In Figure, two identical springs of spring constant$\mathbf{7580}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$are attached to a block of mass$\mathbf{0}\mathbf{.}\mathbf{245}\mathbf{}\mathbf{kg}$. What is the frequency of oscillation on the frictionless floor?

Using Newton’s second law, we find the angular velocityin terms of the spring constant and mass. Then, using this angular velocity; we can find the frequency of oscillations on the frictionless floor.

Formulae:

The force on a body attached to a spring,

$\mathrm{F}=-\mathrm{kx}$ (i)

Force applied on a body in acceleration using Newton’s second law

$\mathrm{F}=\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{d{\mathrm{t}}^2}$ (ii)

The equation of displacement of a body in motion

$\mathrm{x}={\mathrm{x}}_{\mathrm{m}}\cos \left(\mathrm{\omega t}+\mathrm{\varphi }\right)$ (iii)

Frequency of a body in oscillation

$\mathrm{f}=\frac{\mathrm{\omega }}{2\mathrm{\pi }}$ (iv)

The net force exerted by the spring is$-2\mathrm{kx}$ (because of two identical springs) when the block is displaced from equilibrium and returns to its equilibrium. Therefore, using equation (i), we get

$\mathrm{F}=-2{\mathrm{kx}}_{\mathrm{m}}$ (v)

So, from both equations of force (ii) & (v), we get

$$\begin{gathered} \mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{d{\mathrm{t}}^2}=-2{\mathrm{kx}}_{\mathrm{m}} \\ \frac{{\mathrm{d}}^2\mathrm{x}}{d{\mathrm{t}}^2}=\frac{-2{\mathrm{kx}}_{\mathrm{m}}}{\mathrm{m}} \end{gathered}$$(vi)

Again, the acceleration of the body can be written as

$\begin{array}{rcl}\frac{{\mathrm{d}}^2\mathrm{x}}{d{\mathrm{t}}^2}& =& -{\mathrm{\omega }}^2{\mathrm{x}}_{\mathrm{m}}\\ -{\mathrm{\omega }}^2{\mathrm{x}}_{\mathrm{m}}& =& \frac{-2{\mathrm{kx}}_{\mathrm{m}}}{\mathrm{m}}\\ {\mathrm{\omega }}^2& =& \frac{2\mathrm{k}}{\mathrm{m}}\\ \mathrm{\omega }& =& \sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}\end{array}$

Now, substituting the above value in equation (iv), we get

$\begin{array}{rcl}\mathrm{f}& =& \frac{1}{2\mathrm{\pi }}\sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}\\ & =& \frac{1}{2\mathrm{\pi }}\sqrt{\frac{2\left(750{\displaystyle \frac{\mathrm{N}}{\mathrm{m}}}\right)}{0.245\mathrm{kg}}}\\ & =& 39.6\mathrm{Hz}\end{array}$

Therefore, frequency of oscillations on the frictionless floor is $39.6\mathrm{Hz}$.