91影视

Using Newton鈥檚 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鈥檚 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

$\mathrm{x}={\mathrm{x}}_{\mathrm{m}}\cos \left(\mathrm{蝇迟}+\mathrm{蠒}\right)$ (iii)

Frequency of a body in oscillation

$\mathrm{f}=\frac{\mathrm{蝇}}{2\mathrm{蟺}}$ (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

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

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

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

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