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A travelling wave is described by the equation `y = y_(0) sin ((ft - (x)/lambda))`. The maximum particle velocity is equal to four times the wave velocity ifA. `lambda = pi Y_(0)//4`B. `lambda = pi Y_(0)//2`C. `lambda = pi Y_(0)`D. `lambda = 2pi Y_(0)` |
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Answer» Correct Answer - B `v = (dY)/(dt) = Y_(0)cos[2pi(ft-(x)/(lambda))]xx2pif` or `v = 2pi fV_(0) cos[2pi(ft-(x)/(lambda))]` The particle velocity is maximum, when `cos[2pi(ft - x/(lambda))] = 1` `:. v_(max) = 2pifY_(0)"….."(i)` We know that, `Y = a sin (omegat - kx)` The wave velocity V is given by `V = (omega)/(k) = (2pif)/(2pi//lambda) = flambda"......."(2)` Given that `v_(max) = 4V` `:. 2 pif Y_(0) = 4flambda` or `lambda = (piY_(0))/(2)` |
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