In a previous entry, the velocity of a sound-emitting source ( vs ) and a moving receptor of sound ( vr ) were related to the frequency of the sound emitted by the source ( fs ) and the frequency at which the receptor interpreted the emitted sound ( fr ) as follows:
( fr ) = ( fs )[ ( v + vr ) / ( v + vs ) ]
Frequency ( f ) is measured in cycles per second ( 1/s or s^-1 ) in S.I. units of Hertz ( Hz ), and velocity ( v ) is measured in meters per second.
In the derivation above, the frequency ( fs ) of the sound emitted by the source and the relative velocities of source ( vs ) and receptor ( vr ) must be known in order to solve for ( fr ). It would be useful, however, to be able to determine the velocity of a moving target ( vt ) by using a stationary sound source to reflect various types of waves off of such a target. Three known values of interest would make such a task convenient: ( 1 ) The speed of sound ( v = 332 m/s ); ( 2 ) the frequency of the wavefronts emitted by the source ( fs ); and ( 3 ) the frequency at which the source intercepts a wavefront after it has been reflected off of the moving target. Thus, ( fr ) = ( fs )[ ( v + vr ) / ( v + vs ) ] must be reduced to an equation in which three ( rather than four ) unknowns will be needed.
Before deriving an equation that will allow us to determine the speed of a moving target, we must take note of two occurrences that transpire within our frame of reference. First, the frequency ( fs ) of the wavefront emitted by the stationary sound source will be observed differently from the vantage point of the target that is moving towards the source. Therefore, when the emitted wavefront is reflected off of the approaching target, a “ shifted “ waveform will make its way back towards the stationary sound source. For this reason, the stationary sound source will intercept a reflected wavefront whose frequency is different than it was when initially emitted.
The frequency ( fs ) of the wavefront emitted by the stationary source is interpreted by the moving target as follows:
ft = ( fs )[ ( v + vt ) / ( v ) ]
Since the target initially moves towards the wavefronts emitted by the source, ( vt ) is positive, and ft > fs , as we’d expect it to be. After the wavefront has been reflected off of the approaching target at frequency ( ft ), it becomes the source of waves, and the stationary source becomes the observer of the reflected wavefront ( fs = fo ). The equation is rearranged as follows:
( ft )[ ( v ) / ( v + vt ) ] = fo
Remember: fs was changed to fo to reflect that fact that the wavefront emitted by the source has been reflected off the moving target, and it is now returning as a shifted wave. This difference will cause a difference to be observed by the stationary observer. Additionally, the moving target is now trailing the reflected wave, and this change must be reflected by a change of sign of the velocity ( vt ) of the moving target to a negative value. As a consequence, we have the following updated equation:
fo = ( ft )[ ( v ) / ( v – vt ) ]
This form of the equation will give us a value of fo > ft . This is exactly how we’d expect a wavefront reflected from a moving target to be “ seen “ by the sound source as the wave returns to it.
We are now ready to substitute ft = ( fs )[ ( v + vt ) / ( v ) ] into the equation above. Doing so yields the following equation:
fo = ( fs )[ ( v + vt ) / ( v – vt ) ]
Since the frequency ( fs ) originally emitted by the source is known, the velocity ( v ) of sound is known, and the frequency ( fo ) reflected wavefront can be measured, we now have access to all information needed to determine the velocity ( vt ) of a target that moves towards a stationary observer.
Q: A wavefront with a frequency of 1,000 Hz is emitted from a stationary source. Upon being reflected by an approaching target, the wavefront is observed to have been shifted to having a 1,200 Hz frequency. Using the speed of sound to be 332 m/s, how fast is the target moving?
A: The only unknown quantity is vt , and this is the value asked for in the question. If we solve the above derivation for vt , we can plug known values into the equation to obtain our answer.
fo = ( fs )[ ( v + vt ) / ( v – vt ) ]
( fo )( v – vt ) = ( fs )[ ( v + vt )
( fov ) – ( fovt ) = ( fsv ) + ( fsvt )
( fov ) – ( fsv ) = ( fsvt ) + ( fovt )
( v )( fo – fs ) = ( vt )( fs + fo )
Solving for the velocity of the approaching target, we have the following equation:
( vt ) = ( v )[ ( fo – fs ) / ( fo + fs ) ]
( vt ) = ( 332 m/s )[ ( 200 Hz ) / ( 2,200 Hz ) ] = 30.2 m/s
RENAISSANCE PHYSICS 