2019 A/L Physics Past Paper · Question 41 (English Medium)

A rising whistling firecracker — the Doppler effect

A whistling firecracker having a constant whistling frequency is fired vertically upward. It travels initially with an acceleration, then with a deceleration, and finally blasts before coming to the rest. An observer at ground directly below the firecracker listens to the whistling sound of the firecracker.

Consider the following statements regarding the frequency of the sound heard by the observer.

(A) During the acceleration, it is higher than the whistling frequency and is decreasing with time.

(B) During the deceleration, it is lower than the whistling frequency and is increasing with time.

Which of the above statements is/are correct?

(1) Only A
(2) Only B
(3) Only C
(4) Only A and B
(5) Only B and C

Answer and explanation

Answer: (2) Only B

What the question asks

We want:

Which of the three statements about the frequency the observer hears are correct.

A: during the acceleration it is higher than $f_0$ and falls.

B: during the deceleration it is lower than $f_0$ and rises.

C: just before the blast it equals $f_0$ .

Relevant theory

Hence Doppler effect is the apparent change of frequency (pitch) of the sound from a source as heard by an observer when there is any relative motion between the observer and the source emitting the sound.

Suppose the source is moving away from the observer with a velocity u_s.

In this case the apparent frequency of the sound heard by the observer is less than the natural frequency.

The firecracker moves away from the observer with the speed $u_s$ .

The waves behind it are stretched. The frequency heard is:

$f = \dfrac{v}{v + u_s}f_0$

It is always below $f_0$ .

When $u_s$ grows, $f$ falls. When $u_s$ falls, $f$ rises. It reaches $f_0$ only if $u_s = 0$ .

Reading the options

The options list the statements that are correct.

Judge A, B and C first. Then pick the matching option.

Reasoning and calculation

During the acceleration $f > f_0$ , decreasing

The firecracker moves away. So $f < f_0$ always. A says higher. So A is wrong.

During the acceleration $u_s$ grows. So $f$ falls. That part of A is right. But the first part is wrong.

During the deceleration $f < f_0$ , increasing

The firecracker still moves away. So $f < f_0$ .

During the deceleration $u_s$ falls. So $f$ rises.

Both parts of B are right. So B is correct.

Before the blast $f = f_0$

The firecracker blasts before it comes to rest.

It is still moving up. So $u_s > 0$ .

So $f$ is still below $f_0$ . So C is wrong.

A and C both get the motion wrong.

A treats the source as coming closer.

C treats the firecracker as stopped at the blast.

Only B fits a source moving away.

Why the other options fail

A is wrong and B is missing.

C is wrong.

A is wrong.

Conclusion

$f = \frac{v}{v + u_s}f_0$

The source moves away, so $f < f_0$ at every stage.

During the deceleration $f$ rises with time.

Only B is correct.

So the answer is option (2).

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