Set 56 Problem number 20

Problem

A detector is initially 10 meters from a sound source and moving toward it at 37 m/s.  The source emits a sound at a constant frequency of 240 Hz.  Sound ceases when the detector reaches the source.

• How many pulses are emitted as the source approaches the microphone?
• How many pulses received were emitted before the microphone was within 10 meters of the source?
• What sound frequency is therefore received by the microphone?

Assume that sound travels at 340 m/s.

Solution

The detector requires 10 meters /( 37 m/s) = .2702 seconds to reach the source; the first sound requires 10 m / (340 m/s) = .02941 seconds to reach the detector. 

The first pulse received at the 10 meter distance was thus emitted .02941 seconds before its detection, while the last pulse is emitted at the detector and is therefore detected immediately.  So the detector observes all the pulses emitted while it travels to the source, plus those emitted in the .02941 seconds required by the first pulse to reach it:

• time to emit all pulses received by detector = .2702 sec + .02941 sec = .2996 sec.

During the .2996 seconds the number of pulses emitted is .2996 sec * 240 cycles/sec = 71.9 pulses.  Thus 71.9 pulses are received in the .2702 seconds during which the detector approaches the source, and the observed frequency is

• observed frequency = 71.9 pulses / .2702 seconds = 266.1 Hz.

Generalized Solution

If a detector moving toward an observer at velocity vObserver, and if sound travels at velocity vSound, then in any time interval `dt the detector will move distance

• distance moved by detector = vObserver * `dt.

A sound pulse will travel the distance vObserver * `dt in time

• `dtPulse = distance moved by detector / vSound = vObserver * `dt / vSound. 

Let A be the position of the detector at the beginning of the time interval `dt, and B its position at the end of the interval.   The pulses that already exist between A and B, plus all the pulses emitted during the `dt seconds, will all arrive at B during the `dt seconds.  To obtain the observed frequency we therefore need only determine the number of pulses between A and B, and the number emitted during the time interval.

• The pulses between A and B require time `dtPulse to move this distance, so were emitted in this time.  The number of pulses emitted in this time is   f * `dtPulse = f * `dt * vObserver/vSound.
• The number of pulses emitted in the `dt seconds is f * `dt.
• The total number of pulses passing point B in the `dt seconds is therefore

• number of pulses passing B in `dt seconds:  f * `dt + f * `dt * vObserver / vSound = f * `dt (1 + vObserver / vSound).

We therefore have observed frequency equal to the number of pulses divided by `dt:

• observed frequency:  f ' = f * `dt ( 1 + vObserver / vSound) / `dt = f (1 + vObserver / vSound).