The equation of motion of a particle started at t=0 is given by `x=5sin(20t+pi/3)`, where x is in centimetre and t in second. When does the particle
a. first come rest
b. first have zero acceleration
c. first have maximum speed?

Explanation

a. `x = 5 sin (20 t + (pi)/(3))`
`(dx)/(dt) = 100 cos (20 t + (pi)/(3))`
For partical to come into rest
`(dx)/(dt) = 0`
`cos (20 t + (pi)/(3)) = cos (pi)/(2)`
`20 t = (pi)/(2) - (pi)/(3) = (pi)/(6)`
`t = (pi)/(120)`
b. The partical will have zero acceleration when it will be at equilibrium position.
As `omega = 20, 2 pi//T = 20`
`implies T = (pi)/(10) s`
Hence partical will come to equilibrium position.
`t = (pi)/(120) + (pi)/(40) = (pi)/(40)((1)/(3) + 1) = (pi)/(30)s -(pi) s`