Model answers - waves

Question 1

Define amplitude

Answer 1

Maximum displacement from equilibrium position

Question 2

Answer 2

The amplitude of all positions is 0.2m.
o This is because amplitude is the maximum displacement from the equilibrium position
o All of these particles will oscillate up to the maximum displacement of 0.2m as the wave propagates through space

Question 3

Define frequency

Answer 3

Number of complete oscillations per unit time

Question 4

Define period

Answer 4

Time taken for one complete oscillation

Question 5

Define wavelength

Answer 5

Distance between one point on a wave and a point with the same phase on the next wave

Question 6

Define longitudinal wave

Answer 6

where the oscillations of the molecules are parallel to the direction of energy transfer/wave propagation, producing compressions and rarefactions

Question 7

Define transverse wave

Answer 7

oscillations of the molecules/fields are perpendicular to the direction of energy transfer/wave propagation

Question 8

Describe how a displacement – distance graph is made and what it shows

Answer 8

this shows a snapshot of the wave acting on all particles at one instant in time

Question 9

Outline what quantities can and cannot be determined from a displacement – distance graph

Answer 9

We can find the wavelength and amplitude
o Cannot find time period and therefore cannot determine frequency or wave speed

Question 10

Describe how a displacement – time graph is made and what it shows

Answer 10

this shows a single particle’s oscillation over time

Question 11

11. Outline what quantities can and cannot be determined from a displacement – time graph

Answer 11

We can find the time period, T and amplitude
o And hence can calculate the frequency using f = 1/T
o We cannot find the wave speed with this graph alone

Question 12

What does wave propagation direction mean

Answer 12

This is the direction of energy transfer / wave travel

Question 13

Define compression and rarefaction

Answer 13

Compression is a position of maximum pressure
o Rarefaction is a position of minimum pressure

Question 14

Answer 14

Displacement to the right has been regarded as positive, and to the left as negative
o This can be seen from the region on the left hand side of the diagram. From the rarefaction to the compression, all of the particles have been displaced to the right.
o This has been drawn as positive displacement on the displacement, distance graph.

Question 15

State the displacement of a compression / rarefaction

Answer 15

Both compressions and rarefactions are positions of zero displacement

Question 16

Explain what happens to cause a compression and rarefaction at any instant in time and explain how this changes with time

Answer 16

Particles surrounding a rarefaction have been displaced away from the rarefaction
o Particles surrounding a compression have been displaced towards the compression
o The position of the compression and the position of rarefaction propagate with the wave ie. Different particles experience the compression as the wave propagates through space

Question 17

Explain how the pulse-echo technique can be used to determine the distance to an object

Answer 17

A pulse is sent out. When the medium’s density changes, the pulse is partially reflected as at the boundary
b. The reflected wave is picked up and the time between the emission and return is recorded.
c. If the speed in the medium is known, use distance = time for pulse to return x speed / 2
d. The division by 2 is needed as time between emission and reflected wave returning is time taken to travel twice the distance

Question 18

Explain why the pulse length must be less than the time taken for the pulse to return and explain the condition placed on the time between pulses

Answer 18

The pulse length is the time that the pulse lasts for. This must be less than the time taken for the pulse to return after being reflected so that the reflected pulse does not overlap with the emitted pulse.
o The time between each pulse must be long enough that the reflected pulse does not overlap with the emitted pulse.

Question 19

17. A bat approaches a moth. Explain why the time between the pulses sent out by the bat should decrease.

Answer 19

As the distance between the bat and moth decreases, the time taken between emission and receiving the reflected pulse decreases
o So the time between pulses should decrease
o This allows the bat to detect small changes in the moth’s position or speed and have more frequent updates

Question 20

Explain why the wavelength used in the pulse echo technique should be low

Answer 20

Using a low wavelength reduces the diffraction effects
o Resulting in higher resolution images
o And greater detail

Question 21

Define phase

Answer 21

o the fraction of the wave cycle that has been completed relative to the origin

Question 22

Answer 22

X phase = 90 degrees , y phase = O degrees
o X phase = 180 degrees, y phase = 90 degrees
o X phase = 270 degrees, y phase = 180 degrees
o X phase = 360 degrees, y phase = 270 degrees

Question 23

Answer 23

Answers: 1) A is 90 degrees ahead of B
o 2) B is 90 degrees ahead of A
o 3) A and B are in antiphase: phase difference = 180 degrees
o 4) A and B are in phase: phase difference = 0 degrees

Question 24

Answer 24

Answers: 1) B is 90 degrees ahead of A
o 2) A is 90 degrees ahead of B
o 3) A and B are in antiphase: phase difference = 180 degrees (B is ahead)
o 4) A and B are in phase: phase difference = 0 degrees

Question 25

Define path difference

Answer 25

o the difference in the distance travelled by two waves to a single point.

