Question 4: Decibels and intensity Sound intensity I is measured in Watt per met

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Question 4: Decibels and intensity Sound intensity I is measured in Watt per meter squared (w/m2. Intensity drops off from the source according to 1/r2. That is I P,/r where r is the distance from the source to the point of observation and Pis the power (in W of the sound at the source. dB is defined as 10 log10(1/1o) where lo is 10 12 W/m2. It is the most common units measuring the loudness of sound Part a) Consider two microphones, one positioned at 10 m from the source and the other 20 m. The intensity measured at 20 m is 5.10 5w/m2, what is the intensity at 10 m? (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) speaker initially placed 2 m away is employed to test an hearing aid. A sound intensity level of 20 dBreached 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

questino 4:

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))

=45.9784 dB

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