An ambulance is moving in still air towards a mountain at constant speed vA. The frequency of the sound that the driver hears in the echo reflected from the mountain is 20% higher than the frequency emitted by the siren. Find the speed of the ambulance.
Givens: Speed of sound v = 330 m/s.
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In the figure below is shown the ambulance traveling towards the mountain at constant speed vA.
The source emitting the sound waves (the ambulance) is moving, and thus Doppler effect will occur. The frequency (pitch) perceived by the observer (the driver) will be different than that emitted by the source.
The general expression for Doppler shift is:
- |v|: is the speed of the waves with respect to the medium (air in this case).
- vm: is the speed of the medium with respect to Earth.
- vS: is the speed of the source emitting the waves.
- vR: is the speed of the observer (or listener).
- υ0: is the frequency of the waves emitted by the source.
The expression above has been derived assuming that the wavefronts, the source and the receiver are all moving in the same direction:
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In order to apply the general expression to a particular problem, we must first determine what signs we need to use for the various speeds in the equation.
In this problem we are assuming still air, therefore vm = 0. Moreover, the ambulance (driver) is both the emitter and the receiver, because the driver hears the echo of the sound emitted by the ambulance.
The speed of the source of the sound waves vS keeps its positive sign in the expression for the frequency because it is moving in the same direction than the wavefronts. However, we must change the sign of the receiver speed vR because the ambulance moves toward the mountain when the driver hears the echo, and therefore is moving in the opposite direction than the wavefronts. Making these changes, the expression for the frequency becomes:
And since the ambulance is both the emitter and the receiver we can substitute vS = vR = vA. Finally, the frequency perceived by the receiver is:
The frequency perceived by the driver is 20% higher than that emitted by the siren. The speed of sound is also known. Therefore, isolating and substituting the numerical data into the previous formula we get:
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