Waves Ques 60
60. Two sound sources emitting sound each of wavelength $\lambda$ are fixed at a given distance apart. A listener moves with a velocity $u$ along the line joining the two sources. The number of beats heard by him per second is
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(a) $\frac{u}{2 \lambda}$
(b) $\frac{2 u}{\lambda}$
(c) $\frac{u}{\lambda}$
(d) $\frac{u}{3 \lambda}$
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Answer:
Correct Answer: 60.(b)
Solution:
- (b) Frequency received by listener from the rear source,
$n^{\prime}=\frac{v-u}{v} \times n=\frac{v-u}{v} \times \frac{v}{\lambda}=\frac{v-u}{\lambda}$
Frequency received by listener from the front source,
$n^{\prime \prime}=\frac{v+u}{v} \times \frac{v}{\lambda}=\frac{v+u}{\lambda}$
No. of beats $=n^{\prime \prime}-n^{\prime}$
$=\frac{v+u}{\lambda}-\frac{v-u}{\lambda}=\frac{v+u-v+u}{\lambda}=\frac{2 u}{\lambda}$
BETA
