2020 A/L Physics Past Paper · Question 03 (English Medium)

Answer: (1) $0$ dB

Given information

We know:

The sound intensity is $I = 10^{-12}\ \text{W m}^{-2}$.

This is the threshold of hearing itself.

So the reference intensity is $I_0 = 10^{-12}\ \text{W m}^{-2}$.

What the question asks

We want: the intensity level $\beta$ matching this intensity, in decibels (dB).

Relevant theory

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

It's unit is decibel (dB).

Intensity level of sound (β) = 10 log₁₀ ( I/I₀ )

Required equations

$\beta = 10\log_{10}\left(\frac{I}{I_0}\right)$

Applying it to this question

Both intensities are in $\text{W m}^{-2}$.

So the ratio $\dfrac{I}{I_0}$ has no unit.

We can substitute directly.

$\beta = 10\log\left(\frac{10^{-12}}{10^{-12}}\right)$

$\beta = 10\log 1$

$\beta = 10 \times 0$

$0$

$\beta = 0$

Reasoning and calculation

The intensity equals the threshold of hearing.

So the ratio is $1$, and its logarithm is zero.

So the intensity level is $0$ dB.

Option (1).

Checks

The intensity leval corresponding to threshold of hearing β = 10 log₁₀ ( 10⁻¹² / 10⁻¹² )

β = 0

$10 * log(I / I0)$

Does it make sense?

Each ten-fold rise in intensity adds $10$ dB to the intensity level.

From the threshold of hearing ($0$ dB) to the threshold of pain $1\ \text{W m}^{-2}$ is a rise of $10^{12}$ times, that is $120$ dB.

So the quietest audible sound starting at $0$ dB makes sense.

Conclusion

$\beta = 10\log\frac{10^{-12}}{10^{-12}} = 0\ \text{dB}$

Since $I = I_0$, $\beta = 10\log 1 = 0$ dB.

The answer is (1).

$0$ dB does not mean "no sound".

It is the intensity level of the threshold of hearing.

That is where a normal ear begins to hear.

Exam tip