A Particle of mass 2Kg executes k an SHM where displacement x ( in metre ) and time t ( in second) are related as x = 2 sin ` pi ` t, find
( I ) Frequency ,
(ii ) velocity at `t= 1/4 ` second and
(iii ) kinetic energy at ` t= 1/2 ` second

Explanation

Comparing the given equtaion `x=2 sin pi t ` with the equation , `x= a sin omega t ,` we get
amplitude , a= 2m , angular frequency ,` Omega = pi s ^(-1) `
` (I) "frequency" f=(omega)/( 2pi) =(pi)/( 2pi) = 0.5 s^(-1) `
Velocity of the particle , `v=(dx )/(dt) = 2 pi cos pi t `
(ii ) At `t=(1)/(4) s,` velocity `= 2 pi cos (pi . (1)/(4))`
`= 2pi cos (pi)/(4) = sqrt(2) pi m * s^(-1)`
`(iii) At t=1/2s.`
velocity `= 2pi cos (p * 1/2) = 2pi cos (pi)/(2) =0`
` therefore ` at `t=1/2s,` kinetic enery =0