InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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. |
|
Answer» 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)` |
|