A particle performs S.H.M. of amplitude A with angular frequency ω along a straight line. When it is at a distance 2 3 A from mean position, its kinetic energy gets increased by an amount 2 1 m ω 2 A 2 due to an impulsive force. Then its new amplitude becomes :-

( 2 1 k A 2 ) + 2 1 m ω 2 A 2 = 2 1 k A ′ 2 ⇒ A ′2 = A 2 + A 2 ⇒ A ′ = 2 A

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

Options:

Answer:

Solution:

$(\frac{1}{2} k A^{2}) + \frac{1}{2} m ω^{2} A^{2} = \frac{1}{2} k A^{'} 2$

$\Rightarrow A^{'2} = A^{2} + A^{2} \Rightarrow A^{'} = 2 A$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: QB

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