Two test pilots are in identical jets that produce a characteristic tone of 607

Question

Two test pilots are in identical jets that produce a characteristic tone of 607 Hz. One jet is stationary on the airstrip and the second is approaching the airstrip at a speed of 70.6% the speed of sound. The pilot of each jet listens to the sound produced by the other jet. Determine the following. (a) Which pilot hears a greater Doppler shift? the stationery pilot the moving pilot (b) the frequency heard by the pilot in the moving jet Hz (c) the frequency heard by the pilot in the stationary jet Hz

Explanation / Answer

The expression for the observed frequency in the case of Doppler's effect is -

f = [(c+ Vr) / c]*fo

where, c = velocity of sound

Vr = Velocity of receiver (it is negative, if moving towards the observer)

fo = emitted frequency of source.

(b) In this case -

f = [(c - 0.706c) / c]*607 = (0.294/ 1)*607 = 178.4 Hz.

(c) In this case the expression is -

f = [ c / (c + Vs)]*fo

where, Vs = velocity of source (negative if it is moving towards the observer)

So, f = [ c / (c + 0.706c)]*607 = 355.8 Hz.

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