Question

The equation for a wave travelling in x-direction on a string is y =(3.0cm)sin[(3.14 cm^(-1) x - (314s^(-1))t]` (a) Find the maximum velocity of a particle of the string. (b) Find the acceleration of a particle at x =6.0 cm at time t = 0.11 s.

Answer 1

(a) The velocity of the particle at x at time t is `upsilon = (dely)/(delt) = (3.0cm) (-0314s^(-1) cos [3.14 s^(-1)] x - (314s ^(-1)t]`
`=(-9.4 ms^(-1) cos [(3.14 cm^(-1))x - (314 s^(-1)t].` The maximum velocity of a particle will be `upsilon = 9.4 s^(-1)`
(b) The acceleration of the particle at x at time t is `a = (delupsilon)/(delt) = -(9.4 ms^(-1)(314s^(-1)sin[(3.14 cm^(-1))x - (314 s^(-1))t]`
`=- (29052 ms^(-2) sin[3.14 cm^(-1)x -(314s^(-1))t]`.
The acceleration of the particle at x = 6.0 at time
`t = 0.11 s is a = -(2952ms^(-2)) sin[6pi - 11pi] = 0.`