DP Physics (last assessment 2024)

Question 20N.2.HL.TZ0.4

Date	November 2020	Marks available	[Maximum mark: 12]	Reference code	20N.2.HL.TZ0.4
Level	HL	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Estimate, Explain, Predict	Question number	4	Adapted from	N/A

[Maximum mark: 12]

20N.2.HL.TZ0.4

Two loudspeakers, A and B, are driven in phase and with the same amplitude at a frequency of $850 Hz$ . Point P is located $22.5 m$ from A and $24.3 m$ from B. The speed of sound is $340 m s^{- 1}$ .

(a)

Deduce that a minimum intensity of sound is heard at P.

[4]

Markscheme

wavelength $= \frac{340}{850} = 0.40 « m »$ ✓

path difference $= 1.8 « m »$ ✓

$1.8 « m » = 4.5 λ$ OR $\frac{1.8}{0.20} = 9$ «half-wavelengths» ✓

waves meet in antiphase «at P»
OR
destructive interference/superposition «at P» ✓

Allow approach where path length is calculated in terms of number of wavelengths; along path A ( $56.25$ ) and
path B ( $60.75$ ) for MP2, hence path difference $4.5$ wavelengths for MP3

Examiners report

This was answered very well, with those not scoring full marks able to, at least, calculate the wavelength.

(b)

A microphone moves along the line from P to Q. PQ is normal to the line midway between the loudspeakers.

The intensity of sound is detected by the microphone. Predict the variation of detected intensity as the microphone moves from P to Q.

[2]

Markscheme

«equally spaced» maxima and minima ✓

a maximum at Q ✓

four «additional» maxima «between P and Q» ✓

Examiners report

Most candidates were able to score at least one mark by referring to a maximum at Q.

(c)

When both loudspeakers are operating, the intensity of sound recorded at Q is $I_{0}$ . Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound with unchanged amplitude and frequency. The intensity of sound recorded at Q changes to $I_{A}$ .

Estimate $\frac{I_{A}}{I_{0}}$ .

[2]

Markscheme

the amplitude of sound at Q is halved ✓
«intensity is proportional to amplitude squared hence» $\frac{I_{A}}{I_{0}} = \frac{1}{4}$ ✓

Examiners report

Most candidates earned 2 marks or nothing. A common answer was that intensity was 1/2 the original.

In another experiment, loudspeaker A is stationary and emits sound with a frequency of $850 Hz$ . The microphone is moving directly away from the loudspeaker with a constant speed $v$ . The frequency of sound recorded by the microphone is $845 Hz$ .

(d(i))

Explain why the frequency recorded by the microphone is lower than the frequency emitted by the loudspeaker.

[2]

Markscheme

speed of sound relative to the microphone is less ✓

wavelength unchanged «so frequency is lower»
OR
fewer waves recorded in unit time/per second «so frequency is lower» ✓

Examiners report

HL only. The majority of candidates answered this by describing the Doppler Effect for a moving source. Others reworded the question without adding any explanation. Correct explanations were rare.

(d(ii))

Calculate $v$ .

[2]

Markscheme

$845 = 850 \times \frac{340 - v}{340}$ ✓

$v = 2.00 « m s^{- 1} »$ ✓