𝑘 𝑘 1

2

3

If U1, U2, U3 be the elastic potential energies of the three springs then 𝑈1 = ½ 𝑘1 𝓍12 = 𝑈2 = ½ 𝑘2 𝓍22 = 𝑈3 = ½

𝑘3 𝓍32

=

𝑀 2𝑔

2

2𝑘 1 𝑀 2𝑔

2

2𝑘 2 𝑀 2𝑔

2

2𝑘 3

The time period T = 2π

M k

= 2π M

1 k1

1

1

2

3

+k +k

23. The string, the spring and the pulley shown in the figure are

light. Find the time period of mass m.

.. 16 ..

12

Simple Harmonic Motion Let the elongation of the spring be 𝓍. Then for equilibrium we have 𝑚𝑔 = 𝑘𝓍 = 𝑚𝜔2 𝓍 ⇒

𝜔 =

⇒

T=

2π ω

𝑘

= 2π

m

k

𝑚 m k

24. Solve the previous problem if the pulley has a moment of

inertia I about its axis and the string doesn’t slip over the it. In this case let 𝑙, 𝑣, 𝜔1 , 𝑎 be the elongation, velocity and angular velocity and acceleration I of the spring. Let E be the total energy with K and U as the kinetic and potential energies of the earth, block, pulley, spring system. Let the k m position when elongation of the spring is 𝑙 be the refrence position for potential energy and the final postion be when elongation of the spring is 𝑙 + 𝑥. Let the angular frequency and time period of the motion of the block be ω and T respectively. Then we have Δ𝐾 = ½ 𝑚𝑣 2 + ½ 𝐼𝜔12 Δ𝑈 = ½ 𝑘 𝑙 + 𝓍 2 – ½ 𝑘 𝑙 2 – 𝑚𝑔𝓍 ⇒ Δ𝐸 = Δ𝐾 + Δ𝑈 = ½ 𝑚𝑣 2 + ½ 𝐼𝜔12 + ½𝑘 𝑙 + 𝓍 2 – ½𝑘 𝑙 2 – 𝑚𝑔𝓍 𝐼 = ½ 𝑚 + 𝑟2 𝑣 2 + ½ 𝑘𝓍 2 + 𝑘𝑙𝓍 – 𝑚𝑔𝓍 But force at equilibrium position is zero ∴ 𝑘𝑙 – 𝑚𝑔 = 0 𝐼 ⇒ Δ𝐸 = ½ 𝑚 + 𝑟 2 𝑣 2 + ½ 𝑘𝓍 2 Differentiating w.r.t. t and noting that ΔE is zero, we have 𝐼 𝑑𝓍 𝑚 + 𝑟 2 𝑣𝑎 + 𝑘𝓍 𝑑𝑡 = 0 ∴

𝐼

𝑚 + 𝑟 2 𝑣 – 𝜔2 𝓍 = – 𝑘𝓍𝑣 .. 17 ..

12

Simple Harmonic Motion ⇒

⇒

k

ω =

T =

I

m+2 r 2π ω

I

m+ 2 r

= 2π

k

25. Consider the situation shown in the figure. Show that if the

blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period. As the blocks are displaced by k say 𝓍 in opposite directions the m m center of mass remains unchanged. Both blocks will move towards this center with zero initial velocity and equal acceleration as the force is same which is proportional to the displacement from the equilibrium position of each block. Thus the blocks will execute simple harmonic motion. Let k1 be the spring constant of the part of the spring between the center of mass and the either block. And since these two springs are joined serially the spring constant k of the joint spring is given by 1 1 1 2 = 𝑘 +𝑘 =𝑘 𝑘 1

⇒

𝑘1 = 2𝑘

∴

T = 2π

1

m k1

1

= 2π

m 2k

26. A rectangular plate of sides 𝑎 and 𝑏 is suspended from a ceiling by two parallel strings of length 𝑙 each. The separation

between the strings is d. The plate is displaced slightly in its plane keeping the strings tight. Show that it will execute simple harmonic motion. Calculate the time period.

.. 18 ..

12

Simple Harmonic Motion The motion of the plate, it being a rigid body, will be the motion of d L L its center of mass. As the disturbance imparted to the plate 𝒶 𝒷 is in its plane with strings kept tight each string’s lower end with move as a pendulum does and the plate will move as if acting as a pendulum suspended from a string tied at center of mass and parallel to either string. Let θ be the angle either string makes with the vertical. Then the force on the center of mass of the plate is 𝐹 = 𝑚𝑔 sin 𝜃 For small θ, sinθ = θ. Therefore acceleration 𝐹 𝓍 𝑔 𝑎 = 𝑚 = 𝑔 sin 𝜃 = 𝑔𝜃 = 𝑔 𝐿 = 𝐿 𝓍 Since g/L is constant the motion is simple harmonic motion and from 𝑎 = 𝜔2 𝑥 we get 𝑇 =

2𝜋 𝜔

= 2𝜋

𝐿 𝑔

27. A 1 kg block is executing a simple harmonic motion of

amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 Nm -1. A block of mass 3 kg is gently placed on it at the instant it is passing through the mean position. Assuming the two blocks move together, find the frequency and amplitude of motion. For motion of 1 kg block if A is amplitude we have 𝐴 = 0.1 𝑚 also 𝑘 = 100 𝑁𝑚−1 and 𝑚 = 1 𝑘𝑔 Now acceleration is 𝑑2 𝓍 𝑑𝑡2 𝑑2 𝑥 𝑑𝑡2

3 kg 100 N/m

1 kg

𝑘

= – 𝑚 𝑥 = – 100𝓍 =

𝑑𝑣 𝑑𝑡

=

𝑣

𝑑𝑣

𝑑𝑥 𝑑𝑡

𝑑𝑡

𝑑𝑣

= 𝑣 𝑑𝑥 = −100𝑥 .. 19 ..

