Wave Motion 4 Question 1
1. Two sources of sound $S _1$ and $S _2$ produce sound waves of same frequency $660 Hz$. A listener is moving from source $S _1$ towards $S _2$ with a constant speed $u m / s$ and he hears 10 beats/s. The velocity of sound is $330 m / s$. Then, $u$ equal to
(2019 Main, 12 April II)
(a) $5.5 m / s$
(b) $15.0 m / s$
(c) $2.5 m / s$
(d) $10.0 m / s$
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Answer:
Correct Answer: 1. (c)
Solution:
- When observer moves away from $S _1$ and towards $S _2$,
then due to Doppler’s effect observed frequencies of sources by observer are
$$ f _{S _1}^{\prime}=\frac{v-v _o}{v} \cdot f _{S _1} $$
(observer moving away from source) and
$$ f _{S _2}^{\prime}=\left(\frac{v+v _o}{v}\right) \cdot f _{S _2} $$
(observer moving towards source)
(where, $v=$ speed to sound, $v _o=$ speed of observer)
So, beat frequency heard by observer is
Here,
$$ \begin{aligned} & f _b=f _{S _2}^{\prime}-f _{S _1}^{\prime} \\ & v _o=u, v=330 ms^{-1} \\ & f _b=10 Hz, f _{S _1}=f _{S _2}=660 Hz \end{aligned} $$
On putting the values, we get
$$ \begin{aligned} & f _b=f _{S _2}^{\prime}-f _{S _1}^{\prime} \\ &=\left(\frac{v+v _o}{v}\right) \cdot f _{S _2}-\left(\frac{v-v _o}{v}\right) f _{S _1} \\ &=f _{S _1}{\frac{v+v _o}{v}-\frac{v-v _o}{v} }=f _{S _1} \cdot \frac{2 v _o}{v} \\ & \Rightarrow \quad \quad \quad\left[\because v _o=u\right] \\ & \Rightarrow \quad 10=\frac{660 \times 2 u}{330} \quad u=2.5 ms^{-1} \end{aligned} $$
