OSCILLATIONS Flashcards
State the conditions for simple harmonic motion
- The force/acceleration must be proportional to the displacement from
equilibrium - The force/acceleration must be in the opposite direction to displacement from
equilibrium
Sketch a graph of acceleration against displacement for a simple harmonic
oscillator and label the axis
see image
Sketch the graphs of displacement, velocity and acceleration against time for
an oscillating system and explain how to determine one from another
see image
- The velocity, time graph is the gradient of the displacement, time graph
- The acceleration, time graph is the gradient of the velocity, time graph
- Acceleration, time graph is always in antiphase with the displacement,
time graph - Velocity, time graph is always 90 degrees ahead in phase to the
displacement, time graph
Draw a graph showing the kinetic energy, potential energy and total energy
against the displacement from equilibrium
see image
Determine the time period of an oscillator that completes 4 cycles per second
- Frequency = number of cycles per second = 4 Hz
2. Time period = 1/frequency = ¼ = 0.25 s
Describe how the velocity of a simple harmonic oscillator varies across its
cycle
- When the simple harmonic oscillator is at its maximum displacement, the
velocity is zero - The velocity of the oscillator is maximum at the equilibrium position
Describe how the acceleration of a simple harmonic oscillator varies across its
cycle
- When the simple harmonic oscillator is at its maximum displacement, the
acceleration is maximum - The acceleration of the oscillator is zero at the equilibrium position
A mass on a spring oscillates through a total distance of 4.0cm. It makes 40
oscillations in 10.2 seconds.
Calculate:
- The maximum velocity
- The maximum acceleration of the mass on the spring
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
Explain how to accurately determine the time period of a mass on spring/pendulum oscillating
- Time the time taken for the mass/pendulum to complete 20 oscillations and
divide this by 20 - This increases the time measured and hence reduces the % uncertainty in T
- Repeat this measurement 3 times and calculate a mean to reduce the effect
of random error - Use a fiducial marker and get to eye level with this to act as a reference
position - 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
Explain how to determine the maximum velocity of the mass/pendulum
- Determine the time period (as outlined above)
- Use ω = 2π/ T to determine the angular velocity
- Determine the amplitude of the oscillation by measuring the maximum
displacement from equilibrium using a metre ruler - Use Vmax = A ω to determine the maximum velocity
Explain how to determine the acceleration due to gravity using a pendulum
- Measure the time period of oscillation (using the method outlined above) for at
least 6 different values for the length of the pendulum - Plot a graph of T 2 (s 2 ) against L (m)
- T 2 = (4π 2 / g) l
- y = m x
- Determine the gradient of graph = 4π 2 / g
- g = 4π 2 / gradient
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:
- T = 54.2 / 20 = 2.71 s
- T = 2π
- 2.71 = 2π
- l = 1.82m
- 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
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
- 500Hz is the natural frequency of the tuning fork
- The vibration generator causes forced oscillations at the driving frequency of
the signal generator - When the driving frequency is not 500Hz the sound is quiet because less
energy is transferred - When the driving frequency is equal to the natural frequency, 500Hz,
resonance occurs - There is a maximum transfer of energy at this frequency so the amplitude is
maximum
The tuning fork (in the question above) is placed on a sound box.
a) Explain why the volume of the sound heard increases
- The tuning fork forces the sound box into oscillation
- Energy is transferred then from the sound box to the air inside
- There is a larger volume of air set into oscillation than there was with just the
tuning fork, hence the sound is louder
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
- Removing energy from the oscillating system
- 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
Define damping
- A system is damped if energy is removed from the system
- It occurs when a force acts on the oscillator in the opposite direction to the
velocity - Reducing the amplitude of the oscillation
Identify and explain the positions within the oscillation that the damping force
is maximum and zero
- the damping force is maximum when the velocity is maximum ie. at the
equilibrium position - the damping force is zero when the velocity is zero ie. at the maximum
displacement
Compare the maximum velocity of an oscillating system when resonance
occurs to when it is driven at a frequency different from its natural frequency
- When a system is undergoing resonance, its amplitude, A becomes maximum
- Therefore, as v max = ωA , its maximum velocity increases
Explain what happens to the maximum damping force experienced by an
oscillating system when it undergoes resonance
- 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
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.
- Air resistance provides a force in the opposite direction to the velocity,
causing damping - Air resistance is the same provided the surface area, shape and speed is the
same initially - 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) - Therefore the decrease in velocity is less for each swing for the higher mass
pendulum - So the higher mass pendulum will oscillate for a longer time
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.
- When the air temperature is higher in summer the density of air is lower
- So the air resistance acting on the pendulum is lower
- Reducing the damping force, so the pendulum will oscillate for a longer time
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.
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
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.
- The damper absorbs energy from the oscillating system when it deforms
- As the damper oscillates with the natural frequency, there is a maximum
transfer of energy - The dampers should be heavily damped to prevent energy from being
returned to the oscillating object – the energy must be dissipated instead
What happens when the damping is increased
- The maximum amplitude of the oscillation is increased.
2. The frequency that the maximum amplitude occurs at is decreased
