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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) sin [(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 = (0.86 m) sin [(70.5 m^(-1))x - (28562 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 = 2 pi f = 23562 "rad"//s` At `t = 0 , x = 0, (dy)/(dx) = "positive"` and the given curve is a since 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 = (0.06m) sin [(78.5 m^(-1)) xx - (23562 s^(-1))t]m` |
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