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PhysicsSHMKinematicsMedium2 minPYQ_2016

PhysicsMediumsingle choice

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:

Options:

Answer:

Solution:

For Original SHM,

using $v = ω \sqrt{A^{2} - X^{2}}$

$v = ω \sqrt{A^{2} - \frac{4 A^{2}}{9}}$

$= \frac{ω \sqrt{5} A}{3}$

New SHM will be,

$3 v = ω \sqrt{A_{N}^{2} - X_{N}^{2}}$

$\frac{3 ω \sqrt{5} A}{3} = ω \sqrt{A_{N}^{2} - \frac{4 A^{2}}{9}}$

$5 A^{2} = A_{N}^{2} - \frac{4 A^{2}}{9}$

$A_{N}^{2} = \frac{49 A^{2}}{9}$

$A_{N} = \frac{7 A}{3}$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Kinematics

⏱ 2mℹ️ Source: PYQ_2016

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