Question

A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance
(2A)/3 from equilibrium position. The new amplitude of the motion is:

Answer 1

The velocity of a particle executing SHM at any instant, is defined as the time rate of change of its displacement at that instant.

where, ω is angular frequency, A is amplitude and x is displacement of a particle.

Suppose that the new amplitude of the motion be A'.

Initial velocity of a particle performs SHM,

where, A is initial amplitude and ω is angular frequency

Answer 2

Explanation:-

equation of SHM.

=> x = Asin(wt )

=> 2A/3 = A sinwt

=> 2/3 = sin wt

=> wt = sin-¹(2/3)