Question

A particle is performing S.H.M between x = – A and x = + A. The position of particle where K.E. equals to P.E.

• A. x = 0
• B. x = + A
• C. x = + $\frac{A}{\sqrt{2}}$
• D. x = $\frac{\sqrt{3}}{2}$ A

Answer 1

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

K.E. = P.E.

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

$A^2-x^2=x^2$

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