Question

A sound wave of frequency 10 kHz is travellilng in air with a speed of `340 ms^-1`. Find the minimum separation between two points where the phase difference is `60^@`.

Answer 1

The wavelength of the wave is
`lamda=v/v=(340ms^-1)/(10xx10^3s^-1)=3.4cm`
The wave number is `k=(2pi)/lamda=(2pi)/3.4cm^-1`
The phase of the wave is `(kx-omegat)`. At any given instant the phase difference between two points at separatin d is kd. If this phase difference is `60^@` i.e. pi/3` radian,
`pi/3=((2pi)/3.4cm^-1)d or d=3.4/6cm=0.57cm.