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Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
1 
Simple Harmonic Motion & Springs 
Week 13, Lesson 1 
• 
Period & frequency 
• 
Graph of Vibratory Motion 
• 
Displacement, Restoring Force, Hooke’s Law 
• 
Period & Acceleration in SHM 
• 
The Simple Pendulum 
References/Reading Preparation: 
Schaum’s Outline Ch. 11 
Principles of Physics by Beuche – Ch.14
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
2 
Simple Harmonic Motion 
A vibrating system is one which undergoes the same 
motion over and over again. 
Examples are: a pendulum, a weight on a spring, among 
others. 
The time it takes for one complete cycle (the combined 
back and forth motion, or up and down motion) 
is called the period, T. 
It is the total time for the combined back and forth motion. 
The unit is the second, s.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
3 
The number of cycles completed in a unit time is called 
the frequency f. 
f = 1/T 
(cycles/s) 
One cycles/s is equivalent to 1 hertz (Hz). 
The amplitude is the maximum displacement of the 
object from the position the object maintains when it 
is not vibrating.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
4 
For a pendulum, the amplitude is the maximum ‘swing’ 
of the object. In the diagram below, it is the distance AB 
or BC. 
B 
A 
C 
Note the important feature of the interchange of energy between 
potential and kinetic energy. 
At A or C, the ball is at rest – therefore there is no KE, but max. GPE. 
As it swings to B, it loses GPE and gains an equal amount of KE.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
5 
The same goes for a spring system with a mass at the end. 
Consider the set-up shown with the mass sliding on a 
frictionless surface. 
spring 
Equilibrium 
The spring is compressed – therefore there 
is stored PE. The force in the spring wants 
to push the block back to x = 0. 
x = 0 
When released, the block accelerates to the 
right – at x = 0, PE = 0, and KE is max. 
The spring is stretched – therefore 
there is stored PE, and KE = 0 
The force in the spring wants to pull the 
block back to equilibrium.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
6 
When the block is released, it will continue to 
oscillate back and forth if the surface is frictionless. 
This is because the force due to the spring wants to push 
or pull the block back to the equilibrium position. 
This force is called the restoring force. 
A restoring force is necessary if vibration is to occur. 
It is a force that is always directed so as to push or pull 
the system back to its equilibrium (normal resting) position.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
7 
The magnitude of the restoring force depends on the 
‘stiffness’ of the spring – which is called the spring 
constant k. 
Experiments have shown that the force in a spring is directly 
proportional to the displacement. They are related by the formula 
F = -kx, where F is the restoring force exerted by the spring and 
is in the opposite direction to the distortion. 
This relationship is known as Hooke’s Law. 
Hooke’s Law is obeyed by a system if the magnitude of 
The restoring force is proportional to the magnitude of 
the displacement.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
8 
The potential energy stored in the spring, distorted a 
distance x, is ½ kx2, where k is the spring constant. 
If the amplitude is xo , then the energy of the vibrating 
system at all times is ½ kxo2. 
From the law of conservation of energy, 
KE + PE = constant 
or 
½ mv2 + ½ kx2 = ½ kxo2 
(where xo is the amplitude)
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
9 
From this we can determine the velocity of the 
vibrating object at any point during its displacement. 
v= (xo2 + x2)k/m
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
10 
Simple Harmonic Motion (SHM) is the vibratory motion 
that a system that obeys Hooke’s Law undergoes. 
Said another way: 
A vibratory motion that obeys Hooke’s Law is said to be 
in Simple Harmonic Motion. 
This motion is illustrated on the next slide. Because of 
the resemblance of its graph to a sine or cosine curve, 
SHM is frequently called sinusoidal motion.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
11 
The Graph of Vibratory Motion 
The graph of vibratory motion is shown. The motion shown is the 
up-and-down motion of a mass at the end of a spring. 
One complete cycle is from a to b, or from c to d, or from e to f. 
