A particle of mass 0.50 kg executes a simple harmonic motion under a force F = – (50 N/m)x. If it 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 at the centre of oscillation is
`E = (1)/(2) mv^(2) = (1)/(2) (0.50 kg) (10 m//s)^(2) = 2.5 j`.
The potential energy is zero here. At the maximum displacement x = A, the speed is zero and hence the kinetic energy is zero. The potential energy here is `(1)/(2) kA^(2)` . As there is no loss of energy,
`(1)/(2) kA^(2)= 2.5 j `
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) = 2.5 j` or A = `(1)/(sqrt(10))` m.