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JEE Main 20163 April
Exam
JEE Main
Year
2016
Date
3 April
Subject
Physics
Chapter
Oscillations

JEE Main 2016 · 3 April · