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.

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 3 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= 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)`.