3.2 Simple Harmonic Motion (SHM) Flashcards

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1
Q

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

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

State the conditions necessary for SHM to occur.

A

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

  1. Acts in the opposite direction to the displacement x

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

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3
Q

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.

A

F= -kx

5 = k x 0.04 …in magnitude

k = 5/0.04

k = 125 N m-1

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4
Q

Look at the horizontal mass-spring oscillator illustrated.

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

A

The unbalanced restoring tension force T = ma

-ky = m d2y /dt2

d2y /dt2 = -(k/m) y

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5
Q

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.

A
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6
Q

NOTE THAT THIS IS AN SQA DERIVATION AND MUST BE MEMORISED

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

A
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7
Q

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
A

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
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8
Q

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

A

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

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9
Q

NOTE THAT THIS IS AN SQA DERIVATION AND MUST BE MEMORISED

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

A
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10
Q

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

A

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.

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11
Q

State the following equations for a body describing SHM

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

A

a) Ep = ½ mω2y2
b) Ek = ½ mv2 = ½ m ω2 (a2-y2)

c) Etot = Ep + Ek
= ½ mω2y2 + ½ m ω2 (a2 - y2)
Etot = ½ m ω2 a2

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12
Q

The graph below shows how the potential energy, Ep, 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.

A

(a) From graph Ep = 0.10 J
b) Ep = 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

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13
Q

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.

A

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

(a) vmax = ωa = 31.4 x 0.02 = 0.63 m s-1
amax = - 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)

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