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=`

Answer

15

Explanation

`2f = 6, f = 3`, `T = (1)/(3)` sec, `v = 3 m//sec`
`3 = (lambda)/(T) rArr lambda = 1m`
`x = vt`
`3 = 3t rArr t = 1`
total time `= t + (3T)/(4) = 1 + (3)/(4) ((1)/(3)) = (5)/(4) = 1.25` sec.