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A man generates a symmetrical pulse in a string by moving his hand up and down. At `t=0` the point in his hand moves downward. The pulse travels with speed of `3 m//s` on the string & his hands passes `6` times in each second from the mean position. Then the point on the string at a distance `3m` will reach its upper extreme first time at time `t=`A. 1.25 sB. 1sC. `11//12 s `D. 23/24 s |
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Answer» Correct Answer - C `f = 3Hz` `:. Omega = 2pif = (6pi) "rad"//s ` `k = omega/v = (6pi)/3 = (2pi) "rad"//m` ` y = A sin (kx-omegat)` ` = A sin (2pix - 6pit)` Putting `y= +-A` and `x = 3`, we get ` (2pi)(3) - 6pi t = pi/2 ` `:. 6pit = 11/2 pi ` ` :. t = 11/12s `. |
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