A jet ski travels in a straight line at a constant speed of 20 m s^-1 along a marked lane on a lake, continuously sounding its horn at a frequency of 600 Hz. A lifeguard stands at the end of a wooden jetty that lies directly along the extension of the lane, and listens as the je…

AS Level Physics 9702: Doppler Effect Calculation for a Moving Source

A jet ski travels in a straight line at a constant speed of $20\text{ m s}^{-1}$ along a marked lane on a lake, continuously sounding its horn at a frequency of $600\text{ Hz}$ . A lifeguard stands at the end of a wooden jetty that lies directly along the extension of the lane, and listens as the jet ski approaches, passes the jetty, and then continues away in a straight line. The speed of sound in air is $340\text{ m s}^{-1}$ .

(a) State the equation for the frequency $f_\text{o}$ heard by a stationary observer when a source of frequency $f_\text{s}$ moves, at speed $v_\text{s}$ , directly towards or directly away from the observer, where $v$ is the speed of sound. [1]

(b) Calculate the frequency of the horn heard by the lifeguard while the jet ski is approaching the jetty. [2]

(c) Calculate the frequency of the horn heard by the lifeguard after the jet ski has passed the jetty and is moving away. [2]

(d) Calculate the percentage increase between the frequency found in (b) and the frequency $f_\text{s}$ emitted by the horn. [2]

Part (a): The Doppler equation for a moving source

For a source of frequency $f_\text{s}$ moving at speed $v_\text{s}$ directly towards or away from a stationary observer, with the sound travelling at speed $v$ : $f_\text{o} = \frac{f_\text{s}\,v}{v \mp v_\text{s}}$

using the minus sign in the denominator when the source moves towards the observer (giving a higher observed frequency), and the plus sign when the source moves away (giving a lower observed frequency).

Part (b): Frequency heard while the jet ski approaches

The source approaches, so the denominator uses $v - v_\text{s}$ : $f_\text{o} = \frac{f_\text{s}v}{v - v_\text{s}} = \frac{600 \times 340}{340 - 20} = \frac{204\,000}{320} = 637.5\text{ Hz}$

Check (compute numerator and denominator separately, then divide): $600 \times 340 = 204\,000$ ; $340 - 20 = 320$ ; $204\,000 \div 320 = 637.5$ , the same result. So the lifeguard hears a frequency of $\boxed{637.5}\text{ Hz}$ while the jet ski approaches, higher than the emitted $600\text{ Hz}$ , as expected.

Part (c): Frequency heard after the jet ski has passed

The source now recedes, so the denominator uses $v + v_\text{s}$ : $f_\text{o} = \frac{f_\text{s}v}{v + v_\text{s}} = \frac{600 \times 340}{340 + 20} = \frac{204\,000}{360} = 566.6\overline{6}\text{ Hz}$

Check (long division, verified independently): $360 \times 566 = 203\,760$ , leaving a remainder of $240$ ; $240/360 = 0.6\overline{6}$ , giving $566.6\overline{6}$ , consistent with the division above. So the lifeguard hears a frequency of $\boxed{566.7}\text{ Hz}$ (to 3 s.f.) once the jet ski is moving away, lower than the emitted $600\text{ Hz}$ , as expected.

Part (d): Percentage increase relative to the emitted frequency

Using the answer to (b), the increase in frequency compared with the emitted $f_\text{s} = 600\text{ Hz}$ is: $637.5 - 600 = 37.5\text{ Hz}$

As a percentage of the emitted frequency: $\frac{37.5}{600} \times 100\% = 6.25\%$

Check (reverse the calculation): $6.25\%$ of $600\text{ Hz}$ is $0.0625 \times 600 = 37.5\text{ Hz}$ , and $600 + 37.5 = 637.5\text{ Hz}$ , exactly the value found in (b), confirming the percentage increase.

So the percentage increase is $\boxed{6.25}\%$ .

Final answers

(a) $f_\text{o} = \dfrac{f_\text{s}v}{v \mp v_\text{s}}$ (minus for approach, plus for recede)
(b) Approaching: $f_\text{o} = \boxed{637.5}\text{ Hz}$
(c) Receding: $f_\text{o} \approx \boxed{566.7}\text{ Hz}$
(d) Percentage increase $= \boxed{6.25}\%$