InterviewSolution
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InterviewSolution
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A simple harmonic progressive wave in a gas has a particle displacement of y=a at time `t=T//4` at the orgin of the wave and a particle velocity of y =v at the same instant but at a distance `x=lambda//4` from the orgin where T and `lambda` are the periodic time and wavelength of the wave respectively. then for this wave.A. the amplitude `A` of the wave is `A=2a`B. the amplitude `A` of the wave is `A=a`C. the equation of the wave can be represented by `y asin(v)/(a)[(t)-(x)/(V)]`D. The equation of the wave can be represented by `y 2a cos(v)/(a)[t-(x)/(V)]` |
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Answer» Correct Answer - b.,c. Let the equation to the wave be `y=A sin[2pi((t)/(T)-(x)/(lambda))+phi]` where `A` is the amplitude of the wave and `phi`, phase angle. It is given that `y=a`, when `x=0` and `t=T//4` and also that `y=v when `x=lambda//4` and `t=T//4` substituting in (i), `y=a=A sin((pi)/(2)+phi)` `y=v=(2piA)/(T)cos[2pi((1)/(4)-(1)/(4) )+phi]` `v=(2piA)/(T)cosphi` putting `phi=0, y=a=A`, so that amplitude `A=a` Also, `v=(2piA)/(T)[cos0]=(2pia)/(T)` `(2pi)/(T)=(v)/(a)` Hence the equation to the wave is `y=a sin(v)/(a)[t-(Tx)/(lambda)]` ` `y=a sin(v)/(a)[t-(x)/(V)]` where `V=(lambda)/(T)` is the velocity of the wave in the gas. |
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