2024 A/L Physics Past Paper · Question 13 (English Medium)

Answer: (5) $100I$

Given information

This question is about how loud two sounds are at the same place.

Two sound sources, $A$ and $B$, are the same distance $r$ from one point. The intensity level measured at that point is $72\ \text{dB}$ for $A$ and $92\ \text{dB}$ for $B$.

The decibel scale does not grow like the intensity. Each extra $10\ \text{dB}$ means ten times more intensity.

What we must find: the intensity of $B$ at that point, when the intensity of $A$ is $I$.

How we find it: subtract the two levels. The difference tells us how many times stronger $B$ is.

At the point, $\beta_A=72\ \text{dB}$ and $\beta_B=92\ \text{dB}$. The intensity of $A$ there is $I_A=I$. Both readings are taken at the same point, so they can be compared directly.

What the question asks

$\beta_A$, $\beta_B$ ($\text{dB}$): intensity levels of $A$ and $B$ at the point.

$I_A$, $I_B$ ($\text{W m}^{-2}$): sound intensities of $A$ and $B$ at the point.

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

Relevant theory

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

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

What is happening: the measured level is not the intensity itself. It is ten times the $\log$ of the intensity divided by the threshold of hearing $I_0$.

Both levels use the same $I_0$. So when we subtract them, $I_0$ cancels. Only the ratio $I_B/I_A$ is left.

For example, think of a school hall. One student sings and we measure the level at the back.

Ten students sing the same way together. The intensity is ten times, and the measured level is $10\ \text{dB}$ more.

A hundred students sing. The intensity is a hundred times, and the measured level is $20\ \text{dB}$ more.

In words: every $10\ \text{dB}$ added multiplies the intensity by ten.

Required equations

$\beta=10\log\dfrac{I}{I_0}$

$\beta_B-\beta_A=10\log\dfrac{I_B}{I_A}$

Applying it to this question

$20$

$\beta_B-\beta_A=92-72$

$10\log\dfrac{I_B}{I_A}=20$

$2$

$\log\dfrac{I_B}{I_A}$

$\dfrac{I_B}{I_A}=10^2$

$100$

$\dfrac{I_B}{I_A}$

$100I$

$I_B$

Reasoning and calculation

$B$ is $20\ \text{dB}$ louder. That is two steps of $10\ \text{dB}$, so the intensity is multiplied by ten twice. Only $100I$ fits. This is option (5).

$1.3I$ comes from dividing the levels: $92/72\approx1.28$. Decibel levels are not proportional to intensity. Putting $1.3$ back gives $10\log 1.3\approx1\ \text{dB}$, not $20\ \text{dB}$.

$10I$ treats the whole $20\ \text{dB}$ as one factor of ten. A factor of ten adds only $10\log 10=10\ \text{dB}$. The difference here is twice that.

$20I$ uses the difference $20$ itself as the multiplier. The level uses a $\log$ of the ratio. Putting $20$ back gives $10\log 20\approx13\ \text{dB}$, not $20\ \text{dB}$.

$25I$ does not fit the level formula. Putting $25$ back gives $10\log 25\approx14\ \text{dB}$. The two levels differ by $20\ \text{dB}$.

Checks

$10*log(r)$

$\beta_B-\beta_A=10\log 100$

$\beta_B-\beta_A=10\times2$

$\beta_B-\beta_A=20\ \text{dB}$

The ratio $I_B/I_A$ has no unit, because $\text{W m}^{-2}$ cancels. So $I_B=100I$ has the same unit as $I$, which is $\text{W m}^{-2}$.

$B$ has the higher level, so its intensity must be larger than $I$. If the two levels were equal, the ratio would be $10^0=1$. A larger difference gives a much larger ratio, as expected for a $\log$ scale.

Conclusion

$I_B=100I$

The intensity of $B$ at the point is $100I$. The answer is (5).