First slide

Intensity of

Question

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:

Moderate

A

2
B

4
C

5
D

6

Solution

Let a be the amplitude due to $S_{1} and S_{2}$ individually.
Loudness due to $S_{1} = I_{1} = {Ka}^{2}$
Loudness due to $S_{1} + S_{2} = I = K {(2 a)}^{2} = 4 I$
$∴ n = 10 \log_{10} (\frac{4 l_{1}}{l_{1}})$
= $10 \log_{10} (4) = 6$

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