Question

An airplane traveling at speed, 180 m/s, emits sound at a frequency of 2 000 Hz. What is the change in frequency (in Hz) heard by a stationary listener as the plane approaching and passing by

Answer 1

As the plane moves toward the listener, the apparent frequency of the plane would be
`4250\; {\rm Hz}` (
`2250\; {\rm Hz}` higher than the frequency at the source.)

As the plane moves away from the listener, the apparent frequency of the plane would be approximately
`1308\; {\rm Hz}` (approximately
`692\; {\rm Hz}` lower than the frequency at the source.)

Assumption: the speed of sound in the air is
`340\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Crests of this sound wave travel toward the listener at a constant
`v = 340\; {\rm m\cdot s^(-1)}`. Since there is a pause of
`t = 1 / f = (1/2000)\; {\rm s}` between every two consecutive crests of this sound wave, the distance between each pair of consecutive crests would be:

`\begin{aligned}\lambda &= (v)/(f) \\ &= \frac{340\; {\rm m \cdot s^(-1)}}{2000\; {\rm s^(-1)}} \\ &= 0.17\; {\rm m} \end{aligned}`.

Hence, if the aircraft wasn't moving, the first crest would have a head start of
`\lambda = 0.17\; {\rm m}` relative to the second one. This head start would ensure that the first crest arrive
`t = \lambda / v = 0.17\; {\rm m} / (340\; {\rm m \cdot s^(-1)}) = (1/2000)\; {\rm s}` earlier than the second crest.

However, at a speed of
`v_{\text{s}} = 180\; {\rm m\cdot s^(-1)}`, the aircraft would have travelled an additional
`v_\text{s}\, t = 180\; {\rm m\cdot s^(-1)} * (1/2000)\; {\rm s} = 0.09\; {\rm m}` within that
`t = (1 / 2000)\; {\rm s}`.

• If the aircraft was travelling towards the listener, the head start of the first crest over the next one would be reduced to
  `\lambda - v_\text{s}\, t =`
  `0.17\; {\rm m} - 0.09\; {\rm m} = 0.08\; {\rm m}`. The first crest would arrive earlier than the second one by
  `(\lambda - v_{\text{s}}\, t) / (v) = (0.08\; {\rm m}) / (340\; {\rm m\cdot s^(-1)}) \approx 0.000235\; {\rm s}`.
• In contrast, if the aircraft was travelling away from the listener, the head start of the first crest over the next one would be increased to
  `\lambda + v_\text{s}\, t = 0.17\; {\rm m} + 0.09\; {\rm m} = 0.26\; {\rm m}`. The first crest would arrive earlier than the second one by
  `(\lambda + v_{\text{s}}\, t) / (v) = (0.26\; {\rm m}) / (340\; {\rm m\cdot s^(-1)}) \approx 0.000765\; {\rm s}`.

In other words, if the aircraft was moving towards the listener, the period of the sound would appear to the listener to be approximately
`0.000235\; {\rm s}`. in contrast, if the aircraft was moving away from the listener, the period of the sound would appear to the listener as approximately
`0.000765\; {\rm s}`.

Therefore:

• When the aircraft moves toward the listener, the listener would hear a frequency of
  `f = 1 / t \approx 1 / 0.000235\; {\rm s} = 4250\; {\rm Hz}`.
• When the aircraft moves away from the listener, the listener would hear a frequency of approximately
  `f = 1 / t \approx 1 / 0.000765\; {\rm s} \approx 1307\; {\rm Hz}`.