Question

A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350 m s${}^{-1}$.

What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

General expression for pressure is time for sound is

$P=P_{0}sin\omega (t-\frac{x}{v})$

$P=P_{0}sin(\omega t-\frac{\omega x}{v})$

If some phase $\Phi$ is present initially then that gets added up

$P=P_{0}sin(\omega t-\frac{\omega x}{v}+\Phi )$

Lets say $\Phi$ was 0 when particle was at $X_0$ at time to

So equation become $P=P_{0}sin(\omega t_0-\frac{\omega x_0}{v})$

Now the same equation for a particle at a point 10 cm from $x_0$ at same time to

$P=P_{0}sin(\omega t_0-\frac{\omega (x_0+0.1)}{v})$

$\Rightarrow P=P_{0}sin(\omega t_0-\frac{\omega x_0}{v}\frac{\omega 0.1}{v})$

So here phase is $\Phi =\frac{\omega 0.1}{v}$

Given f=100 Hz

$\omega =2\pi f=200\pi$

$v=350m/s$

$\Phi =\frac{200\pi \times 1\times {10}^{-1}}{350}=\frac{2\pi }{35}$

OR

$2\pi$

Phase change is given by $\Delta \Phi =kx=\frac{2\pi x}{\lambda }$

$\lambda =\frac{v}{f}$

$\frac{350}{100}=3.5m$

Therefore, $\Delta \Phi =\frac{2\pi \left(0.1\right)}{3.5}=\frac{2\pi }{35}$