Oscillations (DONE)

Question 1

Which 2 systems are used to demonstrate simple harmonic motion?

Answer 1

• A simple pendulum (attached to string).

- A mass-spring system.

Question 2

What does the angle theta represent for a simple pendulum moving with SHM?

Answer 2

• Theta is the angle moved through by the pendulum from the equilibrium point to reach a certain displacement.

Question 3

How do simple pendulums and mass-spring systems accelerate when displaced?

Answer 3

• When a pendulum is displaced by a distance x, it will try to accelerate towards the equilibrium point with acceleration a.
• When looking at the mass-spring system, if the mass is displaced by a distance x from the equilibrium, it will try to accelerate towards the equilibrium point with acceleration a.

Question 4

What is the amplitude and what type of value is it?

Answer 4

• The amplitude is the magnitude of the max displacement and because of this it must always have a positive value.

Question 5

What is the displacement and what type of value is it?

Answer 5

• The displacement is the distance of an object from the equilibrium position and it can have any value from zero up to the amplitude for both sides of the equilibrium, for this reason it can be positive or negative.

Question 6

What is frequency?

Answer 6

• The frequency is the number of oscillations per unit time of pendulum/mass-spring.

Question 7

What is the time period?

Answer 7

• The period is the time for one complete oscillation.

Question 8

What is the angular velocity?

Answer 8

• The angular velocity is the rate of angular rotation and therefore the number of radians which something moves through per second.
• It has the symbol omega w, and is measured in radians per seconds.

Question 9

How many radians does a pendulum move through in a complete oscillation?

Answer 9

• Oscillating objects don’t move through a complete circle but we can still draw a graph of displacement x, against time t.
• The graph will show a sin shaped graph and we can say that in a full oscillation the pendulum moves through an angle of 2pie radians.

Question 10

How does the length of string on the pendulum impact angular frequency?

Answer 10

• An oscillating pendulum will oscillate faster with a shorter string.
• It will move through more complete oscillations per second and therefore it has a greater angular frequency.

Question 11

What is angular frequency?

Answer 11

• Angular frequency is the rate of change of phase.
• It is also the product of 2pie*f or alternatively, w = (2pie)/T.
• Has the symbol omega w, and is measured in radians per second.

Question 12

How can you prove that 1 complete oscillation of a pendulum is equal to 2pie rads using angular frequency?

Answer 12

• If the period of an oscillation is constant for all oscillations then we can cancel the T form the equation w = (2pie)/T and therefore say that 1 oscillation = 2pie rads.

Question 13

What is the relationship between frequency of an oscillation and angular frequency?

Answer 13

• The higher the frequency of the pendulum, the greater the angular frequency.
• This is because of the equation w = 2pie*f.

Question 14

What types of quantity are angular velocity and angular frequency?

Answer 14

• Angular velocity is a vector quantity.

- Angular frequency is a scalar quantity.

Question 15

How can you tell whether 2 points in an oscillation are in or out of phase?

Answer 15

• If we draw out a displacement against time graph we get a sin shaped curve and we can say that 1 oscillation is 2pie radians.
• If we look at 2 points on the curve and they are 2pie radians apart they are out of phase and therefore they are in phase.
• If we look at another 2 points and they are pie radians apart they are out of phase and in anti phase.

Question 16

What is phase difference?

Answer 16

• Phase difference with symbol phi, is the fraction of a complete oscillation between 2 oscillating points measured in radians.

Question 17

What forces are acting on a pendulum at an angle of theta to the equilibrium position?

Answer 17

• If we draw a force diagram for the pendulum we firstly have 2 different forces.
• Firstly we have the weight w = mg which acts vertically down, this is counteracted by the second force which is the tension force in the string.
• There is a component of the weight which acts at an angle theta to the weight and parallel to the tension.
• There is also a component acting perpendicular to the tension – this force causes the acceleration of the pendulum.
• The size of this component of the force is mgsintheta and so as the angle of theta gets bigger, the size of the force also increases.
• (As the angle of theta from the equilibrium position increases, the angle of theta in the triangle of forces for w = mg will also increase).

