Obtain an equation for the frequency of oscillation of spring of force constant k to which a mass “m” is attached.

Consider an ideal spring of spring constant “k” of one end is attached to a mass “m” and its free end is fixed to a rigid support “s” and the system is in equilibrium at “O” as shown in fig.

The mass “m” pulled down through a distance “y” and released,; Due to restoring force in a spring, the body made by an vertical oscillations between y and –y about its mean position “o”.

$\therefore$The restoring force F $\propto -$ y here –ve sign indicate that F and y are in opposite direction

$\therefore \mathrm{F} =-\mathrm{ky}$ here k is spring constant

But F = ma

$\therefore \mathrm{ma}=-\mathrm{ky}\left(\text{ or }\right)\mathrm{a}=-\left(\frac{\mathrm{k}}{\mathrm{m}}\right)\mathrm{y}-\left(1\right)$

Here k/m is constant 

$\therefore a\propto -\mathrm{y}\text{ its gives the S.H.M }$

But in S.H.M $\mathrm{a}=-{\mathrm{\omega }}^2\mathrm{y} \to \left(2\right)$

From equ (1) and (2)

$-{\mathrm{\omega }}^2\mathrm{y}=-\left(\frac{\mathrm{k}}{\mathrm{m}}\right)\mathrm{y}\left(\text{ or }\right)\mathrm{\omega }=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

$\therefore$The time period of loaded spring

$\mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=\frac{2\mathrm{\pi }}{\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}}\text{ (or) }\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

Now the frequency of loaded spring

$\mathrm{n}=\frac{1}{\mathrm{T}}=\frac{1}{2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}}\Rightarrow \mathrm{n}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$