Physics - SHM Question with Solution | TestHub

For particle undergoing SHM,

$\mathrm{V}=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}^2}\Rightarrow {\mathrm{V}}^2={\mathrm{\omega }}^2\mathrm{ }\left({\mathrm{A}}^2-{\mathrm{x}}^2\right)$

${\mathrm{V}}_1=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}_1^2}\Rightarrow {\mathrm{V}}_1^2={\mathrm{\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}_1^2\right)$...(i)

${\mathrm{V}}_2=\mathrm{\omega }\sqrt{{\mathrm{A}}^2-{\mathrm{x}}_2^2\mathrm{ }}\Rightarrow {\mathrm{V}}_2^2={\mathrm{\omega }}^2\left({\mathrm{A}}^2-{\mathrm{x}}_2^2\right)$...(ii)

${\mathrm{V}}_1^2-{\mathrm{V}}_2^2={\mathrm{\omega }}^2\left({\mathrm{x}}_2^2-{\mathrm{x}}_1^2\right)$

$\mathrm{\omega }=\sqrt{\frac{{\mathrm{V}}_1^2-{\mathrm{V}}_2^2}{{\mathrm{x}}_2^2-{\mathrm{x}}_1^2}}$

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{x}}_2^2-{\mathrm{x}}_1^2}{{\mathrm{V}}_1^2-{\mathrm{V}}_2^2}}$