OSCILLATIONS

Question 1

State the conditions for simple harmonic motion

Answer 1

1. The force/acceleration must be proportional to the displacement from
   equilibrium
2. The force/acceleration must be in the opposite direction to displacement from
   equilibrium

Question 2

Sketch a graph of acceleration against displacement for a simple harmonic
oscillator and label the axis

Answer 2

see image

Question 3

Sketch the graphs of displacement, velocity and acceleration against time for
an oscillating system and explain how to determine one from another

Answer 3

see image

1. The velocity, time graph is the gradient of the displacement, time graph
2. The acceleration, time graph is the gradient of the velocity, time graph
3. Acceleration, time graph is always in antiphase with the displacement,
   time graph
4. Velocity, time graph is always 90 degrees ahead in phase to the
   displacement, time graph

Question 4

Draw a graph showing the kinetic energy, potential energy and total energy
against the displacement from equilibrium

Answer 4

see image

Question 5

Determine the time period of an oscillator that completes 4 cycles per second

Answer 5

1. Frequency = number of cycles per second = 4 Hz

2. Time period = 1/frequency = ¼ = 0.25 s

Question 6

Describe how the velocity of a simple harmonic oscillator varies across its
cycle

Answer 6

1. When the simple harmonic oscillator is at its maximum displacement, the
   velocity is zero
2. The velocity of the oscillator is maximum at the equilibrium position

Question 7

Describe how the acceleration of a simple harmonic oscillator varies across its
cycle

Answer 7

1. When the simple harmonic oscillator is at its maximum displacement, the
   acceleration is maximum
2. The acceleration of the oscillator is zero at the equilibrium position

Question 8

A mass on a spring oscillates through a total distance of 4.0cm. It makes 40
oscillations in 10.2 seconds.

Calculate:

1. The maximum velocity
2. The maximum acceleration of the mass on the spring

Answer 8

ω = 2π/T

T = 10.2 / 40 = 0.255 s

ω = 2π/ 0.255 = 24.6 rads -1

A = 0.020m (amplitude is maximum displacement from equilibrium so we must
halve the distance given)

So a) V max = ωA = 24.6 X 0.02 = 0.49ms -1 – this occurs at the equilibrium position

b) a max = ω 2 A = 24.6 2 x 0.02 = 12.1ms -2 – this occurs at the maximum displacement from equilibrium

Question 9

Explain how to accurately determine the time period of a mass on spring/pendulum oscillating

Answer 9

1. Time the time taken for the mass/pendulum to complete 20 oscillations and
   divide this by 20
2. This increases the time measured and hence reduces the % uncertainty in T
3. Repeat this measurement 3 times and calculate a mean to reduce the effect
   of random error
4. Use a fiducial marker and get to eye level with this to act as a reference
   position
5. The fiducial marker should be at equilibrium as the mass/pendulum has the
   highest velocity at this point so the time it spends in this position is smallest,
   reducing the uncertainty in starting/stopping the stopwatch

Question 10

Explain how to determine the maximum velocity of the mass/pendulum

Answer 10

1. Determine the time period (as outlined above)
2. Use ω = 2π/ T to determine the angular velocity
3. Determine the amplitude of the oscillation by measuring the maximum
   displacement from equilibrium using a metre ruler
4. Use Vmax = A ω to determine the maximum velocity

Question 11

Explain how to determine the acceleration due to gravity using a pendulum

Answer 11

1. Measure the time period of oscillation (using the method outlined above) for at
   least 6 different values for the length of the pendulum
2. Plot a graph of T 2 (s 2 ) against L (m)

• T 2 = (4π 2 / g) l
• y = m x

3. Determine the gradient of graph = 4π 2 / g
   - g = 4π 2 / gradient

Question 12

A sphere of diameter 30cm is hanging from a string. It is set into oscillation and completes 20 oscillations in 54.2 seconds. Calculate the length of the string:

Answer 12

1. T = 54.2 / 20 = 2.71 s
2. T = 2π
3. 2.71 = 2π
4. l = 1.82m
5. this is the length to the centre of mass of sphere, so length of string = 1.82 – radius of sphere = 1.82 – 0.15 = 1.67 m

