Question

Two soucres $P$ and $Q$ produce notes of frequency $660 \,Hz$ each. A listener moves from $P$ to $Q$ with a speed of $1 \,ms^{-1}$. If The speed of sound is $330\, m/s$, 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, $v_{L}=1 \,ms ^{-1}$
Speed of sound, $v=330 \,ms ^{-1}$
Frequency of each source, $v=660\, Hz$
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} $
$=\frac{2 v v_{L}}{v}=\frac{2 \times 660 \times 1}{330}=4$