DP Physics (last assessment 2024)

Question 20N.2.SL.TZ0.4a

Date	November 2020	Marks available	[Maximum mark: 4]	Reference code	20N.2.SL.TZ0.4a
Level	SL	Paper	2	Time zone	TZ0
Command term	Deduce	Question number	a	Adapted from	N/A

[Maximum mark: 4]

20N.2.SL.TZ0.4a

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

Syllabus sections

Core

Core » Topic 4: Waves » 4.2 – Travelling waves

Core » Topic 4: Waves » 4.4 – Wave behaviour

Core » Topic 4: Waves

Question 20N.2.SL.TZ0.4a

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