Wave Motion 5 Question 26
26. A train approaching a hill at a speed of $40 km / h$ sounds a whistle of frequency $580 Hz$ when it is at a distance of $1 km$ from a hill. A wind with a speed of $40 km / h$ is blowing in the direction of motion of the train. Find
$(1988,5 M)$
(a) the frequency of the whistle as heard by an observer on the hill,
(b) the distance from the hill at which the echo from the hill is heard by the driver and its frequency.
(Velocity of sound in air $=1200 km / h$ )
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Answer:
Correct Answer: 26. (a) $599.33 Hz \quad$ (b) $0.935 km, 621.43 Hz$
Solution:
- Given $v _s=v _w=40 km / h$ and $v=1200 km / h=$ speed of sound.
(a) Frequency observed by observer $f^{\prime}=f\left(\frac{v+v _w}{v+v _w-v _s}\right)$
Substituting the values, we have
$$ f^{\prime}=580\left[\frac{1200+40}{1200+40-40}\right]=599.33 Hz $$
(b) Let $x$ be the distance of the source from the hill at which echo is heard of the sound which was produced when source was at a distance $1 km$ from the hill. Then, time taken by the source to reach from $s$ to $s^{\prime}=$ time taken by the sound to reach from $s$ to hill and then from hill to $s^{\prime}$. Thus,
$$ \frac{1-x}{40}=\frac{1}{1200+40}+\frac{x}{1200-40} $$
Solving this equation, we get $x=0.935 km$
Frequency heard by the driver of the reflected wave
$f^{\prime \prime}=f\left[\frac{v-v _w+v _o}{v-v _w-v _{s^{\prime}}}\right]=580\left[\frac{1200-40+40}{1200-40-40}\right]=621.43 Hz$
