A particle of mass 0.50 kg executes a simple harmonic motion under a force `F=-(50Nm^-1)x`. If it crosses the centre of oscillation with a speed of `10ms^-1`, find the amplitude of the motion.

Explanation

The kinetic energy of the partical when it is mean position. It is also its total energy as the potential energy is zero here .
`E = (1)/(2) m v^(2) = (1)/(2) (0.50 kg) (10 m//s)^(2) = 25 J`
As the maximum displacement `x = A`, the speed is zero and hence the kinetic energy here is `1//2 kA^(2)`. At this position it will also its total energy . Hence
`(1)/(2) kA^(2) = 25J` (i)
The force on the partical is gives by `F = - (50N//m)x`
Thus, the spring constant is `k = 50 N//m`
Equation (i) gives
`(1)/(2) ( 50 N//m) A^(2) = 25 J implies A = 1m`