Question

A spring of force constant k of 5Nm (F=-kx) is placed horizontally on a smooth table .One end of the spring is fixed and a mass X of 0.20kg is attached to the free end. X is displaced at a distance of 4mm and released .Show that the motion of X is simple harmonic

Answer 1

$$ma=F=-kx\\\frac{d^2x}{dt^2}+\frac{k}{m}x=0$$

For the simple harmonic motion:

$$\frac{d^2x}{dt^2}+\omega^2x=0$$

Thus, the motion of X is simple harmonic with angular frequency

$$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{5}{0.2}}=5\frac{rad}{s}$$