Date | May 2022 | Marks available | 3 | Reference code | 22M.2.SL.TZ2.3 |
Level | Standard level | Paper | Paper 2 | Time zone | 2 |
Command term | Explain | Question number | 3 | Adapted from | N/A |
Question
A loudspeaker emits sound waves of frequency towards a metal plate that reflects the waves. A small microphone is moved along the line from the metal plate to the loudspeaker. The intensity of sound detected at the microphone as it moves varies regularly between maximum and minimum values.
The speed of sound in air is 340 m s−1.
Explain the variation in intensity.
Adjacent minima are separated by a distance of 0.12 m. Calculate .
The metal plate is replaced by a wooden plate that reflects a lower intensity sound wave than the metal plate.
State and explain the differences between the sound intensities detected by the same microphone with the metal plate and the wooden plate.
Markscheme
«incident and reflected» waves superpose/interfere/combine ✓
«that leads to» standing waves formed OR nodes and antinodes present ✓
at antinodes / maxima there is maximum intensity / constructive interference / «displacement» addition / louder sound ✓
at nodes / minima there is minimum intensity / destructive interference / «displacement» cancellation / quieter sound ✓
OWTTE
Allow a sketch of a standing wave for MP2
Allow a correct reference to path or phase differences to identify constructive / destructive interference
wavelength = 0.24 «m» ✓
= «=» 1.4 «kHz» OR 1400 «Hz» ✓
Allow ECF from MP1
relates intensity to amplitude ✓
antinodes / maximum intensity will be decreased / quieter ✓
nodes / minimum will be increased / louder ✓
difference in intensities will be less ✓
maxima and minima are at the same positions ✓
OWTTE
Examiners report
ai) On most occasions it looked like students knew more than they could successfully communicate. Lots of answers talked about interference between the 2 waves, or standing waves being produced but did not go on to add detail. Candidates should take note of how many marks the question part is worth and attempt a structure of the answer that accounts for that. At SL there were problems recognizing a standard question requiring the typical explanation of how a standing wave is established.
3aii) By far the most common answer was 2800 Hz, not doubling the value given to get the correct wavelength. That might suggest that some students misinterpreted adjacent minima as two troughs, therefore missing to use the information to correctly determine the wavelength as 0.24 m.
b) A question that turned out to be a good high level discriminator. Most candidates went for an answer that generally had everything at a lower intensity and didn't pick up on the relative amount of superposition. Those that did answer it very well, with very clear explanations, succeeded in recognizing that the nodes would be louder and the anti-nodes would be quieter than before.