Two superimposing waves are represented by equation $y_1=2 \sin 2 \pi(10 t-0.4 x)$ and $y_2=4 \sin 2 \pi(20 t-0.8 x)$. The ratio of $I_{\max }$ to $I_{\min }$ is ........
Diffcult
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(b)

Formula of intensity is given by

$I =\frac{1}{2} \rho \nu \omega^2 A ^2 \text { or, } I = 2 \pi^2 f ^2 \rho \nu A ^2$

Hence, $I \propto f ^2 A ^2$

Here $f$ is the frequency, $A$ is the amplitude of wave, $v$ is the speed of wave and $\rho$ is the Density of medium.

Now, Given,

$y_1=2 \sin (20 \pi t-0.8 \pi x) \Rightarrow f_1=10, A_1=2[\because y=A \sin (2 \pi f \pm k x)]$

$y_2=4 \sin (40 \pi t-1.6 \pi x) \Rightarrow f_2=20, A_2=4$

Now, $\frac{ I _{\max }}{ I _{\min }}=\frac{\left( f _1 A _1+ f _2 A _2\right)^2}{\left( f _2 A _2- f _1 A _1\right)^2}[I$ didn't give calculations how to find this formula, you should memorize ]

put the values,

Imax/Imin $=(10 \times 2+20 \times 4)^2 /(20 \times 4-10 \times 2)^2$

$=(20+80)^2 /(80-20)^2$

$=(100 / 60)^2$

$=(5 / 3)^2=25: 9$

Hence $\frac{I_{\max }}{I_{\min }}=\frac{25}{9}$

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