Equilibrium 
Position 
Displacement (y) 
time 
y = 0 
y = yo 
Amplitude = yo 
T 
a 
b 
c 
d 
e 
f 
The time for one cycle is T, the period.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
12 
Acceleration in SHM 
Acceleration in SHM is given by Hooke’s Law, F = -kx, 
and F = ma. Equating these two expressions for F gives 
a = 
- 
k 
m 
x 
The minus sign indicates that the direction of a (and F) 
is always opposite to the direction of the displacement x.
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
13 
Period in SHM 
The period T of a SHM is defined as: 
T = 
2m/k 
Acceleration in terms of T 
a = - 
42 
T2 
x
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
14 
The Simple Pendulum 
The simple pendulum very nearly undergoes SHM if its 
angle of swing is not too large. 
The period of vibration for a pendulum of length L at a 
location where the gravitational acceleration is g is given 
by: 
T = 
2L/g
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
15 
Worked examples 
A 50 g mass vibrates in SHM at the end of a spring. The amplitude 
of the motion is 12 cm, and the period is 1.70 s. 
Find: 
a) 
The frequency, f. (ans. 0.588 Hz) 
b) 
The spring constant. (ans. 0.68 N/m) 
c) 
The maximum speed of the mass. (ans. 0.44 m/s) 
d) 
The maximum acceleration of the mass. (ans. 1.63 m/s2) 
e) 
The speed when the displacement is 6 cm. (ans. 0.38 m/s) 
f) 
The acceleration when x = 6 cm. (ans. -0.82 m/s2)
Week 13, Lesson 1 Simple Harmonic Motion & 
Springs 
16 
Worked example 
The 200 g mass shown is pushed to the left against the spring and 
compresses the spring 15 cm from its equilibrium position. The 
system is then released, and the mass shoots to the right. If friction 
can be ignored, how fast will the mass be moving as it shoots away? 
Assume the mass of the spring to be very small. 
200 g 
spring 
k = 400 N/m 
(ans. 6.7 m/s)

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Physics 16

  • 1. Week 13, Lesson 1 Simple Harmonic Motion & Springs 1 Simple Harmonic Motion & Springs Week 13, Lesson 1 • Period & frequency • Graph of Vibratory Motion • Displacement, Restoring Force, Hooke’s Law • Period & Acceleration in SHM • The Simple Pendulum References/Reading Preparation: Schaum’s Outline Ch. 11 Principles of Physics by Beuche – Ch.14
  • 2. Week 13, Lesson 1 Simple Harmonic Motion & Springs 2 Simple Harmonic Motion A vibrating system is one which undergoes the same motion over and over again. Examples are: a pendulum, a weight on a spring, among others. The time it takes for one complete cycle (the combined back and forth motion, or up and down motion) is called the period, T. It is the total time for the combined back and forth motion. The unit is the second, s.
  • 3. Week 13, Lesson 1 Simple Harmonic Motion & Springs 3 The number of cycles completed in a unit time is called the frequency f. f = 1/T (cycles/s) One cycles/s is equivalent to 1 hertz (Hz). The amplitude is the maximum displacement of the object from the position the object maintains when it is not vibrating.
  • 4. Week 13, Lesson 1 Simple Harmonic Motion & Springs 4 For a pendulum, the amplitude is the maximum ‘swing’ of the object. In the diagram below, it is the distance AB or BC. B A C Note the important feature of the interchange of energy between potential and kinetic energy. At A or C, the ball is at rest – therefore there is no KE, but max. GPE. As it swings to B, it loses GPE and gains an equal amount of KE.
  • 5. Week 13, Lesson 1 Simple Harmonic Motion & Springs 5 The same goes for a spring system with a mass at the end. Consider the set-up shown with the mass sliding on a frictionless surface. spring Equilibrium The spring is compressed – therefore there is stored PE. The force in the spring wants to push the block back to x = 0. x = 0 When released, the block accelerates to the right – at x = 0, PE = 0, and KE is max. The spring is stretched – therefore there is stored PE, and KE = 0 The force in the spring wants to pull the block back to equilibrium.
