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 (a) 5cm, (b) 3cm, (c) 0 cm.

Explanation

A = 5 cm = 0.05 m, T = 0.2 s
`omega = (2pi)/(T) = (2pi)/ (0.2) = 10 pi` rad/s
(a) When y = 5 cm = 0.05m
Acceleration ` = a = -omega^(2)y`
` =-(10pi)^(2) xx 0.05 = -5pi^(2)m//s^(2)`
velocity `= V = omegasqrt((A^(2) - y^(2))`
` = 10 pisqrt((0.05)^(2) = (0.05)^(2)) =0`
(b) When y = 3 cm = 0.03 m
`A = -omega^(2)y = -(10pi)^(2) xx 0.03 = -3pi^(2) m//s^(2)`
`V = omegasqrt(A^(2) - y^(2)) = 10 pisqrt((0.05)^(2) - (0.03)^(2))`
` = 0.4 pi m//s`
(c ) When y = 0, `A = -omega^(2)y = -(10pi)^(2) xx 0 = 0`
`V = omega(A^(2) - y^(2)) = 10 pi sqrt((0.05)^(2) -0)`
` = 10 xx pi xx 0.05 = 0.5 pi m//s`