Question
Sound of frequency f = 2500 Hz is emitted from an aircraft that moves with speed v = 280 ms–1 away from a stationary observer.
The speed of sound in still air is c = 340 ms–1.
frequency heard by the observer. [2]
wavelength measured by the observer. [1]
Answer/Explanation
Ans
i
f’ = 2500 × \(\frac{340}{340+280}\)
f ′ = 1371 ≈1400 «Hz »
ii λ = \(\frac{340}{1371}\) ≈ 0.24 / 0.25 «m »
This question is about waves.
The diagram represents a standing (stationary) wave in air in a pipe which is open at both ends.
Two points in the pipe are labelled P and Q.
a.
(i) State the direction of oscillation of an air molecule at point P.
(ii) Compare the amplitude of oscillation of an air molecule at point P with that of an air molecule at point Q.
A hollow pipe open at both ends is suspended just above the ground on a construction site.
Wind blows across one end of the pipe. This causes a standing wave to form in the air of the pipe, producing the first harmonic. The pipe has a length of 2.1 m and the speed of sound in air is \(330{\text{ m}}\,{{\text{s}}^{ – 1}}\).
Estimate the frequency of the first harmonic standing wave.
The pipe is held stationary by the crane and an observer runs towards the pipe. Outline how the frequency of the sound measured by the observer is different from the frequency of the sound emitted by the pipe.
Answer/Explanation
Markscheme
a.
(i) along axis of pipe / horizontal / left to right;
Do not allow “right” or “left” alone. Allow correct answers that appear on the diagram.
(ii) P has greater/maximum amplitude;
Q does not move/has zero amplitude;
\(\lambda = 4.2{\text{ m}}\);
79 or 78.6 (Hz); (do not allow 78 Hz)
Award [2] for a bald correct answer.
wavefronts emitted by the pipe at regular time intervals;
observer crosses the wavefronts more often than if stationary;
so frequency is higher;
Award [0] for a bald correct answer.
or
observer moving towards the source measures the emitted wavelength;
but (relative) wave speed measured is higher than that emitted;
so frequency is higher;
Award [0] for a bald correct answer.
or
quotes Doppler equation correctly;
deduces higher frequency correctly;
Award [2 max] for this approach.
Examiners report
a.
(i) Most candidates were imprecise in their descriptions of the direction of motion of the air molecule. Very many think that the conventional diagram for the standing sound wave in a gas shows that the molecules move between the drawn lines of the wave. Answers such as “vertical” were common. Other inadequate responses featured “to the right” without the candidate realising that this implies that a molecule moves continually to the right eventually leaving the tube.
(ii) Descriptions of the amplitudes at P and Q were better but for a minority of candidates there was confusion between the meaning of node and antinode.
Calculations were generally well done.
This question was an “outline” and required a more sophisticated approach than “it is caused by the Doppler effect”. Examiners were looking for a description of the causes of the change in wavelength as perceived by the moving observer. Explanations that quoted a Doppler equation (it had to be the correct one) and deduced an increase in frequency were not awarded full marks as this is little more than copying by rote from the data booklet.
Part 2 The Doppler effect and optical resolution
The Doppler effect can be used to deduce that a particular star X is moving towards Earth.
f.
Describe what is meant by the Doppler effect.
One of the lines in the spectrum of atomic hydrogen has a frequency of 4.6×1016Hz as measured in the laboratory. The same line in the spectrum of star X is observed on Earth to be shifted by 1.3×1012Hz.
(i) State the direction of the observed frequency shift.
(ii) Determine the speed at which X is moving towards Earth stating any assumption that you have made.
The star X has a companion star Y. The distance from Earth to the stars is 1.0×1018m. The images of X and Y are just resolved according to the Rayleigh criterion by a telescope on Earth with a circular eyepiece lens of diameter 5.0×10–2m.
(i) State what is meant by the statement “just resolved according to the Rayleigh criterion”.
(ii) The average wavelength of the light emitted by the stars is 4.8×10–7m. Determine the separation of X and Y.
Answer/Explanation
Markscheme
f.
the observed change in frequency/wavelength of a wave;
emitted by a source moving away from or towards/relative to the observer;
(i) a blue-shift / towards the blue end of the spectrum / to a higher frequency / OWTTE;
(ii) \(v = \left( {\frac{{c\Delta f}}{f} = } \right)\frac{{3 \times {{10}^8} \times 1.3 \times {{10}^{12}}}}{{4.6 \times {{10}^{16}}}}\);
8.5×103ms-1;
assume that the speed is very much less than speed of light;
(i) the two stars are (just) seen as separate images;
if the central maximum of the diffraction image of one star coincides with the first minimum of the diffraction image of the other star / OWTTE;
Accept an appropriate diagram for second marking point.
(ii) \(\theta = \left( {\frac{{1.22\lambda }}{b} = } \right)\frac{{1.22 \times 4.8 \times {{10}^{ – 7}}}}{{5.0 \times {{10}^{ – 2}}}}\) or 1.17×10-5rad;
\(\theta = \frac{d}{{1.0 \times {{10}^{18}}}}\);
(d=)1.2×1013m;
Award [2 max] if 1.22 is missing, giving an answer of 0.98×1013.