The graph shows the variation with time of the displacement X of a gas molecule as a continuous sound wave passes through a gas.

Question 9

The graph shows the variation with time of the displacement X of a gas molecule as a continuous sound wave passes through a gas.

The velocity of sound in the gas is 330 m s^-1. All the graphs below have the same zero time as the graph above.

What is the displacement-time graph for a molecule that is a distance of 0.165 m further away from the source of the sound?

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q25

Solution:

Answer: D.

From the graph, we can find the period of the sound wave.

Period T = 500 μs = 500×10^-6 s

Frequency f = 1 / T = 1 / (500×10^-6) = 2000 Hz

Speed v of sound in gas = 330 m s^-1

v = fλ

Wavelength λ = v / f = 330 / 2000 = 0165 m

The other molecule is at a distance of 0.165 m further from the source of sound. This means that it is at a distance equal to the wavelength from the source.

 

Since the path difference is equal to the wavelength, both molecules are in phase. So, the displacement-time graphs are similar.