Question

A source of sound having frequency $300 \, Hz$ and a receiver are located along the same line normal to the wall as shown in the figure. Both the source and the receiver are stationary and the wall recedes from the source with a velocity of $20 \, m \, s^{- 1}$ . If the beat frequency registered by the receiver is $\frac{240}{x} \, Hz$ , then what is the value of $x$ ? [velocity of sound = $330 \, m \, s^{- 1}$ ]

Answer 1

As ; $f_{1}=f$ (for direct sound)
Now for reflected sound $f_{2}=\left(\frac{V - 20}{V + 20}\right)f$
If $b$ is the beat frequency,

$\therefore b=f_{1}-f_{2}$
$\therefore f-\left(\frac{V - 20}{V + 20}\right)f=\frac{f \times 40}{V + 20}$
$=\frac{300 \times 40}{350}=\frac{240}{7}Hz$