Question 4: Decibels and intensity Part a) (i) What are the sound intensity leve

Question

Question 4: Decibels and intensity

Part a)

(i) What are the sound intensity levels measured in dB at the two locations?

(ii) Compare the relative difference in intensity and in dB.

Part b)

A speaker initially placed 2 m away is employed to test an hearing aid. A sound intensity level of 20 dB reached the hearing aid before amplification. The hearing aid will increase the sound level by 32 dB. In order to test the sensitivity of the hearing aid, you moved the speaker to 4 m away from the hearing aid. What is the sound level in dB after amplification with the speaker at 4 m?

Explanation / Answer

part a:

intensity is inversely proportional to square of the distance

hence intensity at 20 m/intensity at 10 m=10^2/20^2=1/4

==>intensity at 10 m=4*5*10^(-5)=2*10^(-4) W/m^2

i) sound level at 10 m=10*log10(2*10^(-4)/10^(-12))=83 dB

sound level at 20 m=10*log10(5*10^(-5)/10^(-12))=76.9897 dB

ii)difference in intensity=1.5*10^(-4) W/m^2

difference in dB=6.0103 dB

dB being a logarithimic value, the differences are close.

part b:

let initial power to be P.

then intensity at 2 m=P/2^2=P/4

now given in dB, this intensity value is 20 dB

so 10*log10(P/(4*10^(-12))=20

==>P=4*10^(-10)

the sound level is amplified by 32 dB

so total output=32+20=52 dB

if output intensity is I1,

then 10*log10(I1/10^(-12))=52

==>I1=1.58489*10^(-7) W/m^2

so amplification factor=I1/(input intensity)

=I1/(P/4)=1584.893

now, when speaker is 4 m away, input intensity=P/4^2

=P/16=2.5*10^(-11) W/m^2

amplified output=2.5*10^(-11)*1584.893=3.9622*10^(-8) W/m^2

then output intensity in dB=10*log10(3.9622*10^(-8)/10^(-12))

=45.9794 dB

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