2022 A/L Physics Past Paper · Question 12 (English Medium)

Answer: (3) $20\ \text{m s}^{-1}$

Given information

We know:

The alarm is at rest and gives out $f_0 = 510\ \text{Hz}$.

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

The person hears $f = 480\ \text{Hz}$ while moving directly away from the car.

What the question asks

We want: the speed $u_0$ of the person on the motorcycle.

What each letter means:

$f_0$ ($\text{Hz}$): frequency of the sound from the alarm.

$f$ ($\text{Hz}$): frequency heard by the person.

$v$ ($\text{m s}^{-1}$): speed of sound in air.

$u_0$ ($\text{m s}^{-1}$): speed of the person, the observer.

Relevant theory

Apparent frequency of sound from the stationary A as heard by B leaving A,

f₂ = ( (v − u_B)/v ) f_A

For example, think of a swimmer who swims away from the beach while waves roll in from the shore.

The swimmer meets fewer waves each second than someone standing still, because he is moving away from them.

It is the same with sound. A listener moving away from the source meets fewer sound waves each second. So the sound seems lower.

The listener moves away at $u_0$, so the sound reaches him at only $v - u_0$.

$$f = \dfrac{v - u_0}{v}\,f_0$$

In words: the frequency heard is the natural frequency times the fraction of the sound speed that is left.

Required equations

$f = \dfrac{v - u_0}{v}\,f_0$

$u_0 = v\left(1 - \dfrac{f}{f_0}\right)$

Applying it to this question

Every speed is in $\text{m s}^{-1}$ and every frequency is in $\text{Hz}$. The ratio $\dfrac{f}{f_0}$ has no unit, so $u_0$ comes out in $\text{m s}^{-1}$.

$0.941$

$\dfrac{f}{f_0} = \dfrac{480}{510}$

$0.0588$

$1 - \dfrac{480}{510} = \dfrac{30}{510}$

$20$

$u_0 = 340 \times \dfrac{30}{510}$

Reasoning and calculation

The fraction $\dfrac{v - u_0}{v}$ must equal $\dfrac{f}{f_0} = \dfrac{480}{510}$.

So $v - u_0 = 340 \times \dfrac{480}{510} = 320$.

So $u_0 = 340 - 320 = 20\ \text{m s}^{-1}$. It is option (3).

Checks

$v * (1 - f / f0)$

The unit: $\text{m s}^{-1} \times \dfrac{\text{Hz}}{\text{Hz}} = \text{m s}^{-1}$. A speed, as it should be.

Does it make sense?

Put the answer back: $f = 510 \times \dfrac{340 - 20}{340} = 480\ \text{Hz}$. That is what the person hears.

The person hears a lower note than $510\ \text{Hz}$, as expected for a listener moving away. A speed of $20\ \text{m s}^{-1}$ is about $72\ \text{km h}^{-1}$, which is sensible for a motorcycle.

Conclusion

$u_0 = 20\ \text{m s}^{-1}$

The person moves away at $u_0 = v\left(1 - \dfrac{f}{f_0}\right) = 340 \times \dfrac{30}{510} = 20\ \text{m s}^{-1}$.

The answer is (3).

When the listener moves away from a source at rest, use $f = \dfrac{v - u_0}{v}\,f_0$. The sound is lower.

The drop in frequency, $30\ \text{Hz}$, is not the speed. Put your answer back to check it.

Exam tip