Question

A particle is performing S.H.M. Its total energy is $E$. When the displacement of the particle is half of its amplitude, its K.E will be

• A. $E$
• B. $\frac{3E}{4}$
• C. $\frac{E}{2}$
• D. $\frac{E}{4}$

Answer 1

The correct option is B $\frac{3E}{4}$

$K.E=\frac{1}{2}m{\omega }^2(A^2-y^2)$
$K.E=\frac{E}{A^2}(A^2-y^2)$ $(\because E=\frac{1}{2}m{\omega }^{2}A)$
$K.E=\frac{E}{A^2}(A^2-\frac{A^2}{4})=\frac{3E}{4}$