A particle performs simple harmonic motion with amplitude A. It’s speed is trebled at the instant that it is at 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 $$=v=wA2-x2$$ given $$x=2A3$$;A is initial amplitude, w is angular frequency $$v=wA2-2A32$$ squaring both sides $$v2=w2A2-2A32---(1)$$ speed is trebled at the instant that it is at a distance $$2A/3$$  let new amplitude be $$A1$$ (3v)2=w2A12-2A32---(2) dividing $$eqn(2)$$ with $$eqn(1)$$ $$(3v)2v2=w2A12-2A32w2A2-2A32$$ $$9=A12-2A32A2-2A32$$ $$A1=7A3$$

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

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