A particle performs simple harmonic motion with amplitude A When it is at distance 2A/3

Question

A particle performs simple harmonic motion with amplitude A When it is at distance \(\frac{2A}{3}\) from equilibrium position its speed is tripled. The new amplitude of the motion is

(a) 3A

(b) \(A\sqrt{3}\)

(c) \(\frac{7A}{3}\)

(d) \(\frac{A}{3}\sqrt{41}\)

Answer 1

Correct option is : (c) \(\frac{7A}{3}\)

\(\Rightarrow\ 5A\omega = \omega\sqrt{A_{new}^2 - \left(\frac{2A}{3}\right)^2}\)

\(A_{new} = \frac{7A}{3}\)