Set 56 Problem number 20

Problem

A microphone is initially 20 meters from a sound source and moving toward it at 31 m/s.  The source emits a sound at a constant frequency of 270 Hz.

What sound frequency is therefore received by the microphone?

Assume that sound travels at 340 m/s.

Solution

The detector requires 20 meters /( 31 m/s) = .6451 seconds to reach the source; the first sound requires 20 m / (340 m/s) = .05882 seconds to reach the detector. 

The first pulse received at the 20 meter distance was thus emitted .05882 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 .05882 seconds required by the first pulse to reach it:

• time to emit all pulses received by detector = .6451 sec + .05882 sec = .7039 sec.

During the .7039 seconds the number of pulses emitted is .7039 sec * 270 cycles/sec = 190 pulses.  Thus 190 pulses are received in the .6451 seconds during which the detector approaches the source, and the observed frequency is

• observed frequency = 190 pulses / .6451 seconds = 294.5 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).