Question 26

24. State the relationship between phase difference, time difference and path difference between two points

Answer 26

Imagine two points on a wave. Their phase difference is Δθ, path difference is Δs, time difference between two points is Δt
o Each of those differences can be expressed as a fraction of the whole wave:
o Δθ / 2π = Δs /λ = Δt/ T

Question 27

Describe what an oscilloscope graph is plotting

Answer 27

An oscilloscope plots the potential difference against time

Question 28

26. Explain how to determine the frequency of a signal using an oscilloscope

Answer 28

Measure the number of divisions along the x axis for one full wave cycle, N
o Time period, T = N x timebase
o Frequency, f = 1/T

Question 29

Answer 29

Use a metre rule to record the position of the microphone when signals are in antiphase
o Move the microphone back until the signals are in antiphase again
o Record new position of microphone and calculate distance moved by microphone – this is the wavelength
o (Repeat this and calculate a mean)
o Calculate time period = number of divisions for one full wave cycle x timebase
o (ensure that timebase has been set to its minimum value for one full wave cycle to be seen – this increases the resolution to reduce the absolute uncertainty in the measurement of T)
o Determine the frequency of the signal using f = 1/T
o Wave speed = frequency x wavelength
o (OR - repeat steps a-d for a different frequency
o Plot a graph where the y-axis is the wavelength (m) and the x-axis is 1/frequency (s)
o This will produce a straight line because c = λf can be rearranged to
▪ λ = c x 1/f
▪ y = m x
▪ The gradient will therefore be a constant – the speed of sound m = c
o Calculate the gradient of the line of best fit and determine the speed of sound )

Question 30

Explain why the rule is appropriate instrument to measure length

Answer 30

The rule is appropriate as its resolution is 1mm and the average reading will be no smaller than 10cm so this gives a % uncertainty of 0.05/100 x 100 = 0.05 % which is very low.

Question 31

Define coherence

Answer 31

when two waves have a constant phase relationship and the same frequency

Question 32

0. Define superposition

Answer 32

when two or more waves meet at a point the resulting displacement is equal to the vector sum of the individual displacements

Question 33

31. Define interference

Answer 33

when two coherent waves meet at a point and undergo superposition – the resultant amplitude is equal to the vector sum of the individual amplitudes

Question 34

2. Explain why coherence is required for constructive or destructive interference to produce a constant amplitude oscillation of a particle

Answer 34

When waves meet at a point and are coherent, their phase difference will be constant at that point
o Constructive interference requires a consistent phase difference of 0
o Destructive interference requires a consistent phase difference of 180 degrees
o If the phase difference were to change over time then the type of superposition occurring would change over time and the particle would not have a constant amplitude oscillation

Question 35

Define constructive interference

Answer 35

o when the resultant amplitude is maximum (as the waves meet in phase)

Question 36

34. Define destructive interference

Answer 36

when the resultant amplitude is zero (as the waves meet pi out of phase)

Question 37

35. Explain how a standing wave is set up

Answer 37

Note that when explaining this in a particular context, you should refer specifically to the contextual information!)​
o Two waves travelling in opposite directions
o Of equal frequency and similar amplitude
o Undergo superposition when they meet
o Constructive interference produces antinodes – positions of maximum amplitude
o Where waves meet in phase
o Destructive interference produces nodes – positions of zero amplitude
o Where waves meet in antiphase

Question 38

Define node

Answer 38

Positions of zero amplitude

Question 39

Define anti node

Answer 39

Position of maximum amplitude

Question 40

Explain why the antinode is defined as a position of maximum amplitude and not position of maximum displacement

Answer 40

The antinode oscillates through zero displacement in its cycle
o So it does not always have the maximum displacment
o It does however have the maximum amplitude – its maximum displacement is higher than any other particle on the standing wave

Question 41

State the conditions for a standing wave to be set up between a speaker and a wall

Answer 41

The emitted and reflected wave must have similar amplitudes in order to completely destructively interfere and produce nodes
o So the wall should not absorb the wave
o The distance between the speaker and wall must be a whole number of half wavelengths (as there must be a node at the wall)
o There must not be multiple reflections from other parts of the room

Question 42

State the distance between nodes in terms of the wavelength

Answer 42

Distance between nodes = λ/2

Question 43

41. State the distance between antinodes in terms of the wavelength

Answer 43

Distance between antinodes = λ/2

Question 44

Draw the standing wave pattern for the first, second and third harmonic for a wave on a a string (two closed ends)

Answer 44

Question 45

State and justify the equation for the wavelength in each of the harmonics for the wave on the string. Explain how the frequencies compare to the fundamental.

