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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 mw^{2}A^{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,

$E\text{'}=E+\frac{1}{2}m{\omega}^{2}{A}^{2}=\frac{1}{2}m{\omega}^{2}{A}^{2}+\frac{1}{2}m{\omega}^{2}{A}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}m{\omega}^{2}A{\text{'}}^{2}=\frac{1}{2}m{\omega}^{2}\left(2{A}^{2}\right)\Rightarrow A{\text{'}}^{2}=2{A}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A\text{'}=\sqrt{2}A$

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