Question

Q 18 shm:

18. A particle performs S.H.M. of amplitude A with angular frequency $\omega$ along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}$A form 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-

$\left(A\right)\frac{\sqrt{5}}{2}A\left(B\right)\frac{\sqrt{3}}{2}A\left(C\right)\sqrt{2}A\left(D\right)\sqrt{5}A$

Answer 1

Dear Student,
$$\begin{gathered} Initiallytotalenergyoftheoscilation=\frac{1}{2}m{\omega }^2{A}^2 \\ Accordingtoquestionnowtotalenergy=\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}m{\omega }^2{A}^2=2\frac{1}{2}m{\omega }^2{A}^2=\frac{1}{2}m{\omega }^2\times 2A^2=\frac{1}{2}m{\omega }^2(\sqrt{2}A{)}^2 \\ sonewamplitudeis\sqrt{2}A \end{gathered}$$
Regard