Question

A particle performs S.H.M of amplitude $A$ along a straight line. When it is at 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

Answer: (C)

$\sqrt{2}A$

Given,

A S.H.M have amplitude A.

So, its total Initial Kinetic Energy of the system is $\frac{1}{2}m{\omega }^{2}A$

Energy given to the system is $\frac{1}{2}m{\omega }^{2}A$. This energy gets added to system energy.

Total final Kinetic energy is = Total initial kinetic energy + added kinetic energy

$\frac{1}{2}m{\omega }^{2}A{{}^{\prime }}^2=\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}m{\omega }^2{A}^2$

$A^{\prime }=A\sqrt{2}$

Hence, New amplitude is $A\sqrt{2}$