A C-130 Hercules is traveling toward a cliff at the speed of 150 m/s. Its engines emit a tone at frequency f_{0}. We consider that the speed of sound is 340 m/s. What frequency is received by a pilot after reflection from the cliff?

Image Source: U.S. Air Force

###### Solution.

The Doppler effect is the change in frequency of a wave for an observer (detector). The classical Doppler effect formula is given by

f_{d} = f_{s}*(v±v_{d})/(v±v_{s}) (1)

where f_{d} is observed (detected) frequency, f_{s} is emitted (source) frequency, v is the velocity of waves in the medium, v_{d} is the velocity of the detector relative to the medium and v_{s} is the velocity of the source relative to the medium.

Remember!

If the detector moves toward the source, use the plus sign. +v_{d}

If the source moves toward the detector, use the minus sign. -v_{s}

Now we will apply the Doppler effect formula to our problem.

First, The C-130 Hercules is traveling toward the cliff. The Hercules’s engines are the source of sound waves and the cliff is the detector. The cliff is not moving. Hence, v_{d}=0. The source moves toward the detector, so we use the minus sign.

f_{1} = f_{0}*v/(v-v_{s}) (2)

Secondly, the sound waves reflect from the cliff. The cliff becomes the source and consequently, the C-130 Hercules becomes the detector. The detector moves toward the source, so we use the plus sign. The cliff (source) is not moving. Hence, v_{s}=0. It is important to mention that the wave frequency doesn’t change during reflection f_{s}=f_{1}.

f_{2} = f_{1}*(v+v_{d})/v (3)

We can rewrite the equation (3) using the equation (2) and we get

f_{2} = f_{0}*(v+v_{d})/(v-v_{s}) (4)

Thirdly, when two sound waves of different frequency approach, we get a phenomenon called in physics “beating” or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves.

We have the engines emit a tone at frequency f_{0} and the sound waves reflect from the cliff f_{2}. Hence,

f = |f_{2} – f_{0}|

We rewrite the last equation using the equation (4)

f = |f_{0}*(v+v_{d})/(v-v_{s}) – f_{0}|

or

f = |f_{0}*((v+v_{d})/(v-v_{s}) – 1)| (5)

Finally, we substitute the values and we get

f = |f_{0}*((340+150)/(340-150) – 1)|

f ≈ 1.58f_{0}