Two radio station broadcast their programmes at the same amplitude $A$ and at slightly different frequency $\omega_{1}$ and $\omega_{2}$ respectively where$\omega_{2} - \omega_{1} = 10^{3}\ Hz$. A detector is receiving signals from the two stations simultaneously. It can only detect signals of intensity $2A^{2}$. The time interval between successive maxima of the intensity of the signal received by the detector is

Question

Two radio station broadcast their programmes at the same amplitude $A$ and at slightly different frequency $\omega_{1}$ and $\omega_{2}$ respectively where$\omega_{2} - \omega_{1} = 10^{3}\ Hz$. A detector is receiving signals from the two stations simultaneously. It can only detect signals of intensity >$2A^{2}$. The time interval between successive maxima of the intensity of the signal received by the detector is

Dr. Abubacker · Accepted Answer

(b) Here , $A_{1} = A_{2} = A;\ \ n_{1} = \omega_{1},\ n_{2} = \omega_{2}$
$	herefore y_{1} = A\sin{2\pi\omega_{1}}t$, $y_{2} = A\sin{2\pi\omega_{2}}t$
$y = y_{1} + y_{2}$
$= 2A\frac{\cos{2\pi(\omega_{2} - \omega_{1})t}}{2}\sin\frac{2\pi(\omega_{2} + \omega_{1})t}{2}$
${= A}^{'}\sin{\pi(\omega_{2} - \omega_{1})t}$
Where $A^{'} = 2A{cos\pi}{(\omega_{2} - \omega_{1})t}$
Sound heard will be of max- intensity $( > 2A^{2})$
When $\cos\pi\left( \omega_{2} - \omega_{1} ight)t = max = \pm 1$
$\pi\left( \omega_{2} - \omega_{1} ight)t = 0,\pi,2\pi,$
$t = 0,\frac{1}{\omega_{2} - \omega_{1}};\frac{2}{\omega_{2} - \omega_{1}};\ldots\ldots\ldots\ldots\ldots\ldots$
Time interval between two successive maxima
$= \frac{1}{\omega_{2} - \omega_{1}} = \frac{2}{10^{3}} = 10^{- 3}s$

Explanation

Reviewed by Dr. Abubacker

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