JJ311 MECHANICAL OF MACHINE Ch 2 Simple Harmonic Motion

SIMPLE HARMONIC MOTION

PREPARED BY: NAZIHAH BINTI MOHD NOOR

CONTENT

2.1 Concept of simple harmonic motion(SHM) 2.1.1

Define the terminologies in SHM

2.1.2

SHM by using a suitable diagram

2.1.3 Calculate time period, amplitude, frequency, velocity, maximum velocity, acceleration and maximum acceleration of the SHM 2.2 Linear motion of an elastic motion 2.2.1

Elastic system(spring and mass)

TERMINOLOGIES OF SHM Motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.

Equilbrium Position

FORMULA SHM

a  x 2

x  A cos 

 

v   A2  x 2 Vmax  A amax   A 2

Ar Inertia  mamax  m( mass)

  2f  f  2 1 T  f 2 T 



FUNDAMENTAL DEFINITIONS Displacement that measured from the equilibrium point

x

Time Velocity or speed Acceleration Mass Force angular velocity Maximum Velocity/Speed Maximum acceleration

T v a m F ω Vmax amax

EXERCISE 1 Points moving with simple harmonic motion have acceleration 9m/s2 and velocity 0.92m/s when it was in 65mm from the centre position.   Find i. Amplitude  i. the Amplitude ii. Periodic time movement SOLUTIONS  

 

ii. Periodic time the movemen  

EXERCISE 2 (i) A particle moving with simple harmonic motion has a periodic time of 0.4s and it was back and forth between two points is 1.22m. Determine i. Frequency and amplitude of the oscillation ii. Velocity and acceleration of the particle when it is 400mm from the center of oscillation iii. The maximum velocityAmplitude and acceleration of the i. Frequency SOLUTIONS movement  

 

 

EXERCISE 2 (ii) ii. Velocity and acceleration of the particle when it is 400mm from the center of oscillation  

     

     

EXERCISE 2 (iii) iii. The maximum velocity and acceleration of the movement  

   

 

EXERCISE 3 (i) A body of mass 1.5kg moving with simple harmonic motion is towards one end of the swing. At the time it was in A, 760mm from the center of oscillation, velocity and acceleration is 9m/s and 110m/s2 respectively. Amplitude i. Frequency and amplitude of the oscillation   of the body ii.SOLUTIONS Maximum acceleration and the inertia i. Frequency when it to the edge of swing    

 

 

=

EXERCISE 3 (ii) ii. Maximum acceleration and the inertia of the body when it to the edge of swing  

   

 

The piston of a steam engine moves with simple harmonic motion. The crank rotates at 120r.p.m with a stroke of 2 metres. Find the velocity and acceleration of the piston when it is at a distance of 0.75metre from i. Velocity of the piston the centre. SOLUTIONS

 

   

ii. Acceleration of the piston  

SHM DIAGRAM

Max. Displacement = A

Max. Velocity = Vmax = ωA

Max. Acceleration = amax = ω

EXERCISE 4 • Figures (a) and (b) are the displacement-time graph and acceleration-time graph respectively of a body in simple harmonic motion. What is the frequency of the motion?

 

SOLUTIONS Frequency  

 

EXERCISE 5 The following graphs show the variation of displacement, x and velocity, v with time, t for a body in simple harmonic motion. What is the value of T?

SOLUTIONS  

 

 

EXERCISE 6 A particle moves in simple harmonic motion along a straight line about point x=0.40cm and the velocity is zero. The frequency of the motion is 2.5Hz. Calculate the; i. Period ii. Angular velocity iii. Amplitude iv. Displacement at time t v. Maximum velocity vi.SOLUTIONS Maximum acceleration  

i.

Period

iii.

 

ii. Angular velocity iv.  

Amplitude

v.

Maximum velocity  

 

Displacement at time t  

vi.

Maximum acceleration  

LINEAR MOTION OF AN ELASTIC MOTION A spring resists being stretched or compressed.

stretched compressed

Hooke’s Law When a spring is stretched, there is a restoring force that is proportional to the displacement.

x m F = -kx

F

HOOKE'S LAW The restoring force of an ideal spring is given by,

where k is the spring constant and x is the displacement of the spring from its unstrained length. The minus sign indicates that the restoring force always points in a direction opposite to the displacement of the spring.

The force described by Hooke’s Law is the net force in Newton’s Second Law FHooke  FNewton  kx  ma  k  a    x  m

k = stiffness of the spring (N/m) = spring constant (N/m)

 k  a    x  m a  x 2

k   m k  m 2

SHM

  2f  f  2 1 T  f 2 T 



Mass & Spring System  



Simple Pendulum

k &   2f   m

k  2f m

g &   2f l

g  2f l

k f  m 2

g f  l 2

1 T  f

1 T  f

T  2

m k

T  2

l g

'

T1  k (e  x ) F  ma '

mg  T1  ma

e T1  mg & T1  ke mg  ke

x

ke  ke  kx  ma  kx  ma  k  a    x  m mg  ke m e  k g

e=static deflection g=gravity

T  2

m k

T  2

e g

DISPLACEMENT IN SHM x

m x = -A

x=0

x = +A

• Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left. • The maximum displacement is called the amplitude A.

VELOCITY IN SHM v (-)

v (+)

m x = -A

x=0

x = +A

• Velocity is positive when moving to the right and negative when moving to the left. • It is zero at the end points and a maximum at the midpoint in either direction (+ or -).

Acceleration in SHM +a

-x

+x

-a

m x = -A

x=0

x = +A

• Acceleration is in the direction of the restoring force. (a is positive when x is negative, and negative when x is positive.)

F  ma  kx • Acceleration is a maximum at the end points and it is zero at the center of oscillation.

Acceleration vs. Displacement x

a

v

m x = -A

x=0

x = +A

Given the spring constant, the displacement, and the mass, the acceleration can be found from:

F  ma   kx

or

kx a m

e: Acceleration is always opposite to displacement.

EXERCISE 7 A body of mass 14kg being hung with springs straight from one end attached to a rigid support. The body produced 25mm static deflection. It was pulled down 23mm and then released. Find i. ii. iii. iv.

The acceleration began to the body Periodic time The spring maximum force Velocity and acceleration the body when it is 12mm SOLUTIONS from the equilibrium position

 

The acceleration began to the body

ii.

Periodic time  

   

 

.

The spring maximum force  

iv.

Velocity and acceleration the body when it is 12mm from the equilibrium position      

THE SIMPLE PENDULUM In order to be in SHM, the restoring force must be proportional to the negative of the displacement. Here we have: which is proportional to sin θ and not to θ itself. However, if the angle is small, sin θ ≈ θ.

Therefore, for small angles, we have:

where The period and frequency are:

EXERCISE 7 A mass is suspended from a string 60mm long. It is nudged so that it makes a small swinging oscillation. Determine the frequency and periodic time. SOLUTIONS

 

Thank you !