InterviewSolution
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InterviewSolution
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A detector at `x=0` receives waves from three sources each of amplitude `A` and frequencies `f + 2,f` and `f-2`. The equation of waves are : `y_(1)=A sin[2 pi(f+2),y_(2)=A sin 2 pift` and `y_(3)=A sin[2pi (f-2)t]`. The time at which intensity is minimum isA. `t=0,1//4,1//2,3//4,…sec`B. `t=1//6,1//3,2//3,5//6,…sec`C. `t=0,1//2,3//2,5//2,..sec`D. `t=1//2,1//4,1//6,1//8,…sec` |
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Answer» Correct Answer - B `y=y_(1)+y_(2)+y_(3)=A sin 2 pift+A sin[2 pi(f-2)t]+Asin[2 pi(f+2)t]=A sin 2pi ft+2 A sin 2 pi ft cos 4 pi t` `= A[1+2 cos4 pi t]sin 3 pift=A_(0) sin 2pit` [where `A_(0)`= Amplitude of the resultant oscillation `=A[1+2 cos 4 pi t`] Intensity `prop A_(0)^(2) :. I prop(1+2 cos 4 pi t)^(2)` For maxima or minima of the intensity `(dI)/(dt)=0rArr 2(1+2 cos 4pit)(2)(-sin 4pit)4 pi=0 rArr 1+2 cos 4 pit =0` or `sin 4 pit =0` `:. cos 4pit=-(1)/(2)rArr 4pi t=2pin pm (2pi)/(3) :. t=(n)/(2)+(1)/(6)rArr t=(1)/(6),(1)/(3),(2)/(5),(5)/(6)....` (point of minimum intensity). |
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