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Q.

Two progressive waves y₁ = 4 sin 400 $π$ t and y₂ = 3 Sin 404 $π$ t , moving in the same direction superpose on each other producing beats. Then the number of beats per second and the ratio of maximum to minimum intensity of the resultant waves are respectively

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a

4 and 49/1

b

2 and 5/1

c

2 and 49/1

d

4 and 16/9

answer is D.

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Detailed Solution

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y₁ = 4sin(400 $π$ t) ---(1) y₂ = 3sin(404 $π$ t) ----(2)
from 1, a₁ =4, $ω_{1}$ = 2 $π$ n₁ =400 $π$ ; n₁ =200 Hz
From 2, a₂ = 3 units w₂ = 2 $π$ n₂ = 404 $π$ ; n₂= 202Hz
Number of beats per sec N = n₁-n₂
N = 202 - 200 = 2
$\frac{I_{\max}}{I_{\min}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = {(\frac{7}{1})}^{2} = \frac{49}{1}$

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