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A man generates a symmetrical pulse in a...

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 s`

B

`1s`

C

`11/12 s `

D

`23/24 s`

Text Solution

Verified by Experts

The correct Answer is:
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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