2021 A/L Physics Past Paper · Question 39 (English Medium)

Frequency heard by an observer who passes a sound source

A sound source $S$ emits sound of constant frequency $f_0$ . An observer $O$ travels in the direction shown at a speed of $0.1u$ , where $u$ is the speed of sound in air. Which of the following gives the correct relationship between $f_0$ and the frequency $f$ of the sound heard when the observer is travelling towards the source?

(1) $f = 1.1 f_0$
(2) $f_0 < f < 1.1 f_0$
(3) $f_0 < f \le 1.1 f_0$
(4) $f = 0.9 f_0$
(5) $f_0 > f > 0.9 f_0$

Answer and explanation

Answer: (2) $f_0 < f < 1.1 f_0$

What the question asks

Find how $f$ is related to $f_0$ .

The observer $O$ moves at $0.1u$ along the dashed line. The source $S$ is at rest, below that line.

$O$ is travelling towards $S$ .

Relevant theory

Suppose the source emitting sound of frequency f₀ is stationary while the observer is approaching the source with a velocity u₀.

the velocity of sound relative to the observer changes due to his motion

Required equations

$f = \dfrac{u + u_o}{u} f_0$

$u_o = 0.1u\cos\theta$

Reading the options

The options say what $f$ is: equal to $1.1 f_0$ , between $f_0$ and $1.1 f_0$ , or below $f_0$ .

First write down what we know. The source is at rest. So the wavelength is not changed. Only the speed of the sound relative to $O$ changes.

The speed of sound relative to $O$ is $u$ plus the part of $O$ 's speed that points at $S$ .

Reasoning and calculation

Test the options.

$O$ is moving towards $S$ , so $u_o$ is positive. So $f$ is greater than $f_0$ . This removes (4) and (5).

$S$ is not on the dashed line. So $\theta$ is not $0^\circ$ , and $\cos\theta$ is less than $1$ . So $u_o$ is less than $0.1u$ . This removes (1) and (3).

$S$ is ahead of $O$ , so $\theta$ is less than $90^\circ$ and $\cos\theta$ is more than $0$ .

$f = \dfrac{u + 0.1u\cos\theta}{u} f_0$

$f = (1 + 0.1\cos\theta) f_0$

$f_0 < f < 1.1 f_0$

$f = 1.1 f_0$ needs $\cos\theta = 1$ , a head-on path.

The sign $\le$ allows $1.1 f_0$ , which needs a head-on path.

$f$ is above $f_0$ when $O$ goes towards $S$ .

Checks

$a + b * c$

Do the two ends make sense?

If $S$ were straight ahead, $\theta = 0^\circ$ and $f = 1.1 f_0$ .

If $S$ were exactly to the side, $\theta = 90^\circ$ and $f = f_0$ . $O$ would be neither coming nor going.

Our $S$ is between these two cases. So $f$ lies strictly between $f_0$ and $1.1 f_0$ .

Conclusion

$f = (1 + 0.1\cos\theta) f_0$ with $0 < \cos\theta < 1$ .

$f_0 < f < 1.1 f_0$ . The answer is (2).

Only the part of the observer's speed along the line to the source changes the frequency.

Ask: is the path straight at the source, to the side, or in between? Then the frequency is at the top, at $f_0$ , or in between.

Exam tip

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