  • 6. Week 13, Lesson 1 Simple Harmonic Motion & Springs 6 When the block is released, it will continue to oscillate back and forth if the surface is frictionless. This is because the force due to the spring wants to push or pull the block back to the equilibrium position. This force is called the restoring force. A restoring force is necessary if vibration is to occur. It is a force that is always directed so as to push or pull the system back to its equilibrium (normal resting) position.
  • 7. Week 13, Lesson 1 Simple Harmonic Motion & Springs 7 The magnitude of the restoring force depends on the ‘stiffness’ of the spring – which is called the spring constant k. Experiments have shown that the force in a spring is directly proportional to the displacement. They are related by the formula F = -kx, where F is the restoring force exerted by the spring and is in the opposite direction to the distortion. This relationship is known as Hooke’s Law. Hooke’s Law is obeyed by a system if the magnitude of The restoring force is proportional to the magnitude of the displacement.
  • 8. Week 13, Lesson 1 Simple Harmonic Motion & Springs 8 The potential energy stored in the spring, distorted a distance x, is ½ kx2, where k is the spring constant. If the amplitude is xo , then the energy of the vibrating system at all times is ½ kxo2. From the law of conservation of energy, KE + PE = constant or ½ mv2 + ½ kx2 = ½ kxo2 (where xo is the amplitude)
  • 9. Week 13, Lesson 1 Simple Harmonic Motion & Springs 9 From this we can determine the velocity of the vibrating object at any point during its displacement. v= (xo2 + x2)k/m
  • 10. Week 13, Lesson 1 Simple Harmonic Motion & Springs 10 Simple Harmonic Motion (SHM) is the vibratory motion that a system that obeys Hooke’s Law undergoes. Said another way: A vibratory motion that obeys Hooke’s Law is said to be in Simple Harmonic Motion. This motion is illustrated on the next slide. Because of the resemblance of its graph to a sine or cosine curve, SHM is frequently called sinusoidal motion.
  • 11. Week 13, Lesson 1 Simple Harmonic Motion & Springs 11 The Graph of Vibratory Motion The graph of vibratory motion is shown. The motion shown is the up-and-down motion of a mass at the end of a spring. One complete cycle is from a to b, or from c to d, or from e to f. Equilibrium Position Displacement (y) time y = 0 y = yo Amplitude = yo T a b c d e f The time for one cycle is T, the period.
  • 12. Week 13, Lesson 1 Simple Harmonic Motion & Springs 12 Acceleration in SHM Acceleration in SHM is given by Hooke’s Law, F = -kx, and F = ma. Equating these two expressions for F gives a = - k m x The minus sign indicates that the direction of a (and F) is always opposite to the direction of the displacement x.
  • 13. Week 13, Lesson 1 Simple Harmonic Motion & Springs 13 Period in SHM The period T of a SHM is defined as: T = 2m/k Acceleration in terms of T a = - 42 T2 x
  • 14. Week 13, Lesson 1 Simple Harmonic Motion & Springs 14 The Simple Pendulum The simple pendulum very nearly undergoes SHM if its angle of swing is not too large. The period of vibration for a pendulum of length L at a location where the gravitational acceleration is g is given by: T = 2L/g
  • 15. Week 13, Lesson 1 Simple Harmonic Motion & Springs 15 Worked examples A 50 g mass vibrates in SHM at the end of a spring. The amplitude of the motion is 12 cm, and the period is 1.70 s. Find: a) The frequency, f. (ans. 0.588 Hz) b) The spring constant. (ans. 0.68 N/m) c) The maximum speed of the mass. (ans. 0.44 m/s) d) The maximum acceleration of the mass. (ans. 1.63 m/s2) e) The speed when the displacement is 6 cm. (ans. 0.38 m/s) f) The acceleration when x = 6 cm. (ans. -0.82 m/s2)
  • 16. Week 13, Lesson 1 Simple Harmonic Motion & Springs 16 Worked example The 200 g mass shown is pushed to the left against the spring and compresses the spring 15 cm from its equilibrium position. The system is then released, and the mass shoots to the right. If friction can be ignored, how fast will the mass be moving as it shoots away? Assume the mass of the spring to be very small. 200 g spring k = 400 N/m (ans. 6.7 m/s)