Question

A subway train has a sound intensity level of 90 dB, and a riveting machine has a sound intensity level of 120 dB. How many times greater is the sound intensity of the machine than that of the train?

Answer 1

ANSWER:

1000 times

STEP-BY-STEP EXPLANATION:

Sound intensity level train = 90 dB

Sound intensity level machine = 120 dB

We can calculate the sound intensity in each case using the following formula:

$\begin{gathered} \beta =10\log _{10}\left(\frac{I}{I_0}\right) \\ \\ I_0=10^{-12}\text{ W/m}^2 \end{gathered}$

We substitute each value and calculate I in each case:

$\begin{gathered} \text{ For the train:} \\ \\ 90=10\cdot\log_{10}\left(\frac{I_1}{10^{-12}}\right) \\ \\ \log_{10}\left(\frac{I_1}{10^{-12}}\right)=\frac{90}{10} \\ \\ \log_{10}\left(\frac{I_1}{10^{-12}}\right)=9 \\ \\ \frac{I_1}{10^{-12}}=10^9 \\ \\ I_1=10^9\cdot\:10^{-12} \\ \\ I_1=0.001\text{ W/m}^2 \\ \\ \\ \text{ For the machine:} \\ \\ 120=10\cdot\log_{10}\left(\frac{I_2}{10^{-12}}\right) \\ \\ \log_{10}\left(\frac{I_2}{10^{-12}}\right)=\frac{120}{10} \\ \\ \log_{10}\left(\frac{I_2}{10^{-12}}\right)=12 \\ \\ \frac{I_2}{10^{-12}}=10^{12} \\ \\ I_2=10^{12}\cdot\:10^{-12} \\ \\ I_2=1\text{ W/m}^2 \\ \\ \text{ Therefore:} \\ \\ \frac{I_2}{I_1}=\frac{1}{0.001}=1000 \\ \\ \text{ It is 1000 times greater is the sound intensity of the machine than that of the train} \end{gathered}$