Question

Two stationary sources A and B are sounding notes of frequency $680Hz$. An observer moves from A to B with a constant velocity u. If the speed of sound is $340m{s}^{-1}$, the value of u so that he hears 10 beats per second is $\frac{x}{2}m/s$. Find $x$ :

Answer 1

Given: Frequency of sources A and B is $f=680Hz$

Also velocity of sound in air $v_{sound}=340m/s$

Let $f^{\prime }$ and $f^{\prime \prime }$ be the frequencies heard by the observer from sources A and B respectively.

Doppler effect when observer moves towards the stationary source B:

$f^{\prime \prime }=f[\frac{v_{sound}+v_{observer}}{v_{sound}}]$

Thus $f^{\prime \prime }=680[\frac{340+u}{340}]$

Doppler effect when observer moves away from the stationary source A:

$f^{\prime }=f[\frac{v_{sound}-v_{observer}}{v_{sound}}]$

Thus, $f^{\prime }=680[\frac{340-u}{340}]$

Now, $f^{\prime \prime }-f^{\prime }=10$ (Given)

$680[\frac{340+u}{340}]$ $-680[\frac{340-u}{340}]=10$

$⟹u=2.5m/s$

Thus, $x=5$