Question
The loudness of a sound, L, measured in decibels, is related to its intensity, I units,
by L = 10 log10 (I × 1012).
Consider two sounds, S1 and S2 .
S1 has an intensity of 10-6
units and a loudness of 60 decibels.
S2 has an intensity that is twice that of S1 .
(a) State the intensity of S2 . [1]
(b) Determine the loudness of S2 . [2]
The maximum loudness of thunder in a thunderstorm was measured to be 115 decibels.
(c) Find the corresponding intensity, I, of the thunder.
▶️Answer/Explanation
Detail Solution
part(a)
Step 1: Determine the intensity of S2.
The intensity of S1 is given as \( I_1 = 10^{-6} \) units. Since S2 has an intensity that is twice that of S1, we can calculate \( I_2 \) as follows:
\( I_2 = 2 \times I_1 = 2 \times 10^{-6} = 2 \times 10^{-6} \) units.
The answer is: $2 × 10^{-6}$
part(b)
Step 1: Use the formula to determine the loudness of S2.
The loudness \( L \) is given by the formula \( L = 10 \log_{10}(I \times 10^{12}) \).
Step 2: Substitute \( I_2 = 2 \times 10^{-6} \) into the formula:
$$ L_2 = 10 \log_{10}((2 \times 10^{-6}) \times 10^{12}) $$
Step 3: Simplify the expression inside the logarithm
$$ L_2 = 10 \log_{10}(2 \times 10^{6}) $$
Step 4: Apply the logarithm properties:
$$ L_2 = 10 (\log_{10}(2) + \log_{10}(10^{6})) $$
Since \( \log_{10}(10^{6}) = 6 \), we have:
$$ L_2 = 10 (\log_{10}(2) + 6) $$
Using \( \log_{10}(2) \approx 0.301 \):
$$ L_2 = 10 (0.301 + 6) = 10 \times 6.301 = 63.01$$ decibels.
The answer is: 63.01
part(c)
Step 1: Use the loudness formula to find the intensity corresponding to 115 decibels.
Given \( L = 115 \) decibels, we can rearrange the formula:
$$ 115 = 10 \log_{10}(I \times 10^{12}) $$
Step 2: Divide both sides by 10:
$$ 11.5 = \log_{10}(I \times 10^{12}) $$
Step 3: Convert from logarithmic form to exponential form:
$$ I \times 10^{12} = 10^{11.5} $$
Step 4: Solve for \( I \):
$$ I = \frac{10^{11.5}}{10^{12}} = 10^{11.5 – 12} = 10^{-0.5} $$
Step 5: Calculate \( 10^{-0.5} \):
$$ 10^{-0.5} = \frac{1}{\sqrt{10}} \approx 0.316 units.$$
The answer is: 0.316
————Markscheme—————–
(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)