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

Step-by-step solution

The velocity of a particle in simple harmonic motion is given by v = A^2 - x^2 At distance x = 2A 3 , the initial velocity is: v₁ = A^2 - ( 2A 3 )^2 = A^2 - 4A^2 9 = 5 3 A The speed is tripled at this instant, so the new velocity is: v₂ = 3v₁ = 3 ( 5 3 A ) = 5 A Let the new amplitude be A' . The position x and angular frequency remain unchanged. Using the velocity formula for the new amplitude: v₂^2 = ^2 (A'^2 - x^2) Substituting the values of v₂ and x : ( 5 A)^2 = ^2 (A'^2 - ( 2A 3 )^2 ) 5A^2 = A'^2 - 4A^2 9 A'^2