Waves: Question 8

Worked solution

Part (a): Why the detected frequency is higher

As the ambulance moves towards the pedestrian, each successive wavefront is emitted from a position slightly closer to the pedestrian than the last. This compresses the wavefronts together in the space between the ambulance and the pedestrian, so the wavelength detected by the pedestrian is shorter than the wavelength that would be detected if the source were stationary. Since the wave still travels through the air at the same speed $v$, and $v = f\lambda$, a shorter detected wavelength corresponds to a higher detected frequency.

Part (b): Speed of the ambulance

For a source approaching a stationary observer: $f_\text{o} = \frac{f_\text{s}v}{v - v_\text{s}}$

Substituting $f_\text{o} = 732\text{ Hz}$, $f_\text{s} = 700\text{ Hz}$ and $v = 340\text{ m s}^{-1}$, and rearranging for $v_\text{s}$: $v - v_\text{s} = \frac{f_\text{s}v}{f_\text{o}} \implies v_\text{s} = v - \frac{f_\text{s}v}{f_\text{o}} = v\left(1 - \frac{f_\text{s}}{f_\text{o}}\right)$

$v_\text{s} = 340 \times \left(1 - \frac{700}{732}\right) = 340 \times \left(1 - 0.95628\right) = 340 \times 0.043716 = 14.86\text{ m s}^{-1}$

Check (substitute back into the original equation): with $v_\text{s} = 14.86\text{ m s}^{-1}$, $v - v_\text{s} = 340 - 14.86 = 325.14\text{ m s}^{-1}$, and $f_\text{o} = \dfrac{700 \times 340}{325.14} = \dfrac{238\,000}{325.14} \approx 732\text{ Hz}$, this reproduces the given observed frequency, confirming $v_\text{s} \approx \boxed{14.9}\text{ m s}^{-1}$ (3 s.f.).

Part (c): Frequency heard once the ambulance is receding

For a source moving directly away from a stationary observer, the sign in the denominator changes to $v + v_\text{s}$: $f_\text{o}' = \frac{f_\text{s}v}{v + v_\text{s}} = \frac{700 \times 340}{340 + 14.86} = \frac{238\,000}{354.86} \approx 670.7\text{ Hz}$

Check (order-of-magnitude sanity check): the receding frequency must be lower than the emitted $700\text{ Hz}$ (since the source is now moving away) and lower than the approaching value of $732\text{ Hz}$; $670.7\text{ Hz}$ satisfies both, and by a similar-sized shift to part (b) since the speed ratio $v_\text{s}/v$ is unchanged. So $f_\text{o}' \approx \boxed{671}\text{ Hz}$ (3 s.f.).

Part (d): Wavelength detected while approaching

Using the wave equation with the speed of sound and the frequency actually detected by the pedestrian while the ambulance approaches, $f_\text{o} = 732\text{ Hz}$: $\lambda_\text{o} = \frac{v}{f_\text{o}} = \frac{340}{732} = 0.4645\text{ m}$

Check (rearranging the other way): $732 \times 0.4645 = 340.0$, which reproduces the speed of sound, confirming $\lambda_\text{o} \approx \boxed{0.464}\text{ m}$ (3 s.f.). Slightly shorter than the wavelength $340/700 = 0.486\text{ m}$ that would be detected from a stationary source, consistent with the compression described in part (a).

Final answers

• (a) Wavefronts are compressed ahead of the approaching source, shortening the detected wavelength and so raising the detected frequency (since $v = f\lambda$ is fixed)
• (b) $v_\text{s} \approx \boxed{14.9}\text{ m s}^{-1}$
• (c) $f_\text{o}' \approx \boxed{671}\text{ Hz}$
• (d) $\lambda_\text{o} \approx \boxed{0.464}\text{ m}$