A particle performs simple harmonic motion with amplitude A. It’s speed is trebled at the instant that it is a distance $$2A/3$$ from equilibrium position. The new amplitude of the motion is

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Infinity Learn · Accepted Answer

initial velocity of a particle performs SHM $$v=wA2-x2given$$  $$x=23Av=wA2-2A/32v2=w2A2-2A32------(1)Where$$ A is initial amplitude, w is angular frequencyFinal $$velocity(3v)2=w2A12-2A32-----(2)From$$ $$equation(1)$$ and (2) we $$get19=A2-4A29(A1)2-4A29$$ $$(A1)2-4A29=9A2-4A29$$                    $$=9A2-9$$ $$4A29$$             $$(A1)2=5A2+4A29$$                       $$=49A29$$                     $$A1=$$ $$7A$$ $$3$$                   .new Amplitude is $$A1$$

A particle performs simple harmonic motion with amplitude A. It’s speed is trebled at the instant that it is a distance 2A/3 from equilibrium position. The new amplitude of the motion is

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