InterviewSolution
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InterviewSolution
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For the wave shown in figure, write the equation of this wave if its position is shown at `t= 0`. Speed of wave is `v = 300m//s`. A. `y=(0.06 m)cos[(78.5 m^(-1))x+(23562 s^(-1))t]m`B. `y=(0.06 m)sin[(78.5 m^(-1))x-(23562 s^(-1))t]m`C. `y=(0.06 m)sin[(78.5 m^(-1))x+(23562 s^(-1))t]m`D. `y=(0.06 m)cos[(78.5 m^(-1))x-(23562 s^(-1))t]m` |
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Answer» Correct Answer - b The amplitude, `A=0.06 m` `(5)/(2)lambda=0.2 m` `:. Lambda=0.08 m` `f=(v)/(lambda)=(300)/(0.08)=3750 Hz` `k=(2pi)/(lambda)=78.5 m^(-1)` and `omega=2pi f=23562 rad//s` At `T=0,x=0, (dy)/(dx)=`positive and the given curve is a sine curve. Hence, equation of wave travelling in positive `x-`direction should have the form, `y(x,t)=A sin (kx-omegat)` substituting the values, we have `y=(0.06 m) sin [(78.5 m^(-1))x-(23562 s^(-1))t] m` |
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