question_answer

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance\[\frac{2A}{3}\]from equilibrium position.                          [JEE MAIN - I 3-4-2016] The new amplitude of the motion is:-

A) \[\frac{7A}{3}\]

B)      \[\frac{A}{3}\sqrt{41}\]

C)  3A                  

D)       A 3

Correct Answer: A

Solution :

[a] Let new amplitude is A' initial velocity

(1)

Where A is initial amplitude & w is angular frequency.

Final velocity

...(2)

From equation & equation (2)