Question 13

A tuning fork is struck and produces a note with a frequency of 500 Hz. A
vibration generator connected to a signal generator is attached to the fork.
The signal generator is adjusted, producing frequencies from 400Hz to
600Hz.
Explain why
a) The tuning fork produces a sound at the applied frequency
b) The sound is quiet for all frequencies except for 500 Hz which is much louder

Answer 13

1. 500Hz is the natural frequency of the tuning fork
2. The vibration generator causes forced oscillations at the driving frequency of
   the signal generator
3. When the driving frequency is not 500Hz the sound is quiet because less
   energy is transferred
4. When the driving frequency is equal to the natural frequency, 500Hz,
   resonance occurs
5. There is a maximum transfer of energy at this frequency so the amplitude is
   maximum

Question 14

The tuning fork (in the question above) is placed on a sound box.
a) Explain why the volume of the sound heard increases

Answer 14

1. The tuning fork forces the sound box into oscillation
2. Energy is transferred then from the sound box to the air inside
3. There is a larger volume of air set into oscillation than there was with just the
   tuning fork, hence the sound is louder

Question 15

The tuning fork (in the question above) is placed on a sound box.
Explain why the sound lasts for a shorter time than without the sound box
- The sound box acts as a damper

Answer 15

1. Removing energy from the oscillating system
2. There is a larger rate of energy transfer to the air when the sound box is
   present and therefore the amplitude is reduced over a shorter period of time

Question 16

Define damping

Answer 16

1. A system is damped if energy is removed from the system
2. It occurs when a force acts on the oscillator in the opposite direction to the
   velocity
3. Reducing the amplitude of the oscillation

Question 17

Identify and explain the positions within the oscillation that the damping force
is maximum and zero

Answer 17

1. the damping force is maximum when the velocity is maximum ie. at the
   equilibrium position
2. the damping force is zero when the velocity is zero ie. at the maximum
   displacement

Question 18

Compare the maximum velocity of an oscillating system when resonance
occurs to when it is driven at a frequency different from its natural frequency

Answer 18

1. When a system is undergoing resonance, its amplitude, A becomes maximum
2. Therefore, as v max = ωA , its maximum velocity increases

Question 19

Explain what happens to the maximum damping force experienced by an
oscillating system when it undergoes resonance

Answer 19

1. As the maximum velocity of the oscillator is maximum at resonance (as v max =
   ωA and amplitude is maximum when resonating), the damping force will be
   maximum, as the damping force increases with velocity

Question 20

Two pendula are oscillating in air with the same initial amplitude. One
pendulum has a mass of 40kg, and the other has a mass of 20kg. Explain
which pendulum will oscillate for longer.

Answer 20

1. Air resistance provides a force in the opposite direction to the velocity,
   causing damping
2. Air resistance is the same provided the surface area, shape and speed is the
   same initially
3. However, the larger mass has a smaller negative acceleration due to a=F/m
   (note that when the system is in the equilibrium position, this damping force is
   the only force that acts and therefore is the resultant force)
4. Therefore the decrease in velocity is less for each swing for the higher mass
   pendulum
5. So the higher mass pendulum will oscillate for a longer time

Question 21

The Foucault pendulum in the Paris Pantheon has to be ‘reset’ in its
oscillation once every hour in winter, but only once every 2 hours in summer.
Explain why.

Answer 21

1. When the air temperature is higher in summer the density of air is lower
2. So the air resistance acting on the pendulum is lower
3. Reducing the damping force, so the pendulum will oscillate for a longer time

Question 22

An object is oscillating with a high amplitude. Dampers are added to the
system. Describe and explain the material properties of something that is
suitable for damping.

Answer 22

The material should be able to plastically deform (it should be ductile)
- This will ensure that the energy is removed from the oscillating system (to
increase the internal energy of the damper)
- This reduces the amplitude of the oscillations over time

Question 23

An object is oscillating with a high amplitude. Explain why applying a damper
that is tuned to the natural frequency of oscillation of the object would reduce
the amplitude and why the damper should be heavily damped.

Answer 23

1. The damper absorbs energy from the oscillating system when it deforms
2. As the damper oscillates with the natural frequency, there is a maximum
   transfer of energy
3. The dampers should be heavily damped to prevent energy from being
   returned to the oscillating object – the energy must be dissipated instead

Question 24

What happens when the damping is increased

Answer 24

1. The maximum amplitude of the oscillation is increased.

2. The frequency that the maximum amplitude occurs at is decreased