Question 18

Why does a pendulum with a greater displacement x from equilibrium have a greater acceleration towards the equilibrium?

Answer 18

• The component of w = mg which acts perpendicular to tension causes the acceleration towards the equilibrium.
• In order to find the size of the force of this component, the equation F = mgsintheta is used.
• The angle theta will increase as x increases, this means when the pendulum has a greater displacement x, the force creating the acceleration will increase causing a larger acceleration.

Question 19

What is the relationship between displacement x form the equilibrium and acceleration towards the equilibrium?

Answer 19

• Due to the equation F = mgsintheta and the fact that a greater displacement x causes a larger accelerating force, we can say that acceleration is directly proportional to the displacement.
• However when x increases in the positive direction, the value of a increases in the negative direction, this means we must place a minus in the relationship between a and x.
• This results in a being directly proportional to –x.

Question 20

What does an object undergoing SHM need to show in terms of x and a?

Answer 20

• An acceleration proportional to the displacement from the equilibrium position.
• An acceleration in the opposite direction to its displacement.

Question 21

How can we create an equation for acceleration of a pendulum undergoing SHM?

Answer 21

• We need to insert a constant into the relationship between a and x.
• The equation can be written as acceleration a = -cx where c is a constant.
• Because c has to be a positive constant we can say that c = (2pief)^2, therefore we can write the equation as a = -[(2pief)^2]x.
• This equation can be simplified using the fact that angular frequency w = 2pie*f and therefore the final equation is a = -[w^2]x.
• This is the defining equation for SHM.

Question 22

What point should the time period of simple pendulums be measured from?

Answer 22

• There are 2 possible points that you can take the time from, firstly you can take the time for a complete oscillation starting from the amplitude of the pendulum.
• However the best way is to take the time for a complete oscillation starting from the equilibrium position.
• The reason this is better is because the pendulum has a lower velocity at the amplitudes, whereas the velocity is max in the equilibrium position meaning that there will be less error when recording results for the time period.

Question 23

How should you record data when investigating the time period of a simple pendulum?

Answer 23

• When measuring the period you should measure 10 oscillations and divide by 10 to get the mean time for 1 full oscillation.
• You can take the reciprocal of this to find the frequency of the pendulum.

Question 24

How does the time period of a simple pendulum change as amplitude decreases?

Answer 24

• For pendulums undergoing SHM through small angles the time period should remain constant despite the decreasing amplitude.
• This also means the frequency should remain constant.

Question 25

What factors directly impact the time period of a simple pendulum?

Answer 25

• The time period of a simple pendulum doesn’t depend on the amplitude, only the length of the string and the mass of the pendulum impact the period.

Question 26

What equation is used to find the displacement x of a simple pendulum when the time t = 0 is taken from the equilibrium position?

Answer 26

• If time t = 0 is when the simple pendulum is at the equilibrium position, a sin curve is created on the graph of displacement x against time t.
• We therefore use the equation displacement x = Asin wt where w is the angular frequency.
• The equation displacement x = Asin 2pie ft, is another variation of the same equation which breaks down angular frequency.
• The angles need to be calculated in radians in these equations.

Question 27

What equation is used to find the displacement x of a simple pendulum when the time t = 0 is taken from the amplitude position?

Answer 27

• If we start time = 0 at the amplitude of the pendulum, then it will create a cos curve on displacement x against time t graph.
• This means the equation to find the displacement x = Acos wt, or x = Acos 2pie ft.
• The angles need to be calculated in radians in these equations.

Question 28

How can we prove the equations ‘x = Asin 2pieft’ and ‘x = Acos 2pieft’ are solutions to the the SHM equation?

Answer 28

First equation:
x = Asin 2pieft

1. velocity v = dx/dt therefore differentiate the equation:

v = A2pief cos 2pieft.

2. Acceleration a = dv/dt, therefore differentiate the equation again

a = - A(2pief)^2 sin 2pieft.

3. Simplify to get SHM equation:

a = - (2pief)^2 x

• Apply same method to co version of equation.
• Start with x -> differentiate to get v -> differentiate again to get a -> simplify using x term.

Question 29

What is the equation for the velocity of an object undergoing SHM at a certain point?

