Question

A particle performs S.H.M of amplitude $A$ along a straight line. When it is at a distance $\frac{\sqrt{3}A}{2}$ from mean position, its kinetic energy gets increased by an amount $\frac{1}{2}m{\omega }^2{A}^2$ due to an impulsive force. Then its new amplitude becomes $nA$. Find $n$:

Answer 1

Due to impulse, the total energy of the particle becomes :
$\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}m{\omega }^2{A}^2=m{\omega }^2{A}^2$
Let $A^{\prime }$ be the new amplitude.
$\therefore \frac{1}{2}m{\omega }^2(A^{\prime }{)}^2=m{\omega }^2{A}^2\Rightarrow A^{\prime }=\sqrt{2}A.$

Why this question? Tip- Mechanical energy in S.H.M. is conserved only if no external force except the one responsible for S.H.M. acts along the line of motion, however the expression for it is $\frac{1}{2}m{\omega }^2{A}^2$ only, where $A$ adjusts as per the energy changes.