A crying baby emits sound with an intensity of $8.0 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 }$ Calculate a reasonable estimate for the intensity level from a set of quintuplets (five babies), all crying simultaneously at the same place? The lowest detectable intensity is $1.0 \times 10 ^ { - 12 } \mathrm {~W} / \mathrm { m } ^ { 2 }$
A) 49 dB
B) 36 dB
C) 56 dB
D) 69 dB
E) 79 dB

Question

A crying baby emits sound with an intensity of $8.0 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 }$ Calculate a reasonable estimate for the intensity level from a set of quintuplets (five babies), all crying simultaneously at the same place? The lowest detectable intensity is $1.0 \times 10 ^ { - 12 } \mathrm {~W} / \mathrm { m } ^ { 2 }$
A)  49 dB
B)  36 dB
C)  56 dB
D)  69 dB
E)  79 dB

Quizplus · Accepted Answer

The intensity level in decibels (dB) is calculated using the formula: $L = 10 \log \left( \frac{I}{I_0} \right)$, where $I$ is the intensity of the sound and $I_0$ is the reference intensity ($1.0 \times 10^{-12} \mathrm{~W/m}^2$). For one baby, the intensity level is $L_1 = 10 \log \left( \frac{8.0 \times 10^{-8}}{1.0 \times 10^{-12}} \right) = 10 \log (8.0 \times 10^4) = 10 \times 4.903 = 49.03 $ dB. For five babies, the total intensity is $5 \times 8.0 \times 10^{-8} = 4.0 \times 10^{-7} \mathrm{~W/m}^2$. The intensity level for five babies is $L_5 = 10 \log \left( \frac{4.0 \times 10^{-7}}{1.0 \times 10^{-12}} \right) = 10 \log (4.0 \times 10^5) = 10 \times 5.602 = 56.02 $ dB, which is a reasonable estimate for the intensity level of quintuplets crying simultaneously.

A Crying Baby Emits Sound with an Intensity Of 8.0×10−8 W/m28.0 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 }8.0×10−8 W/m2

A Crying Baby Emits Sound with an Intensity Of $8.0 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 }$