Answer 29

• We use the equation velocity v = +-w*root(A^2 – x^2) where w is the angular frequency, A is amplitude and x is displacement..

Question 30

How does the velocity of an object undergoing SHM vary at the equilibrium and amplitude positions?

Answer 30

• When the object reaches the max displacement/amplitude, its velocity must at one point be zero in order for it to move repeatedly in positive and then negative directions.
• When it reaches the equilibrium position, this is when any potential energy is converted to kinetic energy and the object therefore has the maximum velocity.

Question 31

How can you prove that velocity of an object undergoing SHM is 0 at the amplitude?

Answer 31

• Using the equation v = +-wroot(A^2 – x^2), if the displacement x = A this means that the equation will become v = +-wroot(A^2 – A^2) which = 0.
• Therefore the velocity at the max displacement is 0.

Question 32

How can you prove that velocity of an object undergoing SHM is max at the equilibrium?

Answer 32

• Using the equation v = +-wroot(A^2 – x^2), if displacement x = 0 then the equation will be v = +-wroot(A^2 – 0).
• This can be simplified to find the equation for the maximum velocity vmax = +-wA.

Question 33

What 3 graphs can we use to look at the motion of an SHM oscillator?

Answer 33

We can look at the following graphs:

x against t
v against t
a against t

Question 34

How can we create a graph of x against t for an SHM oscillator?

Answer 34

• We can look at a displacement time graph where time t = 0 begins when the oscillator is in the equilibrium position x = 0.
• ## We will get a sin curve and we can look at the max displacement/amplitude of the oscillator.

Question 35

How can we create a graph of v against t for an SHM oscillator?

Answer 35

• We can draw a velocity time graph as we know that the velocity of an object is equal to the rate of change of its displacement.
• This can be expressed by v = dx/dt which is the gradient at any point on the sin curve in the x against t graph.
• At x = 0 the sin curve has maximum gradient and therefore v is max.
• As x increases the gradient will get smaller and v will decrease until it reaches 0 at the peak of the sin curve.
• We then have the maximum negative v when the displacement once again reaches 0.
• If we plot these points on a v against t graph then we get a cos curve, as dx/dt of the sin curve = a cos curve.

Question 36

How can we create a graph of a against t for an SHM oscillator?

Answer 36

• We can create an acceleration time graph as a = dv/dt.
• a is also proportional to the negative displacement x.
• Therefore when x is zero, the acceleration will be zero and when x is max we have maximum negative a.

Question 37

How do the graphs of x, v and a against t relate to each other for an SHM oscillator?

Answer 37

• The graph of x against t is a sin curve.
• v against t is a cos curve.
• a against t is a –sin curve.
• We can also say that each graph is further displaced to the left by ¼ wavelengths each time.

Question 38

What energy changes does a simple pendulum experience during SHM?

Answer 38

• The kinetic energy Ek is maximum when the pendulum is at the equilibrium, and minimum at the amplitude.
• This is because v is max at the equilibrium and min at amplitude.
• The potential energy Ep is maximum at the amplitude and min at the equilibrium.
• This is because at the amplitude the pendulum reaches it greatest height and at the equilibrium it reaches its min height.
• We can look at the link between the kinetic and potential energy using a number of graphs.

Question 39

How can we look at the energy changes of a simple pendulum on a graph of total energy against x?

Answer 39

• We can create a graph of total energy against x with respect to the positive and negative amplitude.
• When the pendulum is at amplitude A or –A the potential energy will be max and kinetic energy min.
• At the equilibrium x = 0 the potential energy will be min and kinetic energy max.
• We therefore get 2 x^2 shaped graphs where Ep is of form x^2 and Ek is of form –x^2.
• If we look at the total energy, provided there is no energy loss to surroundings the energy will be constant because as Ep decreases Ek increases at the same rate.

Question 40

How can we look at the energy changes of a simple pendulum on a graph of total energy against time?

Answer 40

• We can create a graph of how the energy changes with respect to time.
• The total energy will remain constant and will therefore be a flat line.
• Ek will have the form of a sin curve, increasing and decreasing continuously.
• Ep will also have the form of a sin curve however when Ek is max, Ep is min.

