A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5cm, (b) 3cm , (c) 0 cm. Here, A= 5cm =0.05 cm, T=0.2 s, omega = (2pi)/(T) = (2pi)/(0.2) = 10 pirad/s When displacement is y, then acceleration , a=-omega^2 y , Velocity , V=omega sqrt(A^2 - y^2)

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 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5cm, (b) 3cm , (c) 0 cm. Here, A= 5cm =0.05 cm, T=0.2 s, omega = (2pi)/(T) = (2pi)/(0.2) = 10 pirad/s When displacement is y, then acceleration , a=-omega^2 y , Velocity , V=omega sqrt(A^2 - y^2)
Answer» Solution :Case (a) When y= 5CM = 0.05 m, `A=-(10 pi)^2 xx0.05 =-5pi^2 m//s^2 and V= 10pi SQRT((0.05)^2- (0.05)^2) =0` Case (b) When y= 3 cm = 0.03 m, A `=-(10pi)^2 xx 0.03 =-3pi^2 m//s^2` `V= 10 A sqrt((0.05)^2 - (0.03)^2) = 10 pi xx 0.04 =0.4 pi m//s^(-1)` Case (C) = When `y=0 , A =-(10 pi)^2 xx 0 =0 , V =10pi sqrt((0.05)^2 - (0)^2)) =10pi xx 0.05 =0.5 pi m//s^(-1)`.