In a car, sound signals emitted by the two cars are detected by the detector on the straight track at the end point of the race. Frequencies observed are 330 Hz and 360 Hz and the original frequency is 300 Hz of both the cars. The race ends with the separation of 100m between the cars. Assume that both cars move with constant velocities and the velocity of sound is 330m/s. Find the time taken (in sec) by the wining car

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answer is 4.

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Detailed Solution

Let the velocity of cars 1 and 2 be $v_{1} m / s$ and $v_{2} m / s$ respectively
$∴$ apparent frequencies of sound emitted by car 1 and 2 as detected at end point are
$f_{1} = f_{0} (\frac{v}{v - v_{1}}), f_{2} = f_{0} (\frac{v}{v - v_{2}})$
$330 = 300 (\frac{300}{300 - v_{1}})$ and $360 = 300 (\frac{330}{330 - v_{2}})$
Solving above equation we get
$v_{1} = 30 m / s$ and $v_{2} = 55 m / s$
The distance between both the cars just when the 2nd car reaches end point B
$100 m = v_{2} t - v_{1} t$ or $t = 4 \sec$