2018 A/L Physics Past Paper · Question 36 (English Medium)

Dog whistle — 20 kHz by the Doppler effect

A whistle that is used to train a dog produces a frequency of 22 kHz which is more than the thereshold of hearing of the human. The trainer of the dog wants to make sure that the whistle is working. While the trainer stands by the side of a long straight road, he asks a friend to blow this whistle from a car moving along the same road. The required speed of the car and the direction of motion of the car for the trainer to hear the whistle at his threshold of hearing of 20 kHz is (Speed of sound in air is $340\ \text{m s}^{-1}$ )

(1) $31\ \text{m s}^{-1}$ , away from the trainer.
(2) $32\ \text{m s}^{-1}$ , away from the trainer.
(3) $34\ \text{m s}^{-1}$ , away from the trainer.
(4) $32\ \text{m s}^{-1}$ , towards the trainer.
(5) $34\ \text{m s}^{-1}$ , towards the trainer.

Answer and explanation

Answer: (3) $34\ \text{m s}^{-1}$ , away from the trainer.

Given information

We know: the whistle gives $f = 22$ kHz.

The trainer can hear up to $f' = 20$ kHz.

The speed of sound is $v = 340\ \text{m s}^{-1}$ .

The trainer stands still. The whistle moves with the car.

What the question asks

We want: the speed and the direction of the car, so that the trainer hears 20 kHz.

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.

After the horn passes you, the lower pitch will remain constant as the train moves away from you.

The trainer must hear a lower frequency.

So the source must move away from him.

A source moving away leaves longer waves behind it.

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

Required equations

$f' = f\,\frac{v}{v + u_s}$

$u_s = v\left(\frac{f}{f'} - 1\right)$

Applying it to this question

$u_s = 340\left(\frac{22}{20} - 1\right)$

$34$

$u_s = 340 \times 0.1$

Reasoning and calculation

$f' = 22 \times \frac{340}{340 + 34}$

$20$

$f' = 22 \times \frac{340}{374}$

So the car moves at $34\ \text{m s}^{-1}$ , away from the trainer.

Checks

$v * (f / f2 - 1)$

Check the unit.

$v$ is in $\text{m s}^{-1}$ . $f/f'$ has no unit.

So $u_s$ is in $\text{m s}^{-1}$ .

Does it make sense?

$34\ \text{m s}^{-1}$ is about $122\ \text{km h}^{-1}$ . A car can go that fast.

The frequency changes by 10%. So the car needs 10% of the speed of sound.

Conclusion

$u_s = 340\left(\frac{22}{20} - 1\right) = 34\ \text{m s}^{-1}$

The car must move away from the trainer.

$u_s = 340\left(\frac{22}{20} - 1\right) = 34\ \text{m s}^{-1}$

The answer is option (3).

Using the formula for a moving observer when the source moves.

When the source moves, $v \pm u_s$ is in the denominator.

Moving away: $f' = f\frac{v}{v + u_s}$ .

Explore

The friend blows the whistle in the car. The dog hears it. The trainer does not.

The car drives away from the trainer. The dog runs after the car.

Behind the car the wavelength grows. So the frequency the trainer hears falls.

When the car goes at $34\ \text{m s}^{-1}$ , he hears 20 kHz.

தமிழ் · සිංහල