Question

Curves of two waves propagating in the same medium is shown. Here, $y$ is displacement in $\text{m}$ and $t$ is time in $\text{s}$. Find the ratio of their average intensities?

• A. $25:16$
• B. $9:16$
• C. $25:9$
• D. $4:9$

Answer 1

The correct option is A $25:16$

We know that the intensity of a wave is given by,
$I=\frac{1}{2}\rho {\omega }^2{A}^{2}v$, where,
$\rho \to$ density of medium
$\omega \to$ angular frequency of wave
$A\to$ displacement amplitude of wave
$v\to$ speed of wave
For a given medium,
$I\propto {\omega }^2{A}^2$
So, $\frac{I_1}{I_2}={\left(\frac{{\omega }_1{A}_1}{{\omega }_2{A}_2}\right)}^2$ ............$\left(1\right)$
Here, from figure, we can see that, the time period of wave $1$ is double the time period of wave $2$.
$\Rightarrow T_1=2T_2$
$\Rightarrow {\omega }_1=\frac{{\omega }_2}{2}$ $[T=\frac{2\pi }{\omega }]$
$\Rightarrow \frac{{\omega }_1}{{\omega }_2}=\frac{1}{2}$ .......$\left(2\right)$
​​​​​​​Also, $\frac{A_1}{A_2}=\frac{5}{2}$ .......$\left(3\right)$


On putting $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$,
$\Rightarrow \frac{I_1}{I_2}={\left(\frac{1\times 5}{2\times 2}\right)}^2$
$\Rightarrow \frac{I_1}{I_2}=\frac{25}{16}=25:16$