Question

The K.E and P.E of a particle executing SHM with amplitude A will be equal when its displacement is

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

Answer 1

The correct option is C $\frac{A}{\sqrt{2}}$

Let the displacement be $x$

$KE=\frac{1}{2}K(A^2-x^2)$

$PE=\frac{1}{2}K{x}^2$

Given:

$KE=PE$

$\frac{1}{2}K(A^2-x^2)$$=$$\frac{1}{2}K{x}^2$

$x=\frac{A}{\sqrt{2}}$