Two identical sounds S_1 and S_2 reach at a p | vedclass

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(D) Let $a$ be the amplitude due to $S_1$ and $S_2$ individually.The intensity $I_1$ due to $S_1$ is proportional to $a^2$,so $I_1 = K a^2$.Since $S_1

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$6$

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(D) Let $a$ be the amplitude due to $S_1$ and $S_2$ individually.The intensity $I_1$ due to $S_1$ is proportional to $a^2$,so $I_1 = K a^2$.Since $S_1$ and $S_2$ reach point $P$ in phase,their amplitudes add up: $A_{res} = a + a = 2a$.The resultant intensity $I$ is proportional to $(2a)^2$,so $I = K(2a)^2 = 4 K a^2 = 4 I_1$.The difference in loudness in decibels is given by $\Delta L = 10 \log_{10} \left( \frac{I}{I_1} ight)$.Substituting the values,$n = 10 \log_{10} \left( \frac{4 I_1}{I_1} ight) = 10 \log_{10} (4)$.Since $\log_{10} (4) \approx 0.602$,we have $n = 10 	imes 0.602 = 6.02 \approx 6$.

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

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