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

6

Explanation

Loudness due to `S_(1)=J_(1)=ka^(2)` where a is the amplitude
Loudness due to `S_(1)` and `S_(2)`
`=I=k(2a)^(2)=4J_(1)`
`N=10log_(10)(4J_(1)//J_(1))`
`=10log_(10)(4)=10(0.6)=6`