A particle performs Simple harmonic motion of amplitude A along a straight line. When it is at a distance 3^1/2/2 A from mean position its kinetic energy gets increased by an amount 1/2 mw²A^​2 due to an impulsive force. Then it's new amplitude becomes what?

Dear Student,

Please find below the solution to the asked query:

If the particle performs SHM with amplitude A along the straight line, than at any point the total energy should be,

$E=K+P=\frac{1}{2}K{A}^2=\frac{1}{2}m{\omega }^2{A}^2$.
If the additional kinetic energy of magnitude $\frac{1}{2}m{\omega }^2{A}^2$ by means of an impulsive force, then the total energy of the particle changes to,

$$\begin{gathered} E\text{'}=E+\frac{1}{2}m{\omega }^2{A}^2=\frac{1}{2}m{\omega }^2{A}^2+\frac{1}{2}m{\omega }^2{A}^2 \\ \Rightarrow \frac{1}{2}m{\omega }^{2}A{\text{'}}^2=\frac{1}{2}m{\omega }^2\left(2A^2\right) \Rightarrow A{\text{'}}^2=2A^2 \\ \Rightarrow A\text{'}=\sqrt{2}A \end{gathered}$$

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