Question

The ratio of displacement to amplitude when kinetic energy of a body performing linear S.H.M. is four times its potential energy is :

• A. $\pm \frac{1}{5}$
• B. $\pm \frac{1}{\sqrt{5}}$
• C. $\pm \frac{2}{5}$
• D. $\pm \sqrt{\frac{3}{5}}$

Answer 1

The correct option is B $\pm \frac{1}{\sqrt{5}}$

$\frac{KE}{PE}=4KE+PE=\frac{1}{2}m{\omega }^2{A}^2\Rightarrow KE(1+\frac{1}{4})=\frac{1}{2}m{\omega }^2{A}^2\Rightarrow KE=\frac{2}{5}m{\omega }^2{A}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)\Rightarrow x^2=A^2(1-\frac{4}{5})=\frac{A^2}{5}\Rightarrow \frac{x}{A}=\pm \frac{1}{\sqrt{5}}$