MCQ
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 ........
- A$36: 4$
- ✓$25: 9$
- C$1: 4$
- D$4: 1$
✓
Answer
Correct option: B.
$25: 9$
b
(b)
(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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