2018 A/L Physics Past Paper · Question 18 (English Medium)

Answer: (4) 10 km

What the question asks

We know: at 1 km the intensity is $10^{-10}\ \text{W m}^{-2}$.

The threshold of hearing is $10^{-12}\ \text{W m}^{-2}$.

We want: the distance where the intensity falls to the threshold.

Relevant theory

It depends on the amplitude (A) of the sound wave received by the observer and also on the distance (d) between the source of sound and the observer as given below.

The minimum intensity that can be sensed by the human ear (10⁻¹² Wm⁻²) is called threshold of hearing

Required equations

$I = \frac{P}{4\pi d^2} \Rightarrow I \propto \frac{1}{d^2}$

Reasoning and calculation

What stays the same? The power of the trumpet, $P$.

What changes? The distance $d$. So the intensity $I$ changes too.

The sound spreads over a sphere of area $4\pi d^2$.

Twice as far, the same power covers four times the area.

So $I d^2$ stays the same.

The intensity must fall 100 times. So $d^2$ must grow 100 times, and $d$ grows 10 times.

$\frac{d_2}{d_1} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{10^{-10}}{10^{-12}}}$

$10$

$\sqrt{100}$

$d_2 = 10\times1\ \text{km}$

Reading the options

$$d_{2} = 10\,\text{km}$$

Among the options, 10 km is option (4).

Checks

$d1 * sqrt(I1 / I2)$

A quick check.

$I_{1}/I_{2}$ has no unit.

So $d_{2}$ has the same unit as $d_{1}$: km.

Conclusion

$d_2 = 10\ \text{km}$

The intensity falls 100 times.

So the distance grows $\sqrt{100} = 10$ times.

$$d_2 = 10 \times 1\ \text{km} = 10\ \text{km}$$

The answer is option (4).

Explore

The elephant trumpets. The sound spreads out in every direction.

Twice as far, the same sound spreads over four times the area. So the intensity is a quarter.

The person walks on. At 10 km the intensity equals the threshold of hearing. He only just hears it.

Beyond that he hears nothing.