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

Answer:

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 $340\; {\rm m\cdot s^{-1}}$.

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:

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

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 $v_\text{s}\, t = 180\; {\rm m\cdot s^{-1}} \times (1/2000)\; {\rm s} = 0.09\; {\rm m}$ 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 $f = 1 / t \approx 1 / 0.000765\; {\rm s} \approx 1307\; {\rm Hz}$.