Wave Motion Ques 96
- A source of sound $S$ is moving with a velocity of $50 m / s$ towards a stationary observer. The observer measures the frequency of the source as $1000 Hz$. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take, velocity of sound in air is $350 \mathrm{~m} / \mathrm{s}$ )
(2019 Main, 10 April II)
(a) $807 \mathrm{~Hz}$
(b) $1143 \mathrm{~Hz}$
(c) $750 \mathrm{~Hz}$
(d) $857 \mathrm{~Hz}$
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Answer:
Correct Answer: 96.(c)
Solution:
- Initially,
When source is moving towards stationary observer, frequency observed is more than source frequency due to Doppler’s effect, it is given by
$$ f _{\text {observed }}=f\left(\frac{v}{v-v _s}\right) $$
where, $f=$ source frequency,
$$ f _o=\text { observed frequency }=1000 Hz \text {, } $$
$v=$ speed of sound in air $=350 ms^{-1}$
and $\quad v _s=$ speed of source $=50 ms^{-1}$
So, $\quad f=\frac{f _{obs}\left(v-v _s\right)}{v}=\frac{1000(350-50)}{350}=\frac{6000}{7} Hz$
When source moves away from stationary observer, observed frequency will be lower due to Doppler’s effect and it is given by
$$ \begin{aligned} f _0 & =f\left(\frac{v}{v+v _s}\right)=\frac{6000 \times 350}{7 \times(350+50)} \\ & =\frac{6000 \times 350}{7 \times 400}=750 Hz \end{aligned} $$
BETA
