A particle performs $S.H.M.$ of amplitude $A$ with angular frequency $\omega$ along a straight line. Whenit is at a 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
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Due to impulse force, the total energy of the particle becomes
$\frac{1}{2} m \omega^{2} A^{2}+\frac{1}{2} m \omega^{2} A^{2}=m \omega^{2} A^{2}$
Let $A^{\prime}$ be the new amplitude. (apply energy conservation law)
$\frac{1}{2} m_{\omega}^{2}\left(A^{\prime}\right)=m_{\omega}^{2} A^{2}$
${A}^{\prime}=\sqrt{2} A$
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