A particle is making simple harmonic motion along the X-axis. If at a distances x₁ and x₂ from the mean position, the velocities of the particle are v₁ and v₂ respectively, then the time period of its oscillation is given as

Velocity of particle performing simple harmonic motion,

$\mathrm{v}=\pm \mathrm{\omega }\sqrt{\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)}$           ...(i)

where, A = amplitude,

$\omega$ = angular frequency,

and x = displacement covered by the particle from its mean position.

According to given situation,

${\mathrm{v}}_1=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}_1^2}$               ...(ii)

and ${\mathrm{v}}_2=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}_2^2}$     ...(iii)

On squaring Eq. (ii), we get

${\mathrm{v}}_1^2={\mathrm{\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}_1^2\right)$           ...(iv)

On squaring Eq. (iii), we get

${\mathrm{v}}_2^2={\mathrm{\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}_2^2\right)$           ...(v)

Subtracting Eq. (iv) from Eq. (v), we get

$\begin{array}{l}{\mathrm{v}}_1^2-{\mathrm{v}}_2^2={\mathrm{\omega }}^2\left({\mathrm{x}}_2^2-{\mathrm{x}}_1^2\right)\\ \Rightarrow {\mathrm{v}}_1^2-{\mathrm{v}}_2^2={\left(\frac{2\mathrm{\pi }}{\mathrm{T}}\right)}^2\left({\mathrm{x}}_2^2-{\mathrm{x}}_1^2\right)\\ \Rightarrow \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\left({\mathrm{x}}_2^2-{\mathrm{x}}_1^2\right)}{\left({\mathrm{v}}_1^2-{\mathrm{v}}_2^2\right)}}\end{array}$