A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at distance 2A/3 from equilibrium position. The new amplitude of the motion is.
A
3A
B
A
C
D
Text Solution
Verified by ExpertsThe correct answer is:
C
v = 
v = 
v new = 3v = 
Aꞷ
So the new amplitude is given by
v new = 
⇒ 
Aꞷ = 
A new = 
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