A particle of mass `0.50 kg` executes a simple harmonic motion under a forve `F = -(50 n//m)x`. If crosses the centre of oscillation with a speed of `10 m//s`, find the amplitude of the motion.

Explanation

The kinetic energy of the particle when it is the centre of oscillation is
`E = (1)/(2)mv^(2), = (1)/(2)(0.50 kg)(10 m//s)^(2), = 25 J`.
The potential energy is zero here. At the maximum displacement `x = A`, the speed is zero and hence the kinetic enery is zero. The potential energy here is `(1)/(2) kA^(2)`. As there is no loss of energy.
`(1)/(2) kA^(2) = 25 J.......(i)`
The force on the particle is given by
`F = -(50 N//m)x`.
Thus, the spring constant is `k = 50 N//m`
Equation `(i)` gives
`(1)/(2)(50 N//m)A^(2) = 25J` or, `A = 1m`