2024 A/L Physics Past Paper · Question 10 (English Medium)

Answer: (5) $10000\ \text{Hz}$

Given information

This question is about the Doppler effect: the sound of something coming towards you sounds higher.

A source makes sound of frequency $1000\ \text{Hz}$. It rushes straight at an observer who stands still.

Its speed is $0.9v$, almost as fast as the sound itself.

What we must find: the frequency the observer hears.

How we find it: the moving source squeezes the waves in front of it, so the wavelength gets shorter. A shorter wavelength reaching the ear means a higher frequency.

The source emits $1000\ \text{Hz}$. It moves directly towards the observer at $u_s=0.9v$. The observer is stationary. The symbol $v$ is the speed of sound in air.

What the question asks

$f_0$ ($\text{Hz}$): frequency emitted by the source.

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

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

$u_s$ ($\text{m s}^{-1}$): source speed towards the observer.

$\lambda_0$, $\lambda$ ($\text{m}$): wavelength with the source at rest, then wavelength in front of the moving source.

Relevant theory

Now, if the source is moving towards the observer with a velocity u_s, then f₀ number of waves get packed within a distance v-u_s, thus changing the wave length of the wave to a value.

λ = (v − u_s)/f₀

This would result in the change of the frequency of sound heard by the observer to an apparent value

f = v/λ = v / ((v − u_s)/f₀)

∴ f = ( v/(v − u_s) ) f₀

What is happening: each new wavefront starts closer to the observer because the source has moved forwards. The wavefronts travel through the air at the same sound speed. Their smaller spacing makes them reach the stationary observer more often.

For example, hear the steady horn of a train as it comes towards a platform. The train moves forwards between two successive waves. The next wave begins closer to you.

The waves in front are closer together. Your ear receives them more often, so the pitch is higher.

In words: the moving source reduces the front wave spacing.

$$\lambda=\dfrac{v-u_s}{f_0}$$

Here $u_s$ belongs to the source. The listener does not move.

Required equations

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

$f=\dfrac{f_0}{1-u_s/v}$

Applying it to this question

$0.1$

$\dfrac{\lambda}{\lambda_0}=1-0.9$

$10$

$\dfrac{f}{f_0}=\dfrac{1}{0.1}$

$10000$

$f=10\times1000$

Reasoning and calculation

The source follows almost as fast as the sound it sends forwards. The remaining front spacing is only one tenth of its stationary value. Dividing the sound speed by that spacing gives ten times the emitted frequency.

$1040\ \text{Hz}$ would require $u_s/v=1-1000/1040\approx0.0385$. The source actually moves at $0.9v$. A small increase of $40\ \text{Hz}$ does not fit that speed.

$1100\ \text{Hz}$ would require $u_s/v=1-1000/1100\approx0.0909$. This is about one tenth of the given speed ratio. The denominator must be $1-0.9$, not a small percentage correction.

$1111\ \text{Hz}$ comes from dividing $1000$ by $0.9$. But $0.9$ is the source speed ratio. The front wave spacing depends on the remaining fraction $1-0.9=0.1$.

$1900\ \text{Hz}$ uses $1000(1+0.9)$. Here the observer is stationary. The moving-source equation divides by $1-0.9$. Substituting $1900$ back would require $u_s/v=1-1000/1900\approx0.474$, which does not match the given $0.9$.

Checks

$1000/(1-0.9)$

$\lambda=\dfrac{0.1v}{1000}$

$f=\dfrac{v}{0.1v/1000}$

$f=10000\ \text{Hz}$

The speed $v$ cancels in the ratio. We do not need a numerical value for the speed of sound. The remaining unit is $\text{Hz}$.

The answer is greater than $1000\ \text{Hz}$ because the source approaches. At zero source speed the equation gives $f=f_0$. As the subsonic source speed approaches $v$, the front spacing decreases towards zero and the ideal-model frequency grows.

Conclusion

$f=10000\ \text{Hz}$

The observer hears $10000\ \text{Hz}$. The answer is (5).