A train is traveling at 10.0 m/s as it approaches a tunnel in the smooth, vertic

Question

A train is traveling at 10.0 m/s as it approaches a tunnel in the smooth, vertical face of a rock cliff. The speed of sound is 343. m/s. The engineer sounds a whistle in the locomotive of fundamental frequency 300. Hz. At what frequency would a rider in the caboose hear the locomotive whistle, if there were no echo from the cliff wall? (a) At what frequency would the rider in the caboose hear the echo from the cliff wall, if the whistle at the locomotive had stopped emitting sound by the time the echo arrived? (b) What beat frequency would the rider in the caboose observe, if the whistle continued to sound as the echo arrived? (c) (d) Assuming the situation of part (c), what beat frequency would be observed by someone standing on the ground near the tracks as the caboose passes? At what frequency would the locomotive whistle be heard by someone standing on the ground near the tracks as the caboose passes? (e) (0 At what frequency would the bystander hear the echo from the cliff wall?

Explanation / Answer

Part a)

According to the doppler shift, for stationary source and approching observer, observed frequency is given by

fobserved = [(Vs+Vo)/Vs] * fsource

fobserved = [(343+10)/ 343]* 300

fobserved = 308.74 Hz

Part b)

The formula of echo is given as

S = V/ 2t

Putting the value of V, as we know that V = f*lambda

S = fo*lambda / 2t

fo = 2*St / lambda

Hence, the observed frequency will be doubled.

fo = 2*308.7

fo = 617.4 Hz

Part e)

As source and observer both are stationary the frequency will be the orignal and no shift will be observed.

Part f)

echo will be heard at the same rate.

V = 2s/t

fs = 2s/t*lambda

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