12

Simple Harmonic Motion By integration we have 𝑣 2 = – 100𝓍 2 + 𝑐 But 𝑣 = 0 when 𝓍 = 0.1 𝑚. Therefore c = 1 ⇒ 𝑣 2 = – 100𝓍 2 + 1 ⇒ 𝑣 = 1 𝑚𝑠 −1 when 𝓍 = 0 After 3 kg block is placed on the 1 kg block the new mass m + M is 4 kg. Let 𝑣1 be the speed at this position of the two blocks. At this position the linear momentum is conserved. ∴ 𝑚𝑣 = 𝑚 + 𝑀 𝑣1 𝑚𝑣 ⇒ 𝑣1 = = ¼ = 0.25 𝑚𝑠1 𝑇 = ⇒

𝑚 +𝑀 2𝜋 𝑘 𝑚 +𝑀

2𝜋

=

100 4

=

2𝜋 5

𝑠

5

𝑓 = 2𝜋 𝐻𝑧 A = 𝓍2 + 𝑣

𝑣′

2

𝜔

= 𝜔1 [∵ 𝑤𝑕𝑒𝑛 𝓍 = 0, 𝑣′ = 𝑣1 = 0.25 𝑚𝑠 −1 ] = 0.25/5 = 5 𝑐𝑚 Aliter With the above notation and by consevation of energy we have 1 1 1 1 𝑚𝑣 2 = 2 𝑘𝑥 2 = 2 100 0.1 2 = 2 2 ⇒ 𝑣 = 1 𝑚𝑠 −1 By conservation of linear momentum before the placement of the 3 kg block on the 1 kg block and after that we have 𝑚𝑣 = 𝑚 + 𝑀 𝑣1 1 ⇒ 𝑣1 = 𝑚𝑠 −1 4 Thus we have 2𝜋 2𝜋 2𝜋 𝑇 = = = 𝑠 100 𝑘 5 𝑚 +𝑀

4

.. 20 ..

12

Simple Harmonic Motion ⇒

5

𝑓 = 2𝜋 𝐻𝑧

We note that at the extremity the extension in the spring is A’. Thus by conservation of energy principle we have 1 1 𝑚 + 𝑀 𝑣12 = 2 𝑘𝐴′2 2 ⇒

𝐴′ =

𝑚 +𝑀 𝑘

2

1

𝑣1 = 10 . 4 = 5 𝑐𝑚

28. The left block in the

𝓋

figure moves at a speed k 𝓋 towards the right m m block placed in L equilibrium. All collisions to take place are elastic and the surfaces are frictionless. Show that the motions of the two blocks are periodic. Find the time period of these periodic motions. Neglect the widths of the blocks. When the block A moves with velocity 𝑣 and collides elastically with the block B, it transfers all energy to the block B. The block B will then move a distance 𝓍 against the spring and return to its original position under the restoring force of the spring to complete half of the oscillation. 2π

The time period of motion of block B is

2

m k

= π

m k

The block B then collides with the block A and since the collision is elastic, comes to rest at that point. The block A then moves back a distance L to elastically collide against wall and then travel another distance L before colliding with block B and halting there. The velocity of block B when it collides elastically with block A is same as its initial velocity and therefore it delivers same energy to A that A delivered to it in the first place. Thus now A travels in opposite direction with constant velocity 𝑣. Time taken by block A to move this 2L .. 21 ..

12

Simple Harmonic Motion distance is 2𝐿/𝑣 Hence the total time period of the periodic motion is 2𝐿 𝑣

+ 𝜋

𝑚 𝑘

29. Find the time period of the motion of the particle shown in

figure. Neglect the small effect of the bend near the bottom. Let the time taken to travel AB and BC be t 1 and t2 respectively and A C For part AB, 𝑎1 = 𝑔 𝑠𝑖𝑛 45° = 10 cm 𝑔/ 2 and 60° 45° B 𝑠1 = 0.1/𝑠𝑖𝑛45° = 2/10 𝑣12 = 2𝑎1 𝑠1 = 𝑔/5 𝑣_1 = 𝑔/5 𝑡_1 = 𝑣_1/𝑎_1 = (𝑔/5) /(𝑔/ 2) = (2/5𝑔) = 0.20 𝑠 For part BC 𝑎2 = 𝑔 sin 60 ° = 𝑔

3 2

𝑚𝑠 −2 and

𝑠2 = 0.1/𝑠𝑖𝑛 60° = 0.2/ 3 𝑚 𝑣22 = 2 𝑎2 𝑠2 = 𝑔/5 𝓋2 = g/5 𝑡2 =

𝓋2 𝑎2

=

𝑔 5 𝑔 3 2

=

3 10𝑔

= 0.17 𝑠

So, time period T = 2 (t1 + t2) = 2(0.2 + 0.17) = 0.74 sec 30. All the surfaces shown in the

figure are frictionless. The mass of the car is M, that of the block is m and the spring has a spring constant k. Initially, the car and

k

m M

.. 22 ..

