Wave Motion Ques 101
- A stationary source emits sound waves of frequency $500 Hz$. Two observers moving along a line passing through the source detect sound to be of frequencies $480 Hz$ and $530 Hz$. Their respective speeds are in $ms^{-1}$,
(Take, speed of sound $=300 m / s$ )
(2019 Main, 10 April I)
(a) 12,16
(b) 12,18
(c) 16,14
(d) 8,18
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Answer:
Correct Answer: 101.(b)
Solution:
- Given,
Frequency of sound source $\left(f _0\right)=500 Hz$ Apparent frequency heard by observer $1, f _1=480 Hz$ and apparent frequency heard by observer $2, f _2=530 Hz$.
Let $v _0$ be the speed of sound.
When observer moves away from the source, Apparent
frequency, $f _1=f _0\left(\frac{v-v _0^{\prime}}{v}\right)\cdots(i)$
When observer moves towards the source,
Apparent frequency, $f _2=f _0\left(\frac{v+v _0^{\prime \prime}}{v}\right)\cdots(ii)$
Substituting values in Eq. (i), we get
$$ 480=500\left(\frac{300-v _0^{\prime}}{300}\right) $$
$$ \begin{aligned} \Rightarrow & 96 \times 3 & =300-v _0^{\prime} \\ \Rightarrow & v _0^{\prime} & =12 m / s \end{aligned} $$
Substituting values in Eq. (ii), we get
$$ \begin{aligned} 530 & =500\left(\frac{330+v^{\prime \prime}}{300}\right) \\ \Rightarrow \quad 106 \times 3 & =300+v^{\prime \prime} \\ \Rightarrow \quad v _0^{\prime \prime} & =18 m / s \end{aligned} $$
Thus, their respective speeds (in $m / s$ ) is 12 and 18 .
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