Question

A particle performs S.H.M of amplitude A along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}A$ 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

• A. $\frac{\sqrt{5}}{2}A$
• B. $\frac{\sqrt{3}}{2}A$
• C. $\sqrt{2}A$
• D. $\sqrt{5}A$

Answer 1

The correct option is C $\sqrt{2}A$

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.
$\therefore \frac{1}{2}m{\omega }^2(A^{\prime }{)}^2=m{\omega }^2{A}^2$
$\Rightarrow A^{\prime }=\sqrt{2}A.$