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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