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 0 cm.

A body describes simple harmonic motion with an amplitude of 5 cm and

1.	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 0 cm.
Answer» Solution :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= 0 cm`. acceleration `a= -omega^(2)y` `= -(10pi)^(2) xx 0 = 0` and velocity `v= omega sqrt(A^(2) -y^(2))` `= 10pi (sqrt((0.05)^(2) - (0)^(2)))` `=10pi xx 0.05` `= 0.5pi ms^(-1)`.