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 form measure position, its kinetic energy gets increased by an amount $\frac{1}{2}$ m ${\omega }^2{A}^2$ due to an impulsive force. Then 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)

$\sqrt{2}A$

$\begin{array}{c}Given,\\ Dis\tan cex=\frac{\sqrt{3}}{2}\\ IncreasedkineticenergyK.E=\frac{1}{2}m{\omega }^2{a}^2\\ now,\\ E=K.E+P.E\\ E=\frac{1}{2}m{\omega }^2{x}^2+\frac{1}{2}m{\omega }^2\left(a^2-x^2\right)\\ E=\frac{1}{2}m{\omega }^2{a}^2\\ Whenparticleisatadis\tan ceof\frac{\sqrt{3}}{2}afrommeanposition,itskineticenergygets\\ increasedbyanamount\frac{1}{2}m{\omega }^2{a}^{2}duetoanimpulsiveforce.\\ So,thenewtotalenergyis\\ E^{\prime }=E+\frac{1}{2}m{\omega }^2{a}^2\\ \frac{1}{2}m{\omega }^{2}a{}^{\prime 2}=\frac{1}{2}m{\omega }^2{a}^2+\frac{1}{2}m{\omega }^2{a}^2\\ a^{\prime }=\sqrt{2}a\\ Hence,theoptionCisthecorrectanswer.\end{array}$