2023 A/L Physics Past Paper · Question 13 (English Medium)

Answer: (5) $10^{5}$

Given information

What we know:

The first source gives a sound intensity level of $\beta_1 = 90\ \text{dB}$ at the point.

The second source gives a sound intensity level of $\beta_2 = 40\ \text{dB}$ at the same point.

The point is at the same distance from both sources.

What the question asks

We want: the ratio $\dfrac{I_1}{I_2}$ of the two source intensities.

What each letter means:

$\beta_1$ ($\text{dB}$): the sound intensity level made by the first source.

$\beta_2$ ($\text{dB}$): the sound intensity level made by the second source.

$I_1$ ($\text{W m}^{-2}$): the intensity of the first source.

$I_2$ ($\text{W m}^{-2}$): the intensity of the second source.

$I_0$ ($\text{W m}^{-2}$): the threshold of hearing.

Relevant theory

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

where I - given sound intensity I₀ - threshold of hearing

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.

Intensity (I) ∝ A²

I ∝ 1/d²

The first quote tells us how to turn an intensity into a level in $\text{dB}$. The second quote says the intensity at a point depends on the distance from the source.

Both sources are the same distance away. So the distance changes both intensities by the same amount, and the ratio of the intensities at the point is the ratio of the source intensities.

For example, think of the floors of a tall building. Each floor up makes the sound ten times stronger.

The $\text{dB}$ scale counts the floors. Each $10\ \text{dB}$ is one floor, which is one step of ten times.

To compare two sounds, count how many floors apart they are. Then multiply ten times, once for each floor.

Required equations

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

$\beta_1 - \beta_2 = 10 \log_{10}\left(\dfrac{I_1}{I_2}\right)$

$\dfrac{I_1}{I_2} = 10^{\,(\beta_1 - \beta_2)/10}$

Applying it to this question

Both levels are already in $\text{dB}$, so no change of units is needed.

$50$

$\beta_1 - \beta_2 = 90 - 40$

$5$

$\log_{10}\left(\dfrac{I_1}{I_2}\right) = \dfrac{50}{10}$

$1e5$

$\dfrac{I_1}{I_2} = 10^{5}$

Reasoning and calculation

The levels differ by $50\ \text{dB}$. That is five steps of $10\ \text{dB}$, and each step is ten times.

So the ratio is $10 \times 10 \times 10 \times 10 \times 10 = 10^{5}$. That is option (5).

Checks

$pow(10, d / 10)$

A ratio of two intensities in the same unit has no unit. The answer $10^{5}$ is a plain number, as it should be.

Does it make sense?

Put the answer back: $10 \log_{10} 10^{5} = 50\ \text{dB}$. That is $90 - 40$, the difference we were given.

The first level is much higher than the second, so the first source must be much more intense. A ratio far above $1$ fits.

Conclusion

$\dfrac{I_1}{I_2} = 10^{5}$

The ratio is $\dfrac{I_1}{I_2} = 10^{50/10} = 10^{5}$.

The answer is (5).

Subtract the two levels first. The $I_0$ cancels, so you do not need its value.

Then divide the difference in $\text{dB}$ by $10$ and use it as a power of ten. Check by putting it back: $10 \log_{10}$ of your ratio should give the difference.

Exam tip