A particle performs SHM of amplitude A . At x = A /2 an impulse adds kinetic energy 4 1 m ω 2 A 2 . Find the new amplitude.

New energy = 2 1 m ω 2 A 2 + 4 1 m ω 2 A 2 = 4 3 m ω 2 A 2 = 2 1 m ω 2 A ′ 2 , so A ′ 2 = 2 3 A 2 and A ′ = 3/2 A .

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Physics - SHM Question with Solution | TestHub

PhysicsSHMOscillations and SHMHard2 min

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A particle performs SHM of amplitude $A$ . At $x = A /2$ an impulse adds kinetic energy $\frac{1}{4} m ω^{2} A^{2}$ . Find the new amplitude.

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Answer:

Solution:

New energy = $\frac{1}{2} m ω^{2} A^{2} + \frac{1}{4} m ω^{2} A^{2} = \frac{3}{4} m ω^{2} A^{2} = \frac{1}{2} m ω^{2} A'^{2}$ , so $A'^{2} = \frac{3}{2} A^{2}$ and $A' = 3/2 A$ .

Subject:PhysicsTopic:SHMSubtopic:Oscillations and SHM

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