### Waves - Result Question 61

#### 63. 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) $\frac{u}{3 \lambda}$

## Show Answer

#### Answer:

Correct Answer: 63. (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 $=n^{\prime \prime}-n^{\prime}$

$=\frac{v+u}{\lambda}-\frac{v-u}{\lambda}=\frac{v+u-v+u}{\lambda}=\frac{2 u}{\lambda}$