A particle performs SHM of amplitude $A$ along a straight line. When it is at a distance $\frac{\sqrt{3}}{2} A$ from mean position, its kinetic energy gets increased by an amount $\frac{1}{2} m ω A^{2}$ due to an impulsive force. Then its new amplitude becomes.

\frac{\sqrt{5}}{2} A

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\frac{\sqrt{3}}{2} A

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\sqrt{2} A

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\sqrt{5} A

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Solution

The correct option is C $\sqrt{2} A$
$\begin{matrix} I f t h e p a r t i c l e p e r f o r m s S H M w i t h a m p l i t u d e A a l o n g t h e s t r a i g h t l i n e, t h a n a t a n y p o i n t t h e t o t a l e n e r g y s h o u l d b e, E = K + P = \frac{1}{2} K A^{2} = \frac{1}{2} m w^{2} A^{2} I f t h e a d d i t i o n a l k i n e t i c e n e r g y o f m a g n i t u d e \frac{1}{2} m w^{2} A^{2} b y m e a n s o f a n i m p u l s i v e f o r c e, t h e n t h e t o t a l e n e r g y o f t h e p a r t i c l e c h a n g e s t o, E = E + \frac{1}{2} m w^{2} A^{2} = \frac{1}{2} m w^{2} A^{2} + \frac{1}{2} m w^{2} A^{2} \frac{1}{2} m w^{2} A^{' 2} = \frac{1}{2} m w^{2} (2 A^{2}) \Rightarrow A^{' 2} = 2 A^{2} \Rightarrow A^{'} = \sqrt{2} A \end{matrix}$
Hence,
option $(C)$ is correct answer.

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Similar questions

A

\frac{\sqrt{3}}{2} A

\frac{1}{2} m ω^{2} A^{2}

\frac{\sqrt{3}}{2} A

\frac{1}{2} m ω^{2} A^{2}

\frac{\sqrt{3}}{2}

\frac{1}{2} m ω^{2} A^{2}

\frac{\sqrt{3}}{2}

\frac{1}{2}

ω^{2} A^{2}

\frac{2 A}{3}

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