If a sound intensity level of 0 dB at 1000 Hz corresponds

Problem 25PE Chapter 17

College Physics by Urone | 1st Edition

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Problem 25PE

If a sound intensity level of 0 dB at 1000 Hz corresponds to a maximum gauge pressure (sound amplitude) of 10–9 atm , what is the maximum gauge pressure in a 60-dB sound? What is the maximum gauge pressure in a 120-dB sound?

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Solution 25PE

Sound intensity level in units of decibels (dB) can be written as,

$\beta{}$ (dB) = ${10log}_{10}$ $\left({\frac{I}{{I}_{0}}}\right)$

Here, $\beta{}$ is the sound intensity level, $\ I$ is the intensity of sound and ${I}_{0}$ is the threshold intensity of hearing.

Calculate the ratio of intensity of sound from relation,

${\beta{}}_{2}$ (dB) - ${\beta{}}_{1}$ (dB) = 60

Here, ${\beta{}}_{1}$ is the initial sound intensity level and ${\beta{}}_{2}$ is the final sound intensity level.

Substitute ${10log}_{10}$ $\left({\frac{{I}_{1}}{{I}_{0}}}\right)$ for ${\beta{}}_{1}$ and ${10log}_{10}$ $\left({\frac{{I}_{2}}{{I}_{0}}}\right)$ for ${\beta{}}_{2}$ in equation,

${\beta{}}_{2}$ (dB) - ${\beta{}}_{1}$ (dB) = 60

Substitute ${10log}_{10}$ $\left({\frac{{I}_{2}}{{I}_{0}}}\right)$ for ${\beta{}}_{2}$ (dB) and ${10log}_{10}$ $\left({\frac{{I}_{1}}{{I}_{0}}}\right)$ for ${\beta{}}_{1}$ (dB),

${10log}_{10}$ $\left({\frac{{I}_{2}}{{I}_{0}}}\right)$ - ${10log}_{10}$ $\left({\frac{{I}_{1}}{{I}_{0}}}\right)$ = 60

$\ \ \ \ {10log}_{10}$ $\left({\frac{{I}_{2}}{{I}_{1}}}\right)$ = 60