A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 2 A 3 from equilibrium position. The new amplitude of the motion is:

Step-by-step solution

For Original SHM, using v = ω A 2 - X 2   v = ω A 2 - 4 A 2 9 = ω 5 A 3 New SHM will be, 3 v =  ω A N 2 - X N 2 3 ω 5 A 3 = ω A N 2 - 4 A 2 9 5 A 2 = A N 2 - 4 A 2 9 A N 2 = 49 A 2 9 A N = 7 A 3