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Two radio stations broadcast their programmes at the same amplitude A, and at slightly different frequenciesomega_(1) and omega_(2) respectively , whereomega_(2) - omega_(1) = 10^(3) Hz. A detector receives the signals from the two stations simultaneously . It can emit signals only of intensity ge 2 A^(2). (i). Find the time intervals between successive maxima of the intensity of the signal received by the detector . ii. Find the time for which the detector remains idle in each cycle of the intensity of the signal . |
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Answer» Solution :i. Let the signal waves be given by ` y_(1) A sin 2 pi omega_(1) t,y_(2) = A sin 2pi omega_(2) t`. The RESULTANT disturbance is given by ` y = y_(1) + y_(2) = A sin 2 pi omega_(1) t + A sin 2 pi omega_(2) t` `= 2 A sin ( 2 pi ( omega_(1) + omega_(2)) t)/( 2) cos ( 2 pi ( omega_(2) - omega_(1))t)/(2)` `= 2 A cos pi ( omega_(2) - omega_(1)) t sin 2 pi (( omega_(1) + omega_(2)) t)/(2)` Let `omega_(1) = omega , omega_(2) = omega + Delta omega` Therefore , ` omega_(1) + omega_(2) ~~ 2 omega` ` y = 2 A cos pi ( omega _(2) - omega_(1))t sin 2 pi omega t` Thus , the resultant disturbance has amlitude ` 2A cos ( omega_(2) - omega_(1)) t` For maxima : `cos pi ( omega_(2) - omega_(1))t = +- 1` or ` pi ( omega_(2) - omega_(1)) t = r pi , r = 0 , 1,2 , 3` ` t = (r)/( omega_(2) - omega_(1)) = 0 , (1)/( omega_(2) - omega_(1)) , (2) /( omega_(2) - omega_(1) ,...` CLEARLY time interval between successive maxima ` = (1)/( omega_(2) - omega_(1)) = (1)/( 10^(3)) = 10^(-3) s` ii. The resultant intensity is given by `I = A_(1) ^(2) + A_(2)^(2) + 2 A_(1) A_(2) cos delta` when ` delta = 0 ` , intensity is maximum `I_(max) = 4 A^(2)` When `Delta = pi//2, "intensity" I = 2 A^(2)` When ` Delta = pi , "intensity" I_(min) = 0` When `Delta = 3pi//2, "intensity" I = 2A^(2)` When `Delta = 2pi , "intensity" I_(max) = 4 A^(2)` The DETECTOR remaining idlefrom ` theta = pi//2 to 3 pi //2` or in each half cycle . Hence the required time ` t = (T)/(2) = (10^(-3))/(2) = 5 xx 10^(-4) s` |
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