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The resultant loudness at a point P is `n` dB higher than the loudness of `S_1` which is one of the two identical sound sources `S_1` and `S_2` reaching at that point in phase. Find the value of n.

Answer» Loudness due to `S_1=I_1=ka^2` where a is the amplitude and loudness due to `S_1` and `S_2` both
`I_2=k(2a)^2=4I_1`
`n=10log_(10)((4I_1)/(I_1))=10log_(10)(4)=10(0.6)=6`


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