Two ships are moving along a line due east. The trailing ship has a speed relative to land based observation point of `64.0(km)/(h)`, and the leading ship has a speed of `45.0(km)/(h)` relative to that point. The two ships are in a region of the ocean where the current is moving uniformly due west at `10.0(km)/(h)`. The trailing ship transmits a sonar signal at a frequency of 1200.0 Hz. What frequency is monitored by the leading ship? Use `1520(m)/(s)` as the speed of sound in ocean water.

Explanation

The speeds of source and observer are small fractions of the speed of sound in water. The Doppler shift should be a small fraction of the radiated frequency. The gap between the ships is closing, so the frequency received will be higher than `1.20kHz`. We guess `1.25kHz`. When the observer is moving in front of and in the same direction as the source, the Doppler equaiton becomes
`f'=((v+(-v_1))/(v-v_1))f`
where `v_l` and `v_t` are the speeds of the leading and trailing ships measured relative to the medium in which the sound is propagated. In this case the ocean current is opposite to the direction of travel of the ships and
`v_l=45.0(km)/(h)-(-10.0(km)/(h))=55.0(km)/(h)=15.3(m)/(s)`
`v_t=64.0(km)/(h)-(-10.0(km)/(h))=74.0(km)/(h)=20.6(m)/(s)`
Therefore,
`f'=(1200Hz)((1520(m)/(s)-15.3(m)/(s))/(1520(m)/(s)-20.6(m)/(s)))=12.04Hz`
The Doppler shift is much smaller than we guessed. Note well that sound moves much faster in water than in air. The difference in speed of the ships is only `19(km)/(h)`, and this is only `5.3(m)/(s)`.