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) A√(3) (b) 7A/3 (c) A/3√(41) (d) 3A

Answer: (b). At x = 2A/3: v = ω√(A² - 4A²/9) = √(5)/3ω A. New speed v' = 3v = √(5) ω A; ω is unchanged (same mass and spring). A'² = x² + (v'²)/(ω²) = 4A²/9 + 5A² = (49A²)/(9). A' = 7A/3, so the answer is (b).

A particle performs simple harmonic motion with amplitude A. Its… | JEE Main 2016

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Velocity & Acceleration in SHM

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. The new amplitude of the motion is:

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Answer: (b)

At $x = \frac{2A}{3}$ : $v = \omega\sqrt{A^2 - \frac{4A^2}{9}} = \frac{\sqrt{5}}{3}\omega A$ .
New speed $v' = 3v = \sqrt{5}\,\omega A$ ; $\omega$ is unchanged (same mass and spring).
$A'^2 = x^2 + \frac{v'^2}{\omega^2} = \frac{4A^2}{9} + 5A^2 = \frac{49A^2}{9}$ .
$A' = \frac{7A}{3}$ , so the answer is (b).

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