Question 41

What 3 types of energy are in a spring mass system undergoing SHM?

Answer 41

• Elastic potential
• Gravitational potential
• Kinetic energy

Question 42

How do energies in a spring mass system vary when it is undergoing SHM?

Answer 42

• At the positive and negative amplitude when the mass is at its highest and lowest point the kinetic energy Ek = 0, at the equilibrium when velocity is maximum kinetic energy Ek = max.
• Gravitational potential energy is defined by GPE = mgh, therefore when the mass is at its max height GPE = max, at the equilibrium GPE = ½ max and when the mass is at its lowest point GPE = min.
• Elastic potential energy depends on how degree to which the spring is stretched ,therefore at the positive amplitude when the spring is stretched the least EPE = min, at the equilibrium the spring is slightly stretched so EPE = ½ max and at the negative amplitude when the mass is at its lowest point the spring is most stretched and EPE = max.

Question 43

What is a free oscillation?

Answer 43

• A free oscillation is where if an object is struck and set vibrating it will oscillate at its natural frequency.
• An example of this is striking a wine glass and then leaving it to undergo free oscillation at its natural frequency.

Question 44

What is a forced oscillation?

Answer 44

• A forced oscillation is where a periodic driving force is applied and an object vibrates at this frequency.
• An example of this is someone pushing a swing periodically, acting as the driving force which causes the swing to oscillate at this frequency.

Question 45

What is resonance?

Answer 45

• Resonance is when the driving frequency = natural frequency.
• When objects resonate it causes the amplitude of any oscillations to increase dramatically.
• This is the way we tune radios, and the way an MRI machine works.

Question 46

What graph can be used to investigate resonance?

Answer 46

• We can look at a graph of the amplitude of an object being driven on the y axis, and the driving frequency on the x axis.
• When driving frequency = 0 the object will oscillate at its natural amplitude.
• When the driving frequency is very low the amplitude will only increase by a small amount, however as the driving frequency is increased we will reach resonance and the amplitude will increase dramatically.
• As you increase the driving frequency above the resonant frequency the amplitude will start to decrease and will end up creating a curve where the final amplitude is lower than the initial amplitude.
• The final amplitude will be smaller than the initial amplitude because if you think about a swing as the frequency of the push exceeds the natural frequency of the swing you will be pushing the air rather than the swing so no energy will be transferred causing the amplitude to decrease.
• The point on the x axis where there is max amplitude is reached is the resonant frequency f0.

Question 47

What are examples of where resonance is not useful?

Answer 47

• If you’re driving a car at a certain speed which causes resonance it can cause the car to vibrate and make a lot of noise.
• If the frequency of footsteps on a bridge is equal to the natural frequency of the bridge, this can cause the bridge to shake.

Question 48

What is damped simple harmonic motion?

Answer 48

• This is where the amplitude of an oscillating object will decrease over time.
• This could be due to friction with the air, components in the oscillator e.g. the string in a pendulum, or friction with other objects.

Question 49

What is light damping?

Answer 49

• This is where the energy lost after every oscillation is small.
• If we were to look at the graph of the oscillator with displacement x against time t, we see the amplitude will decrease by a small amount for every next oscillation.

Question 50

What situation would we need heavy damping?

Answer 50

• Sometimes we need heavier damping in a car when you hit a bump, you don’t want to keep bouncing up and down therefore you need to dissipate the energy quickly.
• For suspension in a car you have a spring with an oil pot attached to the bottom, effectively we have a piston moving inside viscous oil and just like if you are walking through water in a swimming pool it takes a lot energy for the piston to move through the oil and therefore the motion is damped quicker.

Question 51

What are the 4 different types of damping?

Answer 51

1. Light damping - only small amount of energy is lost after every oscillation causing amplitude to get closer and closer to equilibrium.
2. Heavy damping - motion will move slightly beyond equilibrium then return to equilibrium in short amount of time.
3. Extremely heavy damping - the motion gradually moves towards the equilibrium without moving beyond it creating a relatively smooth line towards x = 0.
4. Critical damping - the motion moves back to equilibrium point as fast as possible and never moves beyond it.