12

Simple Harmonic Motion the block are at rest and the spring is stretched through a length 𝓍0 when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and the car as seen from road. (b) Find the time period(s) of the two simple harmonic motions. (a)

Let A1 and A2 be the amplitudes of SHM motion of the block and the car. Then at the initial position total extension of the spring 𝓍0 = 𝐴1 + 𝐴2 . By conservation of momentum we have 𝑚𝑣𝐵𝐶 = 𝑀𝑣𝐶 Where 𝑣𝐵𝐶 is the velocity of the block relative to car and 𝑣𝐶 is the velocity of car. By differentiating w.r.t. t 𝑚𝑎𝐵𝐶 = 𝑀𝑎𝐶 where 𝑎𝐵𝐶 is the acceleration of the block relative to car and 𝑎𝐶 is the acceleration of car. The spring exerts equal and opposite force block and car at all the time. Consequently the time period as well angular frequencies of the SHM's are same. Thus at the extreme positions which occur at the same time instant 𝑎𝐵𝐶 = −𝜔2 𝐴1 and 𝑎𝐶 = −𝜔2 𝐴2 . Putting these values in the above equation we have ⇒ – 𝑚𝜔2 𝐴1 = – 𝑀𝜔2 𝐴2 ⇒ 𝑚𝐴1 = 𝑀𝐴2 ⇒ 𝑚𝐴1 = 𝑀(𝓍0 − 𝐴1 ) M𝓍 ⇒ A1 = m + 0M and

m𝓍

A2 = m + 0M

(b) Let 𝓋BC be the velocity of the block relative to car, 𝓋B be the velocity of the block relative to road and 𝓋C is the velocity of car. By conservation of momentum we have m𝓋BC = M𝓋C. By integrating w.r.t. t we get m𝓍BC = M𝓍C + C where 𝓍BC is the displacement of the block relative to car and 𝓍C is the displacement of car. But 𝑥𝐵𝐶 = 𝑥𝐶 = 0 at .. 23 ..

12

Simple Harmonic Motion the initial position. Therefore C is zero ⇒ m𝓍BC = M𝓍C Also by conservation of energy theorem we have 2 ½ 𝑀𝑣𝐶2 + ½ 𝑚𝑣𝐵𝐶 + ½ 𝑘 𝓍𝐵𝐶 + 𝓍𝐶 2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⇒ ½ M𝓋C2 + ½ m constant

M2 m2

𝓋C2 + ½ k 𝓍BC + 𝓍C

M

⇒

½M 1 + m 𝓋C2 + ½ k

⇒

½ M𝓋C2 + ½ k

M m

M

+1 m

2

2

𝓍C2 = constant

+ 1 𝓍C2 = constant

By differentiating w.r.t. t and substituting 𝒶C for get M M𝓋C 𝒶C + k m + 1 𝓍C 𝓋C = 0 ⇒ ⇒

M𝒶C + k 𝒶C 𝓍C

=–

k

M m

M +1 m

M

T = 2π

d𝓋C dt

we

+ 1 𝓍C = 0 =–

Time period T = 2π T = 2π

=

k M+m 𝑚𝑀

Displacement/Acceleration

𝓍C 𝒶C mM k M+m

31. A uniform plate of

mass M stays horizontally and L symmetrically on two wheels rotating in opposite directions. The separation between the wheels is ℓ. The friction coefficient between each wheel and the plate is μ. Find the time period of the oscillation of the plate if it is slightly displaced along its length and released. .. 24 ..

12

Simple Harmonic Motion Let 𝓍 be the displacement of the plate towards left and therefore the centre of gravity is also displaced through 𝓍 towards left. After displacement the sum of the normal forces N1 + N2 = mg. Since the plate does not rotate about the center of gravity the net torque τnet applied on its center of gravity is zero. Therefore τnet = N1 ℓ/2 – 𝓍) – N2 ℓ/2 + 𝓍) = 0 ⇒ N1 ℓ – 2𝓍)/2 = (mg – N1 ℓ + 2𝓍)/2 ⇒ N1ℓ = mg ℓ + 2𝓍)/2 ⇒ N1 = mg ℓ + 2𝓍 /2ℓ Similarly we can show N2 = mg ℓ – 2𝓍 /2ℓ Also F1 = μN1 = μmg ℓ + 2𝓍 /2ℓ and F2 = μN2 = μmg ℓ – 2𝓍 /2ℓ where F1 and F2 are the forces acting on the plate horizontally due to rotation of the wheels. If F be the net horizontal force acting on the plate then F = F1 – F2 μmg ℓ + 2𝓍

μmg ℓ – 2𝓍

⇒ ⇒

= – 2ℓ = 2μmg𝓍/ℓ m𝒶 = 2μmg𝓍/ℓ 𝒶/𝓍 = 2μg/ℓ

⇒

T = 2π

2ℓ

𝓍/𝒶) = 2π

ℓ 2μg

32. A pendulum having time period equal to two seconds is called

a seconds pendulum. Those used in pendulum clocks are this type. Find the length of a seconds pendulum where g = π2 ms2. If ℓ is the length of the pendulum we have T = 2π ℓ/g) .. 25 ..

