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

Here, r= 5cm , 0.05 m, T= 0.2 s `omega=(2pi)/T=(2pi)/0.2 =10 pi rad//s`
when displacement, is y, then
acceleration , `A=-omega^(2) y`
velocity , `V=omegasqrt(r^(2)-y^(2))`
Case (a) when y=5 cm =0.05 m
`A=-(10pi)^(2)xx0.05=-5pi^(2)m//s^(2)`
`v=10 pisqrt((0.05)^(2)-(0.05)^(2))=0`
Case (b) when `y=3 cm =0.03 m `
`A=-(10pi)^(2)xx0.03=-3pi^(2)m//s^(2)`
`V=10pisqrt((0.05)^(2)-(0.03)^(2))=10pixx0.04=0.4 pi m//s`
Case (c) when `y=0, A=-(10pi)^(2) xx0=0`
`V=10pi sqrt((0.05)^(2)-0^(2))=10pixx0.05 =0.5 pi m//s`