# The sound level at the side of a busy road to be 91 dB when eight trucks are passing at the same time. If the trucks are each contributing about the same level of sound, what would the sound level be if only one truck passed at one time?

## How would this be calculated?

Nov 22, 2017

$I {L}_{1} = I {L}_{8} - \left(10 \mathrm{dB}\right) \setminus \times \setminus {\log}_{10} \left({I}_{8} / {I}_{1}\right)$
$I {L}_{1} = 85$ $\mathrm{dB} - \left(10 \mathrm{dB}\right) \setminus \times \setminus {\log}_{10} 8 = 75.9$ $\mathrm{dB}$.

#### Explanation:

The Intensity Level, in $\mathrm{dB}$, of a sound of intensity $I$ is given by:

$I L \setminus \equiv \left(10 \mathrm{dB}\right) \setminus \times \setminus {\log}_{10} \left(\frac{I}{I} _ 0\right)$
where ${I}_{0}$ is the intensity of sound at the threshold of hearing.

${I}_{8}$ : Intensity of sound at roadside when 8 trucks pass by,
${I}_{1}$ : Intensity of sound at roadside when one truck pass by,

Each truck contribute the same level of sound:
\qquad I_8 = 8I_1; \qquad I_8/I_1 = 8

The intensity levels with a single truck and eight trucks are:
$I {L}_{1} = \left(10 \mathrm{dB}\right) \setminus \times \setminus {\log}_{10} \left({I}_{1} / {I}_{0}\right)$
$I {L}_{8} = \left(10 \mathrm{dB}\right) \setminus \times \setminus {\log}_{10} \left({I}_{8} / {I}_{0}\right)$

Comparing the intensity levels,
$I {L}_{8} - I {L}_{1} = \left(10 \mathrm{dB}\right) \setminus \times \left[\setminus {\log}_{10} \left({I}_{8} / {I}_{0}\right) - \setminus {\log}_{10} \left({I}_{1} / {I}_{0}\right)\right]$

But $\setminus q \quad \setminus {\log}_{10} a - \setminus {\log}_{10} b = \setminus {\log}_{10} \left(\frac{a}{b}\right)$

$I {L}_{8} - I {L}_{1} = \left(10 \mathrm{dB}\right) \setminus \times \left[\setminus {\log}_{10} \left({I}_{8} / {I}_{1}\right)\right]$

$I {L}_{1} = I {L}_{8} - \left(10 \mathrm{dB}\right) \setminus \times \setminus {\log}_{10} \left({I}_{8} / {I}_{1}\right)$

Given: $\setminus q \quad I {L}_{8} = 85$ dB; \qquad I_8/I_1 = 8

$I {L}_{1} = 85$ $\mathrm{dB} - \left(10 \mathrm{dB}\right) \setminus \times \setminus {\log}_{10} 8 = 75.9$ $\mathrm{dB}$.