Question

A tug boat horn emits a frequency of 250 Hz in 18 °C air, which has a speed of sound of 342 m/s. If the tug boat is moving away from a stationary observer at 15 m/s, what is the frequency that is observed?

Answer 1

Answer:

Approximately $240\; {\rm Hz}$.

Explanation:

Consider how the motion of the boat (the source) affects the period of the sound wave as it appears to the observer.

The sound wave from this boat travels at the speed of sound $v(\text{sound})$. At the same time, the boat (the source) is moving away from the observer at a speed of $v(\text{boat})$. Let $T_{s}$ denote the period of this wave, and let $f_{s}$ denote the frequency.

The boat emits a wave crest after every interval of $T_{s}$. During that time, the boat would have moved $v(\text{boat})\, T_{s}$ further away from the observer. As a result, each subsequent crest would require an additional time of $\Delta T$ to reach the observer:

$\begin{aligned}\Delta T &= \frac{(\text{extra distance})}{(\text{wave speed})} = \frac{v(\text{boat})\, T_{s}}{v(\text{sound})}\end{aligned}$.

The source continues emitting crests with an interval of $T_{s}$. However, because of the extra distance, it would appear to the observer that crests are arriving with an interval of:

$\begin{aligned}T_{o} &= T_{s} + \Delta T \\ &=T_{s} + \frac{v(\text{boat})\, T_{s}}{v(\text{source})} \\ &= \left(1 + \frac{v(\text{boat})}{v(\text{source}}\right)\, T_{s}\end{aligned}$.

As it appears to the observer, the frequency of this wave would be:

$\begin{aligned}f_{o} &= \frac{1}{T_{o}}\\ &= \frac{1}{\displaystyle \left(1 + \frac{v(\text{boat})}{v(\text{source}}\right)\, T_{s}} \\ &= \frac{f_{s}}{\displaystyle \left(1 + \frac{v(\text{boat})}{v(\text{source}}\right)} \\ &= \frac{250\; {\rm Hz}}{\displaystyle \left(1 + \frac{15\; {\rm m\cdot s^{-1}}}{342\; {\rm m\cdot s^{-1}}}\right)} \\ &\approx 240\; {\rm Hz}\end{aligned}$.