The equetion of a progressive wave travelling along a string is given by
`y=10 sinpi(0.01x-2.00t)`
where x and y are in centimetres and t in seconds. Find the (a) velocity of a particle at x=2 m and `t=5//6` s. (b) acceleration of a particle at x=1 m and `t=1//4` s. also find the velocity amplitude and acceleration amplitude for the wave.

Explanation

The displacement equation for a particle is given by
`y=10sinpi(0.01x-2.00t) cm`
(a). Particle velocity is given by
`(dy)/(dt)=-20.0pi cos(0.01x-2.00t) cm//s`
putting `x=200 cm` and `t=5//6 s`, we get
`(dy)/(dt)=-20.0pi cospi(2-(5)/(3))cm//s`
`=-10.0pi cm//s`
Also.the velocity amplitude
`(dy)/(dt)]_(max) =20.0 pi cm//s`
`(corresponding to`|cos pi(0.01x-2.00 t)|=1)`
(b). Differentiating eq. (ii) w.r.t. time `t`, we get the particle aceleration as
`(d^(2)y)/(dt^(2))=-40.0pi^(2) sinpi(0.01pi-2.00t) cm//s^(2)`
Putting `x=100 cm` and `t =1//4 s`, we get
`(d^(2)y)/(dt^(2))=-40.0pi^(2) sinpi(1-(1)/(2))cm//s^(2)`
`=-40.0pi^(2) cm//s^(2)`
Also, the acceleration amplitude
`(d^(2)y)/(dt^(2)]_(max)=40.0pi^(2) cm//s^(2)`
`(corresponding to`|sinpi(0.01x-2.00t)|=1)`