91Ó°ÊÓ

A small block of mass is attached to a spring with stiffness ${\mathrm{k}}_{\mathrm{s}}$and relaxed length The other end of the spring is fastened to a fixed point on a low-friction table. The block slides on the table in a circular path of radius R > L.

The block's speed v.The spring exerts a tension force on the block, with the tension force in the spring being determined by

${\mathbf{F}}_{\mathbf{T}}\mathbf{=}{\mathbf{k}}_{\mathbf{s}}\mathbf{x}$ ……………………. (1)

Where${\mathbf{k}}_{\mathbf{s}}\mathbf{x}$denotes the spring's stiffness and x denotes the spring's stretched length or elongation. And it's equal to the difference between the spring's original length and its final length after extension, which we can figure out by

x = R - L

The expression will be entered into equation (1) , and it will take the form

${\mathbf{F}}_{\mathbf{T}}\mathbf{=}{\mathbf{k}}_{\mathbf{s}}\mathbf{(}\mathbf{R}\mathbf{-}\mathbf{L}\mathbf{)}$ ………………… (2)

The rate of change here is related to the perpendicular rate of change and equals the centrifugal force${\mathbf{F}}_{\mathbf{c}}$ that exists due to the tension force; thus, the tension force is provided by

${\mathbf{F}}_{\mathbf{T}}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{R}}$

The change in distance over time is the speed. As a result of v = d / t ,the block moves over the circumference of a circle when it completes one circuit. The circumference is calculated using the formula = $2\mathrm{Ï€¸é}$, where R is the circle's radius. As a result, the block's speed is determined by

$\mathrm{v}=\frac{2\mathrm{Ï€¸é}}{\mathrm{t}}$ …………………………. (3)

The expression of have to be put into equation (3),So the time is:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{T}}=\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ {\mathrm{F}}_{\mathrm{T}}=\frac{\mathrm{m}}{\mathrm{R}}{\left(\frac{2\mathrm{Ï€¸é}}{\mathrm{t}}\right)}^2 \\ \mathrm{t}=\sqrt{\frac{{\left(2\mathrm{Ï€}\right)}^2\mathrm{mR}}{{\mathrm{F}}_{\mathrm{T}}}}(\mathrm{Solve}\mathrm{for}\mathrm{it}) \\ \mathrm{t}=2\mathrm{Ï€}\sqrt{\frac{\mathrm{mR}}{{\mathrm{k}}_{\mathrm{s}}(\mathrm{R}-\mathrm{L})}}({\mathrm{F}}_{\mathrm{T}}={\mathrm{k}}_{\mathrm{s}}(\mathrm{R}-\mathrm{L}) \end{gathered}$$

Time taken for the block to go around once is $2\mathrm{Ï€}\sqrt{\frac{\mathrm{mR}}{{\mathrm{k}}_{\mathrm{s}}(\mathrm{R}-\mathrm{L})}}$.