Question

A particle executes SHM with amplitude $0.1 \,m$. The distance from the mean position where the $K.E.$ and $P.E.$ become equal is

• A. 2 m
• B. 0.05 m
• C. 0.05$\sqrt {2} m$
• D. $2 \sqrt {0.05} m.$

Answer 1

Answer: (C)

0.05$\sqrt {2} m$

Answer 2

We know that $E_k = \frac{1}{2} m \omega^2(A^2 = y^2)$ and $E_p = \frac{1}{2}m \omega^2 y^2$ $\therefore$ $\frac{1}{2} m \omega^2(A^2 - y^2) = \frac{1}{2} m \omega^2 y^2$ or, $A^2 - y^2 = y^2$ or $2 \, y^2 = A^2$ $\therefore$ $y = \pm \frac{A}{\sqrt{2}} = \frac{0.1 }{\sqrt{2}} = 0.05\sqrt{2} \, m$