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Question
A person riding a car moving at 72 km h−1 sound a whistle emitting a wave of frequency 1250 Hz. What frequency will be heard by another person standing on the road (a) in front of the car (b) behind the car? Speed of sound in air = 340 m s−1.
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Solution
Given:
Frequency of whistle \[f_0 = 1250 \text { Hz }\]
Velocity of car \[v_s\] = 72 kmh−1 =\[72 \times \frac{5}{18} = 20 {\text { ms }}^{- 1}\]
Speed of sound in air v = 340 ms−1
(a) When the car is approaching the person:
Frequency of sound heard by the person \[\left( f_1 \right)\] is given by :
\[f_1 = \left( \frac{v}{v - v_s} \right) \times f_0 \]
On substituting the given values in the above equation, we have:
\[f_1 = \frac{340}{340 - 20} \times 1250\]
\[ = 1328 \text { Hz }\]
(b) When the person is behind the car:
Frequency of sound heard by the person \[\left( f_2 \right)\] is given by :
\[f_2 = \left( \frac{v}{v + v_s} \right) \times f_0\]
On substituting the given values in the above equation, we have :
\[f_2 = \left( \frac{340}{340 + 20} \right) \times 1250\]
\[ = \frac{340}{360} \times 1250 = 1181 \text{ Hz }\]
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