| Date | May 2014 | Marks available | 4 | Reference code | 14M.2.HL.TZ1.4 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Sketch and State | Question number | 4 | Adapted from | N/A |
Question
This question is about sound.
A source emits sound of frequency \(f\). The source is moving towards a stationary observer at constant speed. The observer measures the frequency of the sound to be \(f'\).
(i) Explain, using a diagram, why \(f'\) is greater than \(f\).
(ii) The frequency \(f\) is 275 Hz. The source is moving at speed 20.0 ms–1. The speed of sound in air is 330 ms–1. Calculate the observed frequency \(f'\) of the sound.
A source of sound is placed in front of a barrier that has an opening of width comparable to the wavelength of the sound.
A sound detector is moved along the line XY. The centre of XY is marked O.
(i) On the axes below, sketch a graph to show how the intensity I of the sound varies as the detector moves from X to Y.
(ii) State the effect on the intensity pattern of increasing the wavelength of the sound.
(i) Outline the difference between a polarized wave and an unpolarized wave.
(ii) State why sound waves cannot be polarized.
Markscheme
(i) diagram showing (circular) wavefronts around source, so that wavefronts are closer together on side of observer;
speed of sound waves for observer is the same (as for stationary case) but observed wavelength is smaller;
since \(f' = \frac{v}{{\lambda '}}\), (observed frequency is larger);
(ii) \(f'\left( { = f\left[ {\frac{v}{{v - {u_s}}}} \right]} \right) = 275\left[ {\frac{{330}}{{330 - 20}}} \right]\);
=293(Hz);
Award [0] for use of moving observer formula.
Award [1] for use of v+us to give 259 (Hz).
Award [2] for a bald correct answer.
(i) central symmetrical maximum;
at least one secondary maximum on each side, no more than one third the height of the central maximum; { (judge by eye)
minima drawn to zero, ie touching axis;
width of the secondary maximum half the width of the primary maximum; { (judge by eye)
(ii) greater distance between maxima/minima / pattern more spread out;
(i) in a polarized wave, the oscillations/vibrations are in one direction/plane only;
in an unpolarized wave, the oscillations/vibrations are in all directions/ planes (perpendicular to the direction of energy transfer);
Must see mention of oscillations or vibrations in first or second marking point.
(ii) sound waves are longitudinal / the oscillations/vibrations are always parallel to direction of energy transfer;
Examiners report
ai) Many candidates scored the first mark for the diagram showing the wavefronts closer on the side of the observer but most of the written explanations just repeated this and didn’t expand further.
aii) This question was very well answered with the majority of candidates choosing the appropriate formula and evaluating correctly.
bi) Most candidates were able to score full marks on this question.
bii) Again this was answered successfully.
ci) Few candidates included the words oscillations or vibrations in their answers and consequently scored zero marks.
cii) Many recognized that sound waves are longitudinal and that is why they cannot be polarized.
