Question

Two soucres $P$ and $Q$ produce notes of frequency $660Hz$ each. A listener moves from $P$ to $Q$ with a speed of $1m{s}^{-1}$. If The speed of sound is $330m/s$, then number of beats heard by the listener per second will be :

• A. 4
• B. 8
• C. 2
• D. zero

Answer 1

Answer: (A)

The situation as shown in the figure.

Here,
Speed of listener, $v_L=1m{s}^{-1}$
Speed of sound, $v=330m{s}^{-1}$
Frequency of each source, $v=660Hz$
Apparent frequency due to $P$,
$v^{\prime }=\frac{v\left(v-v_L\right)}{v}$
Apparent frequency due to $Q$,
$v^{\prime \prime }=\frac{v\left(v+v_L\right)}{v}$
Number of beats heard by the listener per second is
$v^{\prime \prime }-v^{\prime }=\frac{v\left(v+v_L\right)}{v}-v\frac{\left(v-v_L\right)}{v}$
$=\frac{2v{v}_L}{v}=\frac{2\times 660\times 1}{330}=4$