A body describes SHM with an amplitude of 0.05 m and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5cm (b) 3 cm (c ) zero cm.

Explanation

Given, `A=0.05m, T = 0.2s`
We know that `v=omega sqrt(A^(2)-y^(2))`
and `a=-omega^(2)y`
Hence (i) for `y=0.05m, omega=(2pi)/(T)=(2xx3/142)/(0.2)="31.42 rads"^(-1)`
and `omega^(2)=(31.42)^(2)=987.2`
`v=31/42sqrt(5^(2)-5^(2))xx10^(-2)=-49.36ms^(-2)`
(ii) for `y=0.03m, v=31.42xx10^(-2)sqrt(5^(2)-3^(2))=1.2568ms^(-1)`
`a=-987.2xx0.03=-29.616ms^(-2)`
(iii) `y=0, a=0`
`v=31.42sqrt(5^(2))xx10^(-2)=1.571ms^(-1)`.