Question

Sound from $2$ identical sources $S_1$ and $S_2$ reach a point $P$. When both sounds reach in the same phase, the intensity at $P$ due to each source is $I_0$. The power of $S_1$ is now reduced by 64% and the phase difference between $S_1$ and $S_2$ is varied continuously. The ratio of maximum to minimum intensity at $P$ is

• A. $8:1$
• B. $12:1$
• C. $16:1$
• D. $20:1$

Answer 1

The correct option is C $16:1$


The relation between intensity $I$ and power $P$ is
$I=\frac{P}{\text{Area}}$
Initially the intensity of both the sources is $I_0$.
Now we have reduced the power of $S_1$ by $64\%$. So the intensity also reduces by $64\%$, and the resultant intensity of $S_1$ is
$I_1=36\%I_0=0.36I_0$
Now we have to find the resultant intensity
We have $I_2=I_0$
The resultant intensity of is given by
$I_{res}=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
For maximum intensity $\cos \varphi =1$
$I_{\text{max}}=I_0+0.36I_0+2\sqrt{I_0\times 0.36I_0}$
$\Rightarrow I_{\text{max}}=I_0+0.36I_0+2\times 0.6I_0$
$\Rightarrow I_{\text{max}}=1.36I_0+1.2I_0=2.56I_0$

For minimum intensity $\cos \varphi =-1$
$I_{min}=I_0+0.36I_0-2\sqrt{I_0\times 0.36I_0}$
$=I_0+0.36I_0-2\times I_0\times 0.6$
$=1.36I_0-1.2I_0$
$=0.16I_0$

The ratio of maximum intensity to minimum intensity is
$\frac{I_{max}}{I_{min}}=\frac{2.56I_0}{0.16I_0}$
$\frac{I_{max}}{I_{min}}=16$