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Physics
Two waves are represented by y1 = 4 sin 404 π t and y2 = 3 sin 400 π t. Then
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Q. Two waves are represented by $y_1 = 4 \sin \, 404 \pi t$ and $y_2 = 3 \sin \, 400 \pi t.$ Then
Bihar CECE
Bihar CECE 2009
Electromagnetic Waves
A
beat frequency is 4 Hz and the ratio of maximum to minimum intensity is 49 : 1
16%
B
beat frequency is 2 Hz and the ratio of maximum to minimum intensity is 49 : 1
64%
C
beat frequency is 2 Hz and the ratio of maximum to minimum intensity is 1 : 49
12%
D
beat frequency is 4 Hz and the ratio of maximum to minimum intensity is 1 : 49
8%
Solution:
Given, $y_1 = 4 sin \, 404 \pi t , y_2 = 3 sin \, 400 \pi t$
Comparing with equation y = A sin $ \omega t,$
$\therefore \, \, \, \, \, \omega_1 = 404 \pi , \omega_2 = 400 \pi , A_1 = 4 , A_2 = 3$
$\because \omega_1 = 2 \pi v_1 \Rightarrow 404 \pi = 2 \pi v_1$
or $ v_1 = 202 \, Hz$
$ \omega_2 = 2 \pi v_2$
$\Rightarrow 400 \pi = 2 \pi v_2$
or $ v_2 = 200 \, Hz$
Beat frequency = $v_1 - v_2 = 202 - 200 = 2 \, Hz$
$ \frac{I_{max}}{I_{min}} = \big( \frac{A_1 + A_2}{A_1 - A_2}\big)^2 = \big(\frac{4 + 3 }{4 - 3}\big)^2$
$ = \big(\frac{7}{1}\big)^2 = \frac{49}{1}$