A body describes simple harmonic motion with an amplitude of 5 cm and a period of `0.2s`. Find the acceleration and velocity of the body when the displacement is 3 cm.

Explanation

Here, `A= 5 cm =0.05 m, T= 0.2s`
`therefore omega = (2pi)/(T) = (2pi)/(02) = 10pi rad s^(-1)`
When displacement is y, then acceleration `a= -omega^(2) y" and velocity " v= omega sqrt(A^(2)- y^(2))`
When `y= 3 cm = 0.03 m`.
acceleration `a= -omega^(2)y`
`= -(10pi)^(2) xx 0.03 = -3pi^(2) ms^(-2)`
and velocity `v= omega sqrt(A^(2) -y^(2))`
`= 10pi (sqrt((0.05)^(2) - (0.05)^(2)))`
`=10pi (sqrt(0.0025- 0.0009))`
`= 10pi (sqrt(0.0016)) = 10pi xx 0.04`
`therefore v = 0.4pi ms^(-1)`.