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Question
Two electric trains run at the same speed of 72 km h−1 along the same track and in the same direction with separation of 2.4 km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500 Hz and the speed of sound in air is 340 m s−1, find the frequencies heard by the person.
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Solution
Given:
Speed of sound in air v = 340 ms−1
Frequency of whistles \[f_0\]= 500 Hz
Speed of train \[v_s\]= 72 km/h =\[72 \times \frac{5}{18} = 20 \text { m/s }\]
The person will receive the sound in a direction that makes an angle θ with the track. The angle θ is given by :
\[\theta = \tan^{- 1} \left( \frac{0 . 5}{2 . 4/2} \right) = 22 . 62^\circ\]
The velocity of the source will be 'v cos θ' when heard by the observer.
So, the apparent frequency received by the man from train A is
\[f_1 = \left( \frac{v}{v - v_s \cos\theta} \right) \times f_0 \]
\[ \Rightarrow f_1 = \left( \frac{340}{340 - v_s \cos 22 . {62}^\circ} \right) \times 500\]
\[ \Rightarrow f_1 = \left( \frac{340}{340 - 20 \times \cos 22 . 62^\circ} \right) \times 500\]
\[ \Rightarrow f_1 = 528 . 70 \text{ Hz } \approx 529 \text { Hz }\]
The apparent frequency heard by the man from train B is
\[f_2 = \left( \frac{v}{v + v\cos\theta} \right) \times f_0 \]
\[ \Rightarrow f_2 = \left( \frac{340}{340 + 20 \times \cos 22 . 62^\circ} \right) \times 500\]
\[ \Rightarrow f_2 = 474 . 24 \text { Hz } \approx 474 \text { Hz }\]
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