Question

The displacement of a S.H.M. doing particle when K.E.= P.E. (Amplitude = 4 cm) is :

• A. $2\sqrt{2}cm$
• B. $2cm$
• C. $\frac{1}{\sqrt{2}}cm$
• D. $\sqrt{2}cm$

Answer 1

The correct option is A $2\sqrt{2}cm$

Kinetic energy,
$K.E.=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)$
Potential energy
$P.E.=\frac{1}{2}k{x}^2=\frac{1}{2}m{\omega }^2{x}^2$
Putting$,K.E.=P.E.$
$\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2\Rightarrow A^2=2x^2$
$x=\frac{A}{\sqrt{2}}=2\sqrt{2}cm$