Two identical sounds s(1) and s(2) reach...

Two identical sounds `s_(1) and s_(2)` reach at a point P phase. The resultant loudness at a point P is not higher than the loudness of `s_(1)`. The value of n is

` 2`

`4`

`5`

`6`

Text Solution

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The correct Answer is:

Let `a` be the amplitude due to `S_(1) and S_(2)` individually .
Intensity due to `S_(1) = I_(1) = Ka^(2) `
Intensity due to ` S_(1) + S_(2) = I = K( 2a)^(2)`
`= 4 I_(1)`
`:. N = 10 log_(10) ((4 l_(1))/(I_(1)))`
` = 10 log _(10) (4) = 6`

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