Answer 45

o L = λ/2 so λ = 2L. This is the 1st harmonic (fundamental) with frequency f0
o L = λ. This is the 2nd harmonic with frequency 2f0. This is because the wavelength has been divided by 2 compared to the fundamental, so as frequency and wavelength are inversely proportional (as v=f λ and v is the same for each), frequency is multiplied by 2.
o L= 3λ/2 so λ = 2L/3. This is the 3rd harmonic with frequency 3f0

Question 46

Draw the displacement, distance standing wave graphs for the lowest three frequency standing wave in a pipe with one open end and one closed end

Answer 46

Question 47

State the equation for the wavelength and frequency of each of the standing wave patterns in the open ended close ended pipe

Answer 47

L = λ/4 so λ = 4L. This occurs at the fundamental, f0
o L= 3λ/4 so λ = 4L/3. This occurs at a frequency of 3f0. This is because the wavelength has been divided by 3 compared to the fundamental, so as frequency and wavelength are inversely proportional (as v=f λ and v is the same for each), frequency is multiplied by 3.
o L= 5 λ/4 so λ = 4L/5. This occurs at a frequency of 5f0

Question 48

Draw the standing wave pattern for the first, second and third harmonic for a standing wave in a tube with two open ends. State the equations relating the wavelength to the length of the tube in each case.

Answer 48

Question 49

State the equation for the wavelength and frequency of each of the standing wave patterns in the open ended open ended pipe

Answer 49

L = λ/2 so λ = 2L. This occurs at the fundamental, f0
o L= λ so λ = L. This occurs at a frequency of 2f0. This is because the wavelength has been divided by 2 compared to the fundamental, so as frequency and wavelength are inversely proportional (as v=f λ and v is the same for each), frequency is multiplied by 2.
o L= 3 λ/2 so λ = 2L/3. This occurs at a frequency of 3f0

Question 50

Answer 50

o A longitudinal standing wave is formed in the tube
o As the wave from the speaker is reflected from the end of the tube
o Where the waves meet in phase, an antinode is formed due to constructive interference
o The antinode is a position of maximum amplitude oscillations of the air – below these positions the powder will be pushed to either side
o Where the waves meet in antiphase, a node is formed due to destructive interference
o The piles form at the nodes – positions of zero amplitude - as the air molecules above these positions will not be oscillating and therefore the powder will settle here

Question 51

Answer 51

X and Y are in antiphase: phase difference = 180 degrees or π radians

X and Z are in phase: phase difference = 0 degrees or 0 radians

Question 52

State the general phase relationship between any given two points separated by n nodes on a standing wave

Answer 52

If n is even then the points will be in phase, phase difference = 0
o If n is odd then the points will be in antiphase, phase difference = 180 degrees

Question 53

Compare the amplitude variation for a travelling /progressive wave and a standing/stationary wave

Answer 53

Amplitude is the maximum displacement from the equilibrium position
o The amplitude is fixed for a travelling /progressive wave – each particle oscillates up to a maximum displacement that is the same as the particle next to it (assuming no energy is dissipated as the wave progresses through space)
o Whereas the amplitude varies from particle to particle for a standing wave
o The maximum amplitude is at the antinode
o The amplitude decreases moving from the amplitude towards a node, where there is zero amplitude

Question 54

52. A violin string is plucked. It produces a sound wave in the air. Compare the stationary wave produced on the string and the sound wave transmitted through the air.

Answer 54

The waves have the same frequency
o But different speeds - the speed of sound is 330m/s in air, whereas the speed of the stationary wave is v = where T is the tension on the string and μ is the mass per unit length
o The stationary wave has varying amplitude for each point along the string (maximum at the antinode falling to minimum at the node), whereas the sound wave has a fixed amplitude for all points
o The sound wave transfers energy whereas the stationary wave does not
o The wave on the string is transverse whereas the sound wave is longitudinal

Question 55

Answer 55

Once a standing wave in the first harmonic has been set up:
o Measure the distance between nodes of the first harmonic standing wave using a metre rule.
o Measure the mass attached to the end of the string using an electronic mass balance.
o Measure the frequency of the wave on the signal generator (or, more accurate – if oscilloscope is available then determine the time period on the oscilloscope trace (T = number of divisions for one full cycle x timebase, then use frequency = 1/T.)
o Change the tension in the wire by adding extra masses to the end of the wire
o Adjust the signal generator frequency until the fundamental frequency standing wave is observed again, keeping the length the same
o Determine the tension due to each value of the mass added by using T = mg where m is the mass and g is the acceleration due to gravity
o Plot a graph where the y-axis is the frequency2 , f2 (Hz2) and the x-axis is the period T (s)
o This will produce a straight line because can be rearranged to
o f2 x T
o y = m x
o The gradient will therefore be a constant – the speed of sound m =
o Calculate the gradient of the line of best fit and rearrange the equation to find the mass per unit length as
o Extra points:
o Note, this practical could also be done by changing the length of the wire between the notes and finding the fundamental for different lengths, keeping the tension the same.
o The mass balance is a suitable instrument because its resolution is 0.01g and typical masses used in this practical are 100g, giving a % uncertainty of 0.01% which is very low
o A metre rule is a suitable instrument because its resolution is 1mm and typical distances are 1m so this gives a % uncertainty of 0.1% which is very low.

Question 56

Answer 56

Question 57

Answer 57

Question 58

Answer 58