The sound of a bird singing is given as $5 \mathrm{~dB}$. What is the intensity of the sound of a bird singing?
A) $4.25 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}$
B) $3.16 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}$
C) $2.98 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}$
D) $1.95 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}$
E) $2.04 \times 10^{-12} \mathrm{~W} / \mathrm{m}^{2}$

Question

Quizplus · Accepted Answer

The intensity level in decibels (dB) is given by the formula $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$, where $I$ is the intensity of the sound in watts per square meter ($W/m^2$), and $I_0 = 1 \times 10^{-12} W/m^2$ is the reference intensity level, which is the threshold of human hearing. Rearranging the formula to solve for $I$ gives $I = I_0 \times 10^{L/10}$. Substituting $L = 5$ dB into the formula gives $I = 1 \times 10^{-12} \times 10^{5/10} = 3.16 \times 10^{-12} W/m^2$.

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The Sound of a Bird Singing Is Given As $5 \mathrm{~dB}$