A particle performs S.H.M. of amplitude A with angular frequency … - Vidyadip

MCQ

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

A
$\frac{{\sqrt 5 }}{2}A$
B
$\frac{{\sqrt 3 }}{2}A$
✓
$\sqrt 2$ $A$
D
$\sqrt 5$ $A$

✓

Answer

Correct option: C.

$\sqrt 2$ $A$

c
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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