Question

A siren emitting a sound of frequency 1003 Hz moves away from you toward the face of a cliff at a speed of 12.0 m/s. Take the speed of sound in air as 334 m/s. (a) What is the frequency of the sound you hear coming directly from the siren

Answer 1

Answer:

(a) Approximately 968 Hz.

Explanation:

The observed frequency is less than 1003 Hz because of Doppler's Effect. When the source is moving away from an observer that doesn't move, the equation for the observed frequency $f_\text{observed}$ would be:

$f_\text{observed} = \displaystyle \frac{c}{c + v_\text{source}} \cdot f_\text{source}$,

where in the context of this problem,

• $c = 334\; \rm m \cdot s^{-1}$ is the speed of sound in the air.
• $v_\text{source} = 12.0\; \rm m \cdot s^{-1}$ is the speed at which the source moves away from the observer.
• $f_\text{source} = 1003\; \rm Hz$ is the frequency at the source.

Apply this equation to find $f_\text{observed}$:

$\begin{aligned}f_\text{observed} &= \frac{c}{c + v_\text{source}} \cdot f_\text{source} \\ &= \frac{334}{334 + 12.0} \times 1003 \\ &\approx 968\; \rm Hz\end{aligned}$.

Here's an alternative explanation.

The frequency of the siren at the source is $f = 1003\; \rm Hz$. That corresponds to a period of $T = \displaystyle \frac{1}{f} \approx 9.97\times 10^{-4}\; \rm s$.

In other words, at the source, a peak arrives about every $9.97\times 10^{-4}\; \rm s$.

The source is moving away from the observer at a speed of $v = 12.0\; \rm m \cdot s^{-1}$. In the $9.97\times 10^{-4}\; \rm s$ between the first and the second peak, the source moved $9.97 \times 10^{-4} \times 12.0 \approx 0.011964\; \rm m$ away from the observer. It would take an extra $\displaystyle \frac{0.011964\; \rm m}{334\; \rm m \cdot s^{-1}} \approx 3.58 \times 10^{-5}\; \rm s$ for the sound to cover that extra distance.

As a result, the period of the sound would appear to be $9.97\times 10^{-4} + 3.58 \times 10^{-5} \approx 1.03 \times 10^{-3} \; \rm s$ to the observer.

That corresponds to an observed frequency of $\displaystyle \frac{1}{1.03 \times 10^{-3}} \approx 968\; \rm Hz$. (Same as the answer from the formula.)