Gujarati
Hindi
14.Waves and Sound
normal

Two identical sounds $S_1$ and $S_2$ reach at a point $P$ in phase. The resultant loudness at point $P$ is $n\,\, dB$  higher than the loudness of $S_1$. The value of $n$ is

A

$2$

B

$4$

C

$5$

D

$6$

Solution

Let a be the amplitude due to $S_1$ and $S_2$  individually.

Loundness due to $S_1 = I_1 = Ka_2$

Loundness due to $S_1 + S_2 = I = K(2a)^2 = 4I_1$

$\therefore \,n\,\, = 10\,{\log _{10}}\left( {\frac{{4{I_1}}}{{{I_1}}}} \right)$

$\,\, = 10\,{\log _{10}}(4)\, = \,6$

Standard 11
Physics

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