Two sources of sound $A$ and $B$ are moving towards and away from a stationary observer with the same speed respectively. If the frequency of the sound produced by both the sources is equal to $400 \mathrm{Hz}$, then the speed of sources (approximately) if the observer detects 4 beats per second, is [Given, speed of sound $=340 \mathrm{m} {\mathrm{s}}^{-1}$]

Given, frequency of sound produced by sources $A$ and $B$, $f_A=f_B=400 \mathrm{Hz}$

Speed of sound in air, $v=340 \mathrm{m} {\mathrm{s}}^{-1}$

Apparent frequency heard by the observer when source $A$ is moving towards him,

$f_A^{\text{'}}=\frac{v}{v-v_S}\times f_A=\frac{340}{340-v_S}\times 400$

Apparent frequency heard by the observer when source $B$ is moving away from him,

$f_B^{\text{'}}=\frac{v}{v+v_s}\times f_B=\frac{340}{340+v_s}\times 400$

Given, Beats $=f_A^{\text{'}}-f_B^{\text{'}}=4$

$\frac{340}{340-v_s}\times 400-\frac{340}{340+v_s}\times 400=4$

$\Rightarrow \frac{34000\times 2v_S}{{340}^2-v_S^2}=1$

$\frac{34000\times 2v_S}{{340}^2}=1$   $\left[\because v_S^2<<{340}^2\right]$

$\Rightarrow v_s=1.7 \mathrm{m} {\mathrm{s}}^{-1}$