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Question
A train approaching a platform at a speed of 54 km h−1 sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. the train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platform? The speed of sound in air = 332 m s−1.
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Solution
Given:
Speed of sound in air v= 332 ms−1
Velocity of train \[v_s\] = 54 kmh−1 =\[54 \times \frac{5}{18} \text{ m/s } = 15 \text { m/s }\]
Let
\[f_0\] be the original frequency of the train.
When the train approaches a platform, the frequency of sound heard by the observer \[\left( f \right)\] is given by :
\[f = \left( \frac{v}{v - v_s} \right) \times f_0\]
On substituting the values, we have:
\[1620 = \left( \frac{332}{332 - 15} \right) f_0 \]
\[ \Rightarrow f_0 = \left( \frac{1620 \times 317}{332} \right) \text { Hz } .\]
When the train crosses the platform, the frequency of sound heard by the observer \[\left( f_1 \right)\] is given by :
\[f_1 = \left( \frac{v}{v + v_s} \right) \times f_0\]
Substituting the respective values in the above formula, we have:
\[f_1 = \left( \frac{332}{332 + 15} \times \frac{1620 \times 317}{332} \right)\]
\[ = \frac{317}{347} \times 1620 = 1480 \text { Hz }\]
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