Physics - SHM Question with Solution | TestHub

Velocity $v=\omega \sqrt{A^2-x^2}$ and acceleration $={\omega }^{2}x$ Given, $\omega \sqrt{A^2-x^2}={\omega }^{2}x$ or $\sqrt{A^2-x^2}=\omega x$ Given, $$T=\frac{2\pi }{\sqrt{3}}$$ and $$\omega =\frac{2\pi }{T}=\sqrt{3}$$ Substituting the value of $\omega$ in Eq (i), we get $\sqrt{A^2-x^2}=\sqrt{3}x$ $$\Rightarrow A=2x$$ $$\begin{array}{rl}\text{ As amplitude }& =\frac{\text{ path length }}{2}=2\mathrm{cm}\\ \Rightarrow x=1\mathrm{cm}& \end{array}$$