A particle performs linear S.H.M. When the displacement of the particle from mean position is 3 cm and 4 cm , corresponding velocities are \(8 \mathrm{~cm} / \mathrm{s}\) and \(6 \mathrm{~cm} / \mathrm{s}\) respectively. Its periodic time is

We have \(v=\omega \sqrt{A^2-x^2}\)
Substituting the given values in the above equation we get,
\(8=\omega \sqrt{A^2-9}\)
...(i) \(\left(\because \mathrm{x}=3 \mathrm{~cm}, \mathrm{v}=8 \mathrm{cms}^{-1}\right)\)
\(6=\omega \sqrt{\mathrm{A}^2-16}\)
...(ii) \(\left(\because \mathrm{x}=4 \mathrm{~cm}, \mathrm{v}=6 \mathrm{cms}^{-1}\right)\)
Diving equation (i) and (ii),
\(\begin{array}{ll}
& \frac{8}{6}=\frac{\sqrt{\mathrm{A}^2-9}}{\sqrt{\mathrm{~A}^2-16}} \\
& \frac{4}{3}=\frac{\sqrt{\mathrm{A}^2-9}}{\sqrt{\mathrm{~A}^2-16}} \\
\therefore \quad & \frac{16}{9}=\frac{\mathrm{A}^2-9}{\mathrm{~A}^2-16} \\
\therefore \quad & 16 \mathrm{~A}^2-256=9 \mathrm{~A}^2-81 \\
\therefore \quad & 7 \mathrm{~A}^2=175 \\
\therefore \quad & \mathrm{~A}^2=\frac{175}{7}
\end{array}\)
Substituting this value of \(\mathrm{A}^2\) in equation (ii) and solving we get
\(\begin{aligned}
6 & =\omega \sqrt{\frac{175}{7}-16} \\
6 & =\omega \sqrt{\frac{175-112}{7}}=\omega \sqrt{9} \\
6 & =3 \omega \\
\omega \quad & =2 \\
\therefore \quad \omega & =\frac{2 \pi}{T}=2 \\
\therefore \quad T & =\pi \text { seconds }
\end{aligned}\)