A body is executing simple harmonic motion. When the displacements from the mean position is 4 cm an

Question

A body is executing simple harmonic motion. When the displacements from the mean position is 4 cm and 5 cm , the corresponding velocities of the body is $10 \mathrm{~cm} / \mathrm{sec}$ and $8 \mathrm{~cm} / \mathrm{sec}$. Then the time period of the body is\

Professor PC Thomas Classes · Answer

(c) : For simple harmonic motion velocity at\ displacement $x, v=\omega \sqrt{a^{2}-x^{2}}$\ $\begin{array}{lll}10=\omega \sqrt{a^{2}-16} & \ldots\left(	ext { (i) ; } \quad 8=\omega \sqrt{a^{2}-25}ight. \ \frac{100}{\omega^{2}}=a^{2}-16 & \ldots(	ext { iii }) ; & \frac{64}{\omega^{2}}=a^{2}-25\end{array}$\ $	herefore$ Equation(iii)-(iv) gives $\frac{36}{\omega^{2}}=9$\ $\Rightarrow \omega=2 \mathrm{rad} / \mathrm{s}$ or $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi \mathrm{sec}$\