Question

A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Kinetic energy K = $\frac{1}{2}m{\omega }^2(a^2-y^2)$

= $\frac{1}{2}\times 10\times {\left(\frac{2\pi }{1}\right)}^2[{10}^2-5^2]=375{\pi }^2$ ergs