A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $\frac{2 A}{3}$ from equilibrium position. The new amplitude of the motion is.

\frac{A}{3} \sqrt{41}

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3 A

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A \sqrt{3}

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\frac{7 A}{3}

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Solution

The correct option is D $\frac{7 A}{3}$
In SHM, $V = \sqrt{A^{2} - x^{2}}$

$V = \sqrt{A^{2} - \frac{4 A^{2}}{9}} = \sqrt{\frac{5 A^{2}}{9}}$

$3 V = \sqrt{{A^{'}}^{2} - \frac{4 A^{2}}{9}}$

Dividing the above two equations give $A^{'} = \frac{7 A}{3}$

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A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 2A3 from equilibrium position. The new amplitude of the motion is.

The correct option is D 7A3In SHM, V=√A2−x2V=√A2−4A29=√5A293V=√A′2−4A29Dividing the above two equations give A′=7A3

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $\frac{2 A}{3}$ from equilibrium position. The new amplitude of the motion is.

The correct option is D $\frac{7 A}{3}$
In SHM, $V = \sqrt{A^{2} - x^{2}}$

$V = \sqrt{A^{2} - \frac{4 A^{2}}{9}} = \sqrt{\frac{5 A^{2}}{9}}$

$3 V = \sqrt{{A^{'}}^{2} - \frac{4 A^{2}}{9}}$

Dividing the above two equations give $A^{'} = \frac{7 A}{3}$