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

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