InterviewSolution
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InterviewSolution
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A sinusoidal wave travelling in the positive direction on a stretched string has amplitude `2.0 cm`, wavelength `1.0 m` and velocity `5.0 m//s`. At `x = 0` and ` t= 0` it is given that `y = 0` and `(dely)/(delt) lt 0`. Find the wave function `y (x, t)`.A. `Y(x,t)=(0.02 m)sin[(2pi m^(-1))x+(10pis^(-1))t]m`B. `y(x,t)=(0.02 m)cos(10pis^(-1))t+(2pi m^(-1))xm`C. `y(x,t0=(0.02 m)sin[(2pi m^(-1))x-(10pi s^(-1))t]m`D. `y(x,t)=(0.02m)sin[(pim^(-1))x+(pis^(-1))t]m` |
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Answer» Correct Answer - c We start with a general form for a rightward moving wave, `y(x,t)=A sin(kx-omegat+phi)` The amplitude given is `A=2.0 cm=0.02 m`. The wavelength is given as, `lambda=1.0 m` wave number `=k=2pi//lambda=2pi m^(-1)` Angular frequency, `omega =vk =10pi rad //s` `y(x,t)=(0.02) sin [2pi(x-5.0 t)+phi]` we are told that for `x=0, t=0`, `y=0` and `(dely)/(delt)lt0` i.e., `0.02 sin phi =0` `(as y=0)` and `-0.2 pi cos philt0` From these condition, we may conclude that `phi=2npi` where `n=0,2,4,6,......` Therefore, `y(x,t)=(0.02 m) sin [(2pi m^(-1))x-(10pi s^(-1))t] m` |
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