 # Sound level L_P in units of decibles (dB)is determined by: L_P = 20log_(10)(P/(P_(0))) where p is the sound pressure of the sound, and p_0 = 20 * 10^(-6) Pa is a referencesound pressure (the sound pressure when L_P = 0dB). Determine the sound pressure of 90 dB noise (noisegenerated by a passing truck). By how many times the soundpressure of the truck is larger (louder) than the sound pressureduring normal conversation where the loudness is 65 dB? comAttitRize8 2022-07-26 Answered
Sound level ${L}_{P}$ in units of decibles (dB)is determined by:
${L}_{P}=20{\mathrm{log}}_{10}\left(\frac{P}{{P}_{0}}\right)$
where p is the sound pressure of the sound, and ${p}_{0}=20\ast {10}^{-6}Pa$ is a reference sound pressure (the sound pressure when ${L}_{P}=0dB$). Determine the sound pressure of 90 dB noise (noisegenerated by a passing truck). By how many times the soundpressure of the truck is larger (louder) than the sound pressureduring normal conversation where the loudness is 65 dB?
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${L}_{p}=20\mathrm{log}\left(P/20\ast 10-6\right)=90$
so, $\mathrm{log}\left(P/20\ast 10-6\right)=4.5$
so, P = 0.6325 Pa
$\mathrm{△}{L}_{p}={L}_{p}$(truck) - ${L}_{p}$(norm.conv.)
so,
$90-65=20\mathrm{log}\left(Pt/Pnc\right)$
so,
$P/Pc={10}^{1.75}=56.234$ times