InterviewSolution
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InterviewSolution
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The streight shown in figure is driven at a frequency of `5.00 Hz`. The amplitude of the motion is `12.0 cm`, and the wave speed is `20.0 m//s`. Furthmore, the wave is such that `y = 0` and `t = 0`. Determine (a) the angular frequency and (b) wave number for this wave (c) Write an expression for the wave funcation. Calculate. (d) the maximum transverse speed and (e) the maximum transverse acceleration of a point on the string. |
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Answer» Correct Answer - (a) `10 pi rad//s` , (b) `pi//2 pi rad//s` , (c) `y = (12 xx 10^(-2)m) sin ((pi)/(2)x - 10 pit)` , (d) `(6)/(5) pi m//s` (e) `12 pi^(2) m//s^(2)` (a) `omega = 2pif = 10 pi` Red/sec `(b) lambda xx 5 = 20 , lambda = 4 m , K = (2pi)/(lambda) = (pi)/(2) rad//m` (c) ` y = 12 xx 10^(-2) sin (omegat - kx + phi) = 12 xx 10^(2) sin (10 pit - (pi)/(2) + phi)` At `t = 0 , x = 0 , y = 0` `(delx)/(delx) = 12 xx 10^(-2) (-(pi)/(2)) cos "(100 pit - (pi)/(2)x+phi)` At `t = 0 , x = 0` `(dely)/(delx) = -12 xx 10^(-2) ((pi)/(2)) cos phi` `(dely)/(delx)` should be positive `:. phi = pi` `y = 12 xx 10^(-2) sin (10 pit - (pi)/(2) + pi)= 12 xx 10^(-2) sin ((pi)/(2)xx-10 pit)` At `t = 0 , x = 0` `(dely)/(delx) = -12 xx 10^(-2) ((pi)/(2)) cos phi` `(dely)/(delx)` should be positive `y = 12 xx 10^(-2) sin (10pit - (pi)/(2)x + pi) = 12 xx 10^(-2) sin ((pi)/(2) x - 10 pit)` (d) `V_(max) = Aomega = 12 xx 10^(-2) xx 10 pi xx 10 pi = 12pi^(2)` (e) `A_("max") = Aomega^(2)` `= 12 xx 10^(-2) xx 10pi xx 10pi = 12 pi^(2)` |
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