3.2 Simple Harmonic Motion (SHM)

Question 1

Give three examples of systems which oscillate with Simple Harmonic Motion.

Answer 1

1. The simple pendulum,
2. The vertical mass-spring oscillator
3. The projected shadow of a body describing uniform circular motion

Question 2

State the conditions necessary for SHM to occur.

Answer 2

The unbalanced force F acting on the body exhibiting SHM is

1. Proportional to the displacement of the body x from the equilibrium position

and

2. Acts in the opposite direction to the displacement x

This is summarised in the equation F = -kx where k is the Force constant

Question 3

Find the magnitude of the spring constant for a Hooke’s Law spring which exerts a restoring force of 5 N when stretched by 4 cm.

Answer 3

F= -kx

5 = k x 0.04 …in magnitude

k = 5/0.04

k = 125 N m^-1

Question 4

Look at the horizontal mass-spring oscillator illustrated.

Show that the acceleration of the mass is given by a = d²y/dt² = – (k/m) y

Answer 4

The unbalanced restoring tension force T = ma

-ky = m d²y /dt²

d²y /dt² = -(k/m) y

Question 5

Using the auxiliary circle (shown), show that

a) y = A cos ωt
b) y = A sin ωt
c) d²y/dt² = a = -ω² y

are all solutions of the equation for SHM.

Answer 5

Question 6

NOTE THAT THIS IS AN SQA DERIVATION AND MUST BE MEMORISED

Show that v = +/- ω √(A² - y²) for a body describing SHM

Answer 6

Question 7

Show that the only two factors which affect the period T of SHM are

1. The Force (spring) constant, k
2. The mass of the body describing SHM, m

Answer 7

ω = 2π/T

So T = 2π/ω

But ω² = k/m

So T = 2π/√(k/m)

Therefore, the only two factors which affect the period of SHM are

1. The Force (spring) constant, k
2. The mass of the body describing SHM, m

Question 8

a) In SHM what is meant by the tern ω?
b) What is the unit of ω?
c) State the equation for ω

Answer 8

a) Angular frequency
b) Radians per second
c) ω = 2π / T = 2πf

Question 9

NOTE THAT THIS IS AN SQA DERIVATION AND MUST BE MEMORISED

Derive the expression E_k = ½m ω²(A² – y²) for the kinetic energy of a body exhibiting SHM.

Answer 9

Question 10

Describe and explain how damping on an oscillatory system causes the amplitude of oscillation to decay.

Answer 10

Damping is the process by which energy in an oscillating system is lost to the surroundings, often as heat due to frictional forces. This reduces the energy in the oscillating system causing the amplitude of the oscillation to decrease.

Question 11

State the following equations for a body describing SHM

a) Potential energy
b) Kinetic energy
c) Total energy

Answer 11

a) E_p = ½ mω²y²
b) E_k = ½ mv² = ½ m ω² (a²-y²)

c) E_tot = E_p + E_k
= ½ mω²y² + ½ m ω² (a² - y²)
E_tot = ½ m ω² a²

Question 12

The graph below shows how the potential energy, E_p, of an object undergoing SHM, varies with its displacement, y. The object has mass 0.40 kg and a maximum amplitude of 0.05 m.

(a) Find the potential energy of the object when it has a displacement of 0.02 m.
b) Calculate the force constant, k for the oscillating system. (k should have unit N m-1).
c) Find the amplitude at which the potential energy equals the kinetic energy.

Answer 12

(a) From graph E_p = 0.10 J
b) E_p = 1/2 k y²
0. 1 = 1/2 k (0.02)²

k = 0.2 (0.02)²

k = 500 N m^-1

c) Ep = Ek

1/2 k y² = 12 m ω²( a² - y² )

= 1/2 k (a² - y²) since ω² = k/m

y² = a² - y²

So 2 y² = a²

y = a/√2 when Ep = Ek

y = 0.05/√2

y = = 0.035 m

Question 13

Example

An object is vibrating with simple harmonic motion of amplitude 0.02 m and frequency 5.0 Hz. Assume that the displacement of the object, y = 0 at time, t = 0 and that it starts moving in the positive y-direction.

(a) Calculate the maximum values of velocity and acceleration of the object.
(b) Calculate the velocity and acceleration of the object when the displacement is 0.008 m.
(c) Find the time taken for the object to move from the equilibrium position to a displacement of 0.012 m.

Answer 13

Initial conditions:
y = a sin ωt v = aω cos ωt a = - ω²y f = 5 Hz ω = 2πf = 31.4 rad s^-1

(a) v_max = ωa = 31.4 x 0.02 = 0.63 m s^-1
a_max = - ω² a = -(31.4)2 x 0.02 = -19.7 m s^-2

(b) v = ± ω √ (a² - y²) = ± 31.4 x √ (0.02² - 0.008²)= ± 0.58 m s^-1
a = - ω² y = - (31.4)² x 0.008 = - 7.9 m s^-2

(c) use y = a sin ω t
0. 012 = 0.02 sin 31.4t (when y = 0.012 m)

sin 31.4 t = 0.012 / 0.02

= 0.6 giving

31.4 t = 0.644

and t = 0 644 / 31 4
.
.
Thus t = 0.0205 s (Remember that angles are in radians)