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A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is:
- A√3
- 7A/3
- A/3 √41
- 3A
Correct answer: 7A/3
Solution
At x = 2A/3, v^2 = w^2(A^2 - 4A^2/9) = 5w^2A^2/9. Tripling speed gives v'^2 = 9*v^2 = 5w^2A^2. New amplitude: 5w^2A^2 = w^2(A'^2 - 4A^2/9) -> A'^2 = 49A^2/9 -> A' = 7A/3.
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