91Ó°ÊÓ

The expression for the momentum is given by,

$\mathrm{p}=\mathrm{mv}$ ……. (1)

Here $p$is the momentum, $\mathrm{m}$is the mass, $\mathrm{v}$is the velocity.

The rate of change of momentum is known as force and it is expressed as,

$\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}$…… (2)

Here$\mathrm{F}$ is the force.

(a)

The net force is equal to the $2{\mathrm{k}}_{\mathrm{s}}\mathrm{Ó\neg }$,

Substitute $\mathrm{mv}$ for $\mathrm{p}$and $2{\mathrm{k}}_{\mathrm{s}}\mathrm{Ó\neg }$for $\mathrm{F}$into the equation (2),

$\begin{array}{c}2{\mathrm{k}}_{\mathrm{s}}\mathrm{Ó\neg }=\frac{\mathrm{d}\left(\mathrm{mv}\right)}{\mathrm{dt}}\\ =\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}\end{array}$

From the above expression,

$\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}\mathrm{x}$ …… (3)

Here $\mathrm{x}$ is the displacement.

Equation (3) is in the form $\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}={\mathrm{Ó\neg }}^2\mathrm{x}$

Here$\mathrm{Ó\neg }$ is the angular velocity,

$\begin{array}{c}{\mathrm{Ó\neg }}^2=\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}\\ \mathrm{Ó\neg }=\sqrt{\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}\end{array}$

The expression for the time period is given by,

$\mathrm{T}=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }}$

Substitute $\sqrt{\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}$ for $\mathrm{Ó\neg }$in the above equation,

$\mathrm{T}=\sqrt{2\mathrm{Ï€}\left(\frac{\mathrm{m}}{2{\mathrm{k}}_{\mathrm{s}}}\right)}$

Therefore the time period is$\sqrt{2\mathrm{Ï€}\left(\frac{\mathrm{m}}{2{\mathrm{k}}_{\mathrm{s}}}\right)}$ .

(b)

As the expression of time period is independent of length, so on increasing or decreasing the length ${\mathrm{L}}_0$, time period will not change.