Question

Two superimposing waves are represented by the equations $x_1=5\sin 2\pi (20t-0.1x)$ and $x_2=10\sin 2\pi (20t-0.2x)$ where $x$ and amplitude of each wave are given in meters and $t$ is in seconds. If the ratio of maximum intensity to minimum intensity is $x:1$, then $x$ is

Answer 1

Given that,
Amplitude of wave $-1$, $\left(A_1\right)=5\text{m}$
Amplitude of wave $-2$, $\left(A_2\right)=10\text{m}$

and we know that,

Intensity $\left(I\right)\propto \left(\text{Amplitude}\right(A){)}^2$

Let $I_1$ and $I_2$ be the intensities of wave$-1$ and wave $-2$
Then,
Maximum intensity is given by
$I_{max}\propto (A_1+A_2{)}^2$

Minimum intensity is given by
$I_{min}\propto (A_1-A_2{)}^2.$

From this, $\frac{I_{max}}{I_{min}}=|\frac{(A_1+A_2)}{(A_1-A_2)}{|}^2$

From the given data,

$\frac{I_{max}}{I_{min}}=|\frac{(5+10)}{(5-10)}{|}^2$
$\Rightarrow \frac{I_{max}}{I_{min}}={\left(\frac{15}{5}\right)}^2=\frac{9}{1}$
$\Rightarrow I_{max}:I_{min}=9:1$

Hence, option (c) is the correct answer.