Solve the problem.
-Use the formula $\beta = 10 \log \left( \mathrm { I } _ { 0 } \right)$, where the loudness of a sound in decibels is determined by I, the number of Watts/ $\mathrm { m } ^ { 2 }$ produced by the sound wave, and $\mathrm { I } _ { 0 } = 10 ^ { - 12 } \mathrm {~W} / \mathrm { m } ^ { 2 }$. A certain noise measures 101 decibels. If the intensity is multiplied by 100 , how many decibels will the new noise measure? (Round to the nearest unit.)
A) 10,100 decibels
B) 121 decibels
C) 103 decibels
D) 202 decibels

Question

Quizplus · Accepted Answer

The formula for sound intensity level in decibels is $\beta = 10 \log \left( \frac{I}{I_0} \right)$, where $I$ is the intensity in Watts/$m^2$ and $I_0 = 10^{-12} \text{W/m}^2$. If the intensity is multiplied by 100, the new intensity level can be calculated by adding $10 \log(100)$ to the original level because $\log(100I) = \log(I) + \log(100)$, and $\log(100) = 2$. Therefore, the new level is $101 + 20 = 121$ decibels.

Solve the Problem β=10log⁡(I0)\beta = 10 \log \left( \mathrm { I } _ { 0 } \right)β=10log(I0​)

Solve the Problem $\beta = 10 \log \left( \mathrm { I } _ { 0 } \right)$