Question
The loudness of a sound, $L$, measured in decibels, is related to its intensity, $I$ units, by $L = 10 \log_{10}(I \times 10^{12})$.
Consider two sounds, $S_1$ and $S_2$.
$S_1$ has an intensity of $10^6$ units and a loudness of 60 decibels.
$S_2$ has an intensity that is twice that of $S_1$.
(a) State the intensity of $S_2$.
(b) Determine the loudness of $S_2$.
The maximum loudness of thunder in a thunderstorm was measured to be 115 decibels.
(c) Find the thunder’s corresponding intensity, $I$.
▶️Answer/Explanation
Detailed solution
Part (a): We’re given:
The intensity of \( S_1 \) is \( 10^{-6} \) units.
The intensity of \( S_2 \) is twice that of \( S_1 \).
So, the intensity of \( S_2 \):
\[
I_{S_2} = 2 \times I_{S_1} = 2 \times 10^{-6} = 2 \times 10^{-6} \text{ units}
\]
Part (b): Given the loudness formula is:
\[
L = 10 \log_{10}(I \times 10^{12})
\]
First, let’s verify the loudness of \( S_1 \), which is given as 60 decibels with intensity \( I_{S_1} = 10^{-6} \):
\[
L_{S_1} = 10 \log_{10}(10^{-6} \times 10^{12})
\]
\[
10^{-6} \times 10^{12} = 10^{-6 + 12} = 10^6
\]
\[
L_{S_1} = 10 \log_{10}(10^6) = 10 \times 6 = 60 \text{ decibels}
\]
This matches the given loudness, confirming the formula is consistent.
Now, for \( S_2 \), with intensity \( I_{S_2} = 2 \times 10^{-6} \):
\[
L_{S_2} = 10 \log_{10}((2 \times 10^{-6}) \times 10^{12})
\]
\[
2 \times 10^{-6} \times 10^{12} = 2 \times 10^{-6 + 12} = 2 \times 10^6
\]
\[
L_{S_2} = 10 \log_{10}(2 \times 10^6)
\]
\[
= 10 \left( \log_{10} 2 + \log_{10} 10^6 \right)
\]
\[
= 10 \left( \log_{10} 2 + 6 \right)
\]
\[
\log_{10} 2 \approx 0.3010
\]
\[
L_{S_2} = 10 (0.3010 + 6) = 10 \times 6.3010 = 63.01
\]
So, the loudness of \( S_2 \) is approximately 63.01 decibels
Part (c): The thunder has a loudness of 115 decibels. Using the formula:
\[
L = 10 \log_{10}(I \times 10^{12})
\]
\[
115 = 10 \log_{10}(I \times 10^{12})
\]
\[
\log_{10}(I \times 10^{12}) = \frac{115}{10} = 11.5
\]
\[
I \times 10^{12} = 10^{11.5}
\]
\[
I = \frac{10^{11.5}}{10^{12}} = 10^{11.5 – 12} = 10^{-0.5}
\]
\[
10^{-0.5} = \frac{1}{\sqrt{10}} \approx \frac{1}{3.162} \approx 0.3162
\]
\[
I \approx 3.162 \times 10^{-1} \text{ units}
\]
To four significant figures:
\[
I \approx 0.3162 \text{ units}
\]
…………………Markscheme…………………..
Solution: –
(a) $I = 2 \times 10^{-6} = \frac{1}{500000}$ (units)
(b) substitutes their doubled $I$-value from part (a) into $L$
$L = 10 \log_{10}(2 \times 10^{-6} \times 10^{12}) = 63.0102…$
$L = 63.0$ (decibels)
(c) $115 = 10 \log_{10}(I \times 10^{12})$
attempts to solve for $I$
$I = \frac{10^{13.5}}{10^{12}}$ (or equivalent) = 0.316227…
$I = 0.316$ (units)