Q.A body of mass 0.2 kg performs linear S.H.M. It experiences a restoring force of 0.2 N when its displacement from the mean position is 4 cm. Determine

SHM's restoring force gives a second-order linear differential equation.

In linear SHM, the restoring force on a particle is directly proportional to its displacement from the mean position and always directed towards it:

$F = -kx$

By Newton's second law, $F = m\dfrac{d^2x}{dt^2}$, so …

Linear SHM arises when the restoring force (and hence acceleration) is proportional to displacement and directed opposite to it: $F=-kx$.

Consider a particle of mass $m$ performing linear SHM about a mean position, restrained by a linear restoring force

$F = -kx$

where $x$ is the displacement from the mean position and $k$ is the force constant. By Newton's second law,

$m\frac{d^2x}{dt^2} = -kx$

$\frac{d^2x}{dt^2} = -\frac{k}{m}x$

Writing $\omega^2 = k/m$ (a positive constant, $\omega$ being the angular frequency of oscillation), this becomes the standard differential equation of linear SHM:

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

…