Question

A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

• A. 5
• B.
• C.
• D. 3

Answer 1

The correct option is C

K.E=UE

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{1}{2}m{\omega }^2{x}^2$ $(v=(\omega \sqrt{A^2-x^2},K=m{\omega }^2)$

$2x^2=A^2$

$x=\frac{A}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}$