Immediately below this paragraph is a table
with links to the exam questions and answers for all three externals way back to 2013.
Focus on the exams for the last three years. Recognise the type of question e.g This is
a question about..... Know how you are going to set out your working when answering each type of question.
These are discussed at 13:40 of theDrPhysicsA video
At Level 3 we are interested in stringed instruments like guitars (where there is a node at each end), open pipes (where
there is an antinode at each end) and closed pipes (where there is a node at one end and an antinode at the other)
We are interested in the fundamental frequency (also called the first harmonic). This is the lowest frequency a standing
wave can have. For a stringed instrument or an open pipe this is the frequency associated with half a wavelength. For a
closed pipe this is the frequency associated with one quarter of a wavelength.
The second harmonic has twice the fundamental frequency. The third harmonic has three times the fundamental frequency,
and so on.
Q1a from the 2023 exam
v=fλ
v = 196 x 2 x 0.331
v = 130ms-2
Q1b from the 2023 exam
This is a merit question. it is easy
Waves - Doppler Effect
The Doppler effect is discussed at p59 of SciPad
Questions 1c and 1d from the 2023 exam actually explain the Doppler effect well
when we use them as worked examples
Q1c from the 2023 exam
Q1d from the 2023 exam.
This is the excellence question
Waves - Beats
Q2a from the 2023 exam
The string is longer so the wavelength is longer. As v is unchanged and v = fλ
, the fundamental frequency decreases.
Q2b from the 2023 exam
This is a merit question. it is easy
For Merit you need to describe the volume change and say in-phase is loud, out of phase is quiet
Beats are an example of interference. Constructive interference gives a loud beat.
The frequency of a beat is the difference between the two frequencies. So, in this question
if the violin string is out of tune and its first harmonic is 200Hz, the frequency of the beat will
be 200-196 = 4 beats per second.
Question 2c from the 2023 Exam
Together, all four parts of this question make up an excellence question
Sam hear a beat of 2.1Hz
(i) Determine the possible frequencies at which the string is vibrating
Answer: 196 ± 2.1 = 193.9Hz or 198.1Hz
She increases the speed of the wave along the string by increasing the tension in the string and
the beat frequency increases.
(ii) Use this information to determine the frequency at which the string was vibrating before
adjustment.
Answer: The length of the string has not changed
(i.e the wavelength is the same) so the increase in the speed of the wave means frequency has increased.
The increase in beat frequency
tells us that the string is vibrating further from 196 Hz, so the string
must have been at 198.1 Hz.
(iii) Explain what Sam must do to get the string to vibrate at 196 Hz
Answer: Sam must decrease the tension to get the string to 196 Hz.
(decrease v to lower f).
(iv) State how she will know when the string is vibrating at 196 Hz.
Answer:When the string is vibrating at exactly 196 Hz, there is no
beat. She will hear a steady 196 Hz tone.
Q2d from the 2023 exam
This is another excellence question
Watch
this youtube video
about the natural frequency of a system/object, forced oscillations, and resonance
Resonance occurs when the driving frequency is equal to the natural frequency.
And watch
this youtube video
for a demonstration of the interference patterns from Young's double slit experiment
Diffraction is the spreading out of waves when they pass through a gap
For your exam there is one triangle you need to know in order to work out either θ or the distance between the central antinode and another
antinode or node. This has one side (the adjacent) which is the distance from the 'gap' or source to the screen. It has length 'L'. The distance
'x' from the central antinode to whichever other antinode you are interested in is the 'opposite' side and tanθ is L/x. So you
can work out θ if you know 'L' and 'x' and you can work out 'x' if you know 'L' and θ. You will probably need to work out θ so you
can use it in the formula nλ = dsinθ
If 'x' is a lot smaller than 'L' (as happens when there are only 2-slits or two sources) then θ is small and sinθ ≈ tanθ,
so you can use x/L as an approximation for sinθ
With the diffraction grating the antinodes are further apart so 'x' is larger and the approximation sinθ ≈ x/L cannot be used.
Answer: Diffraction is the spreading out of waves as they go
through a gap (or bend around an obstacle).
Q3b from the 2023 exam
This is another excellence question
Answer: Waves from multiple slits overlap and superpose. If
the waves didn’t spread out, they wouldn’t overlap
and interfere.
The waves only arrive in phase and reinforce at the
places where the path difference from successive slits
is a whole number of wavelengths (nλ). The light is
bright in these places.
When n is not a whole number, even if the phase
difference, θ , between two adjacent slits is small, then
the phase difference between subsequent slits will be
increasing multiples of θ. With many sources there
will be waves from different slits arriving in antiphase
and the overall interference will cause many more
points of destructive interference resulting in a wide
dark region between the bright fringes
