The total energy of a particle of mass 0.1 kg performing SHM is 0.2 J . Find its maximum speed and period if the amplitude is 2cm.

Explanation

Data : m= 0.1 kg, E = 0.2 J , A = 2 cm = ` 2xx 10^(-2)` m
(i) The total energy of the particle.
` E = KE_(max) = 1/2mv^(2) _(max)`
`v_(max) = sqrt((2E)/m) = sqrt((2xx0.2)/(0.1) = 2 m//s`
(ii) ` omega = (2pi)/T = omegaA = v_(max) = 2 m//s`
` (2piA)/T= 2m//s`
The period of SHM of the particle .
` T= (2piA)/2 = piA = 3.142 xx 2xx 10^(-2) = 0.06284 s`