Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Kinetic energy =$\frac{1}{2}m{\omega }^2(A^2-Y^2)$
$=\frac{1}{2}\left(10\right)\left(\frac{4{\pi }^2}{4}\right)({10}^2-5^2)=375{\pi }^{2}ergs$