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Question 18M.2.HL.TZ2.3a.iv

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Date May 2018 Marks available [Maximum mark: 2] Reference code 18M.2.HL.TZ2.3a.iv
Level HL Paper 2 Time zone TZ2
Command term Calculate Question number a.iv Adapted from N/A
a.iv.
[Maximum mark: 2]
18M.2.HL.TZ2.3a.iv

A loudspeaker emits sound towards the open end of a pipe. The other end is closed. A standing wave is formed in the pipe. The diagram represents the displacement of molecules of air in the pipe at an instant of time.

X and Y represent the equilibrium positions of two air molecules in the pipe. The arrow represents the velocity of the molecule at Y.

(a.iv)

The speed of sound is 340 m s–1 and the length of the pipe is 0.30 m. Calculate, in Hz, the frequency of the sound.

[2]

Markscheme

wavelength is λ = « 4 × 0.30 3 =» 0.40 «m»

f = « 340 0.40 » 850 «Hz»

 

Award [2] for a bald correct answer

Allow ECF from MP1

[2 marks]

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