Question

Two waves are represented by $y_1 = 4 \sin \, 404 \pi t$ and $y_2 = 3 \sin \, 400 \pi t.$ Then

• A. beat frequency is 4 Hz and the ratio of maximum to minimum intensity is 49 : 1
• B. beat frequency is 2 Hz and the ratio of maximum to minimum intensity is 49 : 1
• C. beat frequency is 2 Hz and the ratio of maximum to minimum intensity is 1 : 49
• D. beat frequency is 4 Hz and the ratio of maximum to minimum intensity is 1 : 49

Answer 1

Answer: (B)

beat frequency is 2 Hz and the ratio of maximum to minimum intensity is 49 : 1

Answer 2

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 \hspace25mm \omega_1 = 2 \pi v_1 \Rightarrow 404 \pi = 2 \pi v_1

or \hspace25mm v_1 = 202 \, Hz
\hspace30mm \omega_2 = 2 \pi v_2
\Rightarrow \hspace25mm 400 \pi = 2 \pi v_2
or \hspace25mm v_2 = 200 \, Hz
Beat frequency = $v_1 - v_2 = 202 - 200 = 2 \, Hz$
\hspace25mm \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
\hspace25mm = \big(\frac{7}{1}\big)^2 = \frac{49}{1}