12

Simple Harmonic Motion ⇒ ⇒

ℓ = g T/2π 2 ℓ = π2 2/2π 2 = 1 m

33. The angle made by the string of a simple pendulum with the

vertical depends on time as θ = π/90 sin[ π s-1)t]. Find the length of the pendulum if g = π2 ms-2. By inspection of the equation of SHM of the pendulum we have angular frequency ω = π s-1 ∴ T = 2π/ω = 2 2𝜋 also T = ℓ 𝑔

⇒

ℓ = g T/2π 2 = π2 2/2π

2

=1m

34. The pendulum of a certain clock has time period 2.04 s. How fast or slow does the clock run during 24 hours? Given that T = 2.04 s. But we know the time period of a normal clock is 2 seconds. ∴ Time lag per oscillation = 2.0 – 2.04 = –0.04 s Number of oscillations per 24 hours = (24x60x60)(0.5) = 43200 ∴ Total time lag = 43200x0.04 = 1728 s = 28.8 m 35

A pendulum clock giving correct time at a place where g = 9.800 ms-2 is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place. ⇒ ⇒

T=2 ℓ = g T/2π 2 = g’ T’/2π 2 g’ = gT2/T’2 = 9.8 x 4/[(86400+24)/43200]2 = 9.795 ms-2

36. A simple pendulum is constructed by hanging a heavy ball

with a 5.0 m string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be .. 26 ..

12

Simple Harmonic Motion the frequency if the system is taken on the moon where acceleration due to gravitation of moon is 1.67 ms-2? (a)

(b)

Here ℓ = 5.0 m. ∴ T = 2π ℓ/g = 2π 5.0/9.8 = 8.976 and f = 1/T = 0.70/π Hz 1

𝑔 ℓ

𝑓𝑚𝑜𝑜𝑛 = 𝑇 = 2𝜋 =

1.67 ½ 5

2𝜋

1

= 2π

3

Hz

37. The maximum tension in the string of an oscillating pendulum

is double the minimum tension. Find the angular amplitude. Let the angular amplitude be θm and the velocity at the mean position be 𝓋m. The tension in the pendulum is the vertical component of the net force acting on the string and is maximum at the mean position and minimum on the extreme position. By conservation of energy theorem we have ½𝑚𝑣𝑚2 + 0 = 0 + 𝑚𝑔𝑙(1 – 𝑐𝑜𝑠 𝜃𝑚) ⇒ 𝑣𝑚2 = 2𝑔𝑙(1 – 𝑐𝑜𝑠 𝜃𝑚) Now 𝑇𝑚𝑎𝑥 = 𝑚𝑔 + (𝑚𝑣𝑚2 /𝑙) = 𝑚𝑔 + 2𝑚𝑔(1 – 𝑐𝑜𝑠 𝜃𝑚 ) = 3𝑚𝑔 – 2𝑚𝑔 𝑐𝑜𝑠 𝜃𝑚 At the extreme position velocity is zero and therefore the contribution of centripetal acceleration to tension is also zero. This gives 𝑇𝑚𝑖𝑚 = 𝑚𝑔 𝑐𝑜𝑠 𝜃𝑚 But 𝑇𝑚𝑎𝑥 = 2𝑇𝑚𝑖𝑛 ⇒ 3𝑚𝑔 – 2𝑚𝑔 𝑐𝑜𝑠 𝜃𝑚 = 2𝑚𝑔 𝑐𝑜𝑠 𝜃𝑚 ⇒ 3𝑚𝑔 = 4𝑚𝑔 𝑐𝑜𝑠 𝜃𝑚 ⇒ 𝑐𝑜𝑠 𝜃𝑚 = ¾ ⇒ 𝜃𝑚 = 𝑐𝑜𝑠– 1 ¾ 38. A small block oscillates back and forth on a smooth concave

surface of radius R. Find the time period of small oscillation. .. 27 ..

12

Simple Harmonic Motion Let R be the radius of the concave surface and m the mass of the block. Then the force driving the oscillations is the normal force N = mg sinθ where θ is the angle the line from center of the surface to the block makes with the vertical. ∴ N = mg sinθ ≈ mgθ ∴ 𝒶 = N/m ≈ gθ ⇒ 𝓍 = R sinθ = Rθ ⇒ ω = acceleration/displacement = gθ/Rθ = g/R T = 2π

R x Free Body diagram

g R

39. A spherical ball of mass m and radius r rolls without slipping

on a smooth concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period. Let

the angular R–r R velocity of the (R–r)(1-cos θ system about the point of suspension at any time be ω1 and 𝜔2 , 𝑣 the rotational velocity and the translational velocity of the ball. Then 𝓋 = (R–r ω1 𝓋 = rω2 ⇒ ω2 = ω1(R-r)/r The fact that total energy in SHM is constant gives us ½m𝓋2+ ½Iω22 +mg(R–r)(1 – cosθ)=constant ⇒ ½ m(R–r)2ω12 + ½(2/5)mr2(R–r)2ω12/r2 + mg(R – r)(1 – cosθ) = constant θ

.. 28 ..

12

Simple Harmonic Motion ⇒

½ m(R–r ω12 + (1/5)m(R–r ω12 + mg(1 – cosθ) = constant Differentiating w.r.t. t we have ½ m(R–r) 2ω1α + (1/5)m(R–r)2ω1α + mg(sin θ)ω1 = 0 ⇒ (R–r α 1+ 2/5) + g(sin θ) = 0 ⇒ α = –(5/7) g(sin θ)/(R–r) But θ is small 5g ⇒ α ≈ – 7 R−r θ So the motion is S.H.M. with T = 2π

7 R–r 𝑚 5g

40. A simple pendulum of length 40 cm is taken inside a deep

mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period there. Radius of Earth = 6400 km. Given that ℓ = 40 cm = 0.4 m If g’ be the gravitational force of the earth 1600 km inside the earth surface then g' = GM’/r2 where M’ is the mass of the sphere with radius r = 6400-1600 km = 4800 km. ⇒ M’ = 4/3 πr3ρ M = 4/3 πR3ρ ⇒ M’ = r/R 3M ⇒ g’ = GM’/r2 = GM(r/R)3/r2 = gR2(r/R)3/r2 = gr/R = 9.8x4800km/6400km = ¾ 9.8 = 7.35 ms–2 ⇒

T = 2π

ℓ g

= 2π

0.4 7.35

= 1.47 s .. 29 ..

12

Simple Harmonic Motion

41. Assume that a tunnel is dug across the earth (radius = R)

passing through its center. Find the time a particle takes to cover the length of the tunnel if (a) if it is projected into the tunnel with a speed of 𝑔𝑅 (b) it is released above the height R above the tunnel (c) it is thrown vertically upward along length of the tunnel with a speed of 𝑔𝑅. (a)

Let T be the time period of the oscillatory motion of the particle, 𝓍t, 𝓋t and 𝒶t the displacement from the center, velocity and acceleration at time t. Then 𝒶t = F/m = GM(𝓍t3/R3)/𝓍t2 = GM𝓍t/R3 T = 2π 𝓍t/𝒶t = 2π R3/GM = 2π R/g ω = g/R If A be the amplitude of the motion then 𝓋t2 = ω2(A2 – R2) = g(A2 – R2)/R But at the surface of the earth we have 𝓋𝑡 = 𝑔𝑅 ∴ gR = g(A2 – R2)/R ⇒ R2 = A2 – R2 ⇒ A2 = 2R2 ⇒ A= 2R Let t = t1 and t = t2 be two successive instants when value of 𝓍t changes from +R to –R and therefore t2 – t1 is the time taken to cover the length of the tunnel. But R = 2R sin ωt1 ⇒ sin ωt1 = 1/ 2 ⇒ ωt1 = 3π/4 And –R = 2R sin ωt2 ⇒ sin ωt2 = –1/ 2 ⇒ ωt2 = 5π/4 ⇒ ωt2 – ωt1 = π/2 ⇒

𝜋

𝑡2 – 𝑡1 = 𝜔 =

π

R

2

g

(b) In this case the particle is released from height R above .. 30 ..

12

Simple Harmonic Motion the tunnel and this being a case of free fall motion till the particle reaches surface of the earth. The velocity of the particle when it reaches the surface of the earth is gR . Therefore it covers the length of the tunnel in the same time as in case (a) above. (c)

In this case again as the particle is thrown up with initial velocity gR it comes back to the surface of the earth with same velocity gR and then covers the length of the tunnel in the same time as in case (a) above.

42. Assume that a tunnel is dug along a chord

𝓍

of the earth, at a perpendicular distance R/2 ℓ R/2 from earth’s center where R is the θ radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational force exerted by the earth on a particle of mass m placed in tunnel at a distance 𝓍 from the center of the tunnel. (b) Find the component of the force along the tunnel and perpendicular to the tunnel. (c) the normal force exerted by the wall on the particle. (d) Find the resultant force on the particle. (e) show that the motion of the particle in the tunnel is simple harmonic and find the time period. (a)

From the figure we have ℓ = 𝓍2+R2/4)½ The mass of the sphere of radius ℓ centered at the center of the earth is M’ =

4 3

πℓ3ρ =

4 πℓ3 M 3 4 πR 3 3

= M

ℓ 3 R

If F be the gravitational force of the earth on the particle then F = GM’

m ℓ2

=

GM

ℓ 3 m R 2 ℓ

=

GMm R3

𝓍2 +

R2

½

4

(b) Using value of F from part (a) we have the component of the force along the tunnel .. 31 ..

12

Simple Harmonic Motion Ft = F𝓍/ℓ = GMm𝓍/R3 And the component of the force perpendicular to the tunnel FR GMm F⊥ = 2ℓ = 2R 2 (c)

The normal force exerted by the particle on the wall is the component perpendicular to tunnel, i.e. FR GMm F⊥ = 2ℓ = 2R 2

(d) The resultant force on the particle is the component of the gravitational force parallel to the tunnel, i.e. F𝓍 GMm 𝓍 Ft = ℓ = R 3 (e)

Acceleration

=

Ft m

=

GM R3

𝓍 which

varies

with

displacement from the center of the tunnel. Thus the motion is simple harmonic and time period T = 2π displacement/acceleration = 2π

R3 GM

43. A simple pendulum of length ℓ is suspended through the

ceiling of an elevator. Find the time period of the oscillations if the elevator (a) is going up with an acceleration 𝒶0 (b) is going down with an acceleration 𝒶0 and (c) is moving with an a uniform velocity. (a)

Let m be the mass of the bob of the pendulum. Then the driving force of motion is F = mg + m𝒶0 =m(g + 𝒶0) ∴ 𝒶 = F/m = g + 𝒶0 ⇒

T = 2π

ℓ g

+ 𝒶0

(b) The driving force of motion in this case is F = mg – m𝒶0 =m(g – 𝒶0) .. 32 ..

12

Simple Harmonic Motion

(c)

∴

𝒶 = F/m = g – 𝒶0

⇒

T = 2π

ℓ g

– 𝒶0

If the elevator is moving with uniform velocity then acceleration is zero and T = 2π

ℓ g

44. A simple pendulum of length 1 feet suspended from the ceiling

of an elevator takes 𝜋/3 seconds to complete one oscillation. Find the acceleration of the elevator. Let 𝒶0 be the acceleration of the elevator. Then (from the previous problem) T = 2π [ℓ/ g+ 𝒶0)] = 𝜋/3 ⇒ ℓ/ g + 𝒶0) = 1/36 ⇒ g+ 𝒶0 = 36ℓ ⇒ 𝒶0 = 36ℓ – g = 36 – 32 = 4 ft/s2 45. A simple pendulum fixed in a car has a time period of 4

seconds when the car is moving with a uniform velocity on a horizontal road. When the accelerator is pressed the time period changes to 3.99 seconds. Making an approximate analysis find the acceleration of the car. When the car moving with uniform velocity we have ℓ

⇒

T = 2π 4 = 2π

ℓ/g) ℓ/g)

When the car accelerates by 𝒶0 we have T’ = 2π [ℓ/(g2 + 𝒶02)½] ⇒ 3.99 = 2π [ℓ/(g2 + 𝒶02)½]

𝒶0 mg

.. 33 ..

12

Simple Harmonic Motion ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

T/T’ = 4/3.99 2 = [ℓ/g / [ℓ/(g2 + 𝒶02)½] 16g = 3.992(g2 + 𝒶02)½ = 15.92(g2 + 𝒶02)½ 256g2 = 15.922 (g2 + 𝒶02) (256–253.45)g2 = 253.45 𝒶02 𝒶02 = 2.55 g2/253.45 = g2/100 𝒶0 ≈ g/10 ms–2

46. A simple pendulum of length ℓ is suspended from the ceiling

of a car moving with a speed 𝓋 on a circular horizontal road of radius r. (a) Find the tension in the string when it is at rest with respect to car. (b) Find the time period of small oscillation. When the car is moving with uniform velocity 𝓋, the driving force F of the SHM is F=

2

mg

= m g2 + 𝒶 = g2 + ∴

T = 2π

⇒

T = 2π

m𝓋 2

+ 𝓋4 r2

𝓋4

2 ½

ℓ m𝓋2/r

mg

r ½ ½

r2

ℓ 𝒶 ℓ 𝓋4

g 2+ 2 r

½

47. The ear-ring of the lady in the figure has a 3 cm long light

suspension wire. (a) Find the time period of small oscillations if the lady is standing on the ground. (b) The lady now sits in a merry-go-round of radius 2 m moving at a 4 ms-1. Find the time period of the oscillations of the ear-ring. (a)

The length ℓ of the pendulum formed by the light suspension wire and the ring is 3 cm. Therefore T = 2π ℓ/g = 2π 0.03/9.8 = 0.348 s .. 34 ..

12

Simple Harmonic Motion (b) Since the lady now sits in the merry-go-round of radius r = 2 m, a centripetal force of m𝓋2/r also acts on the earring. ∴

ℓ

T = 2π

𝓋4

g 2+ 2 r

½

0.03

= 2π

256 ½ 4

9.82 +

= 0.306 s 48. Find the time period of the oscillations of the following

systems (a) A meter stick suspended through 20 cm mark. (b) A ring of mass m and radius r suspended through a point on its periphery. (c) A uniform square plate of edge 𝒶 suspended through a corner. (d) A uniform disk of mass m and radius r suspended through a point r/2 away from the center. (a)

Since the pendulum here is a physical pendulum, the formula for time period is T = 2π I/mgℓ Here I is the moment of inertia of the meter stick about point of suspension and ℓ the distance between the point of suspension and the center of mass. ⇒ ℓ = 0.5 – 0.3 = 0.2 And I = m12/12 + 0.22m = 0.08m + 0.04m = 0.12m ⇒

T = 2π I/mgℓ = 2π 0.12m/0.2m9.8 = 2π 6/98 = 1.55 s

(b) Here I = mr2 + mr2 = 2mr2 And ℓ = r ∴

T = 2π

I

= 2π mg ℓ

2m r 2 mgr

= 2π

2r g

.. 35 ..

12

Simple Harmonic Motion (c)

Here the axis through the corner of the square (taken as z-axis) is perpendicular the plane of the square (containing x-axis and y-axis). 𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦 + 𝑚𝑟 2 =

⇒

(d)

𝒶

And

ℓ =

∴

T = 2π

Here 𝐼 =

𝑚 𝒶2

+

12

𝑚 𝑎2 12

+

𝑚 𝑎2 2

=

2𝑚 𝑎 2 3

2 2m 𝒶2 2

I

= 2π mg ℓ

𝑚𝑟2 2

And

ℓ = r/2

∴

T = 2π

+ 𝑚

𝑟 2

3mg 𝒶

=

2

I

= 2π mg ℓ

= 2π

𝒶 8 3g

3𝑚 𝑟 2 4

3m r 2 4mgr 2

= 2π

3r 2g

49. A uniform rod of length ℓ is suspended by an end and is made

to undergo small oscillations. Find the length of the pendulum having time period equal to that of the rod. The time period of the physical pendulum consisting of rod is given by T = 2π I/mgℓ’ ℓ 2

Here 𝐼 = 𝐼𝑐 + 𝑚 And ℓ’ = ℓ/2 ∴

T = 2𝜋 = 2𝜋 = 2𝜋

𝐼 𝑚𝑔 ℓ

2

=

𝑚 ℓ2 12

+

𝑚 ℓ2 4

=

𝑚 ℓ2 3

’

2𝑚 ℓ2 3𝑚𝑔 ℓ 2ℓ 3𝑔

If L is the length of the string of the simple pendulum, then comparing with the above 2ℓ 𝐿 = 3 50. A uniform disc of radius r is to be suspended through a small .. 36 ..

12

Simple Harmonic Motion hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the center for it to have minimum time period? Let s be the distance of the hole from the center of the disc. Then I = Ic + ms2 = mr2/2 + ms2 ∴

I

T = 2π

mgs

= 2π [

mgd s

r2 2

= 2π

r2 + s2 2

m

+ s2 gs

r2

= 2π

2gs

s

+g dT

The condition for minimality of T is ds vanishes. dT ds

=

2π

r2

2

2gs

⇒ ⇒

s2 = r2/2 s = r/ 2

And

T = 2π = 2π = 2π = 2π

r2 2gs r2 2 2gr r g 2

s

+g

–½

r2

1

– 2gs 2 + g = 0

s

+g +g

+g

r 2

r 2

r 2 g

51. A hollow sphere of radius 2 cm is attached to an 18 cm long

thread to make a pendulum. Find the time period of the oscillation of this pendulum. How does it differ from the time .. 37 ..

12

Simple Harmonic Motion period calculated using the formula for a simple pendulum? Let ℓ be the distance between the center of mass of the sphere and the other end of the thread and T the time period of its motion. Then ℓ = 18 cm +2 cm = 20 cm = 0.2 m ∴ T = 2π I/mgℓ But I = Ic + mℓ2 = 2m0.022/3 + m0.22 = 0.04m(1+0.0067) = 0.0403m ⇒ T = 2π I/mgℓ = 2π 0.0403m/mg0.2 = 0.901 s If the pendulum be taken as simple pendulum and T’ be time period of its motion then T’ = 2π ℓ/g = 2π 0.2/9.8 = 0.8976 s ⇒ T – T’ = 0.9006 – 0.8976 = 0.003 s Thus T is 0.3% more than T’.

52. A closed circular wire hung on a nail in a wall undergoes small

oscillations of amplitude 2° and time period 2 s. Find (a) the radius of the circular wire, (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position. (c) the acceleration of this particle as it goes through its mean position and (d) the acceleration of this particle when it is at an extreme position. Take g = 𝜋2 ms-2. (a)

Given that angular amplitude θm = 2° and time period T = 2 s. Let r be the radius of the wire. Then from the time period formula for a physical pendulum we have T = 2π I/mgr And I = Ic + md2 = mr2 + mr2 = 2mr2 Thus 2 s = T = 2π I/mgr mgr = π2I = π22mr2 .. 38 ..

12

Simple Harmonic Motion ⇒ ⇒

g = π22r r = g/2π2 = 0.50 m = 50 cm

(b) Let ω be the angular frequency of the motion. At the mean position the angular velocity ω’ is maximum ∴ ω’ = ωθm But ω = 2π/T= 2π/2= π ⇒ ω’ = π 2π/180 = π2/90 If 𝓋 be the velocity of the point farthest away from point of suspension then 𝓋 = ω’ 2r = π2/90)2x0.5 = 11 cms–1 (c)

Acceleration at the mean position will be centripetal. If ω’2 be the angular acceleration and 𝒶 the acceleration then 𝒶 = ω’2/(2r) = (0.11)2 /1.00 ms–2 = 1.2 cms–2 The direction 𝒶 of is towards the point of suspension.

(d) At the extreme position the angular acceleration α is maximum ⇒ α = –ω2θm But ω = 2π/T = 2π/2 = π ∴ α = –π2 2π/180 = –π3/90 If 𝒶 be the acceleration of the point farthest away from point of suspension then 𝒶 = α 2r = π3/90)2x0.5 = 0.35 cms–2 53. A uniform disc of mass m and radius r is suspended through a

wire attached to its center. If the time period of the torsional oscillations be T, what is the torsional constant of the wire? The moment of inertia of a uniform disc of mass m and radius r is given by mr 2

⇒ Or

I= 2 T = 2π κ=

4π 2 I T2

I/κ =

4m r 2 π 2 T2

.. 39 ..

12

Simple Harmonic Motion

54. Two small balls, each of mass m are connected by a light rigid

rod of length L. The system is suspended from its center by a thin wire of torsional constant κ. The rod is rotated through and angle 𝜃0 and released. Find the tension in the rod as the system passes through its mean position. The mass of the suspended balls is 2m and ignoring the small radii, the center of mass is at the mid-point, L/2 away from both balls. ∴

I = 2m T = 2π = 2π

L 2 2

=

mL 2

m κ

m

2

𝜃0

L

m

I m

κ mL 2 2κ

Noting that there being no translational motion, the entire energy of the system comes from rotational kinetic energy and the stored potential energy, the conservation of energy theorem gives us KRi + Ui = KRf + Uf where suffix i refers to mean position and f refers to the extreme position. ⇒ ½ Iω2 + 0 = 0 + ½ κ𝜃02 where 𝜃0. is angular amplitude. ∴ (mL2/2 ω2 = κ𝜃02 ⇒ ω2 = 2κ𝜃02/mL2 Let T' be the tension in the rod. This tension has two sources viz., the weight mg on the end point of the rod and the centripetal force mω2L/2 provided by the rotation about the wire. ⇒

T’ = =

mg

2

m2 g 2 +

+

mω 2 L 2 2

κ 2 θ 04 L2

.. 40 ..

12

Simple Harmonic Motion

55. A particle is subjected to two simple harmonic motions of

same time period in the same direction. The amplitude of the first motion is 3.0 cm and that of the second is 4.0 cm. Find the resultant amplitude if the phase difference between the motions is (a) 0° (b) 60° and (c) 90°. (a)

The amplitudes 𝓍1 and 𝓍2 are given as 𝓍1m = 3.0 cm 𝓍2m = 4.0 cm Since time periods and consequently angular frequencies are equal, equations of the SHMs are 𝓍1 = 3.0cm sin ωt 𝓍2 = (4.0cm sin ωt The equation of the resultant SHM is 𝓍 = 𝓍1 + 𝓍2 = 7.0cm sin ωt Thus 𝓍m = 7.0 cm

(b) Here the amplitudes 𝓍1 and 𝓍2 are given as 𝓍1m = 3.0 cm 𝓍2m = 4.0 cm Since time periods and consequently angular frequencies are equal, equations of the SHMs are 𝓍1 = 3.0cm sin ωt 𝓍2 = 4.0cm sin ωt+π/3 The equation of the resultant SHM is 𝓍 = 𝓍1 + 𝓍2 = 3.0cm sin ωt + 3.0cm sin ωt+π/3 2 2 Thus 𝓍m = 𝓍1m + 𝓍2m + 2𝓍1m 𝓍2m 𝑐𝑜𝑠

π 3

= 6.1 cm (c)

Here the amplitudes 𝓍1 and 𝓍2 are given as 𝓍1m = 3.0 cm 𝓍2m = 4.0 cm Since time periods and consequently angular frequencies are equal, equations of the SHMs are .. 41 ..

12

Simple Harmonic Motion 𝓍1 = 3.0cm sin ωt 𝓍2 = 4.0cm sin ωt+π/2 The equation of the resultant SHM is 𝓍 = 𝓍1 + 𝓍2 π = 3.0cm 𝑠𝑖𝑛 ωt + 3.0cm 𝑠𝑖𝑛 ωt + 2 2 2 Thus 𝓍m = 𝓍1m + 𝓍2m = 5.0 cm

56. Three simple harmonic motions of equal amplitude A and

equal time periods in the same direction combine. The phase of the second motion is 60° ahead of the first and the phase of the third motion is 60° ahead of the second. Find the amplitude of the resultant motion. Since time periods and consequently angular frequencies are equal each equal to ω and amplitudes are also equal each being A, equations of the SHMs are 𝓍1 = A 𝑠𝑖𝑛 ωt π 𝓍2 = A 𝑠𝑖𝑛 ωt + 3 2π

𝓍3 = A 𝑠𝑖𝑛 ωt + 3 The equation of the resultant SHM is 𝓍 = 𝓍1 + 𝓍2 + 𝓍3 π 2π = A 𝑠𝑖𝑛 ωt + 𝑠𝑖𝑛 ωt + + 𝑠𝑖𝑛 ωt + π

= A 2 𝑠𝑖𝑛 ωt + 3 𝑐𝑜𝑠 π

3 π 3

π

3

+ 𝑠𝑖𝑛 ωt + 3 π

= A 𝑠𝑖𝑛 ωt + 3 + 𝑠𝑖𝑛 ωt + 3 π

= 2A 𝑠𝑖𝑛 ωt + 3 Thus 𝓍m = 2A 57. A particle is subjected to two simple harmonic motions given

by 𝓍1 = 2.0 sin (100𝜋 t) and 𝓍2 = 2.0 sin (120𝜋 t + 𝜋/3) where 𝓍 is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025. .. 42 ..

12

Simple Harmonic Motion 𝓍1 = 2.0 sin (100𝜋 t) π 𝓍2 = 2.0 𝑠𝑖𝑛 120π t +

3

The equation of the resultant SHM is 𝓍 = 𝓍1 + 𝓍2 π = 2.0 𝑠𝑖𝑛 100π t + 𝑠𝑖𝑛 120π t + 3 (a)

When t = 0.0125 we have 𝓍 = –2.41 cm

(b) When t = 0.025 we have 𝓍 = 0.27 cm 58. A particle is subjected to two simple harmonic motions one

along the 𝓍-axis and the other on a line making an angle 45° with the 𝓍-axis. The two motions are given by 𝓍 = 𝓍0 𝑠𝑖𝑛 𝜔𝑡 and 𝑠 = 𝑠0 𝑠𝑖𝑛 𝜔𝑡 Find the amplitude of the resultant motion. By vector addition, the amplitude 𝓍m of the resultant motion 𝓍m = 𝓍02 + s02 + 2𝓍0 s0 𝑐𝑜𝑠 =

𝓍02 + s02 +

π 4

2 𝓍0 s0 ∎

.. 43 ..