A particle performs simple harmonic motion with amplitude A- Its speed is tripled at the instant it is at a distance 2 A -3 from equilibrium position- The new amplitude of the motion isA- A -3-41B- 3 AC- 7 A -3D- A -3

In SHM, V=√( A ^ 2 x ^ 2 ) V=√( A ^ 2 4 A ^ 2 9 nbsp;) =√( 5 A ^ 2 9 nbsp;) 3V=√( A' ^ 2 4 A ^ 2 9 nbsp;) Dividing the above two equations, ...

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Standard XII

Physics

SHM expression

A particle pe...

Question

A particle performs simple harmonic motion with amplitude $A$ . Its speed is tripled at the instant 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, we get $A^{'} = \frac{7 A}{3}$

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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, we get $A^{'} = \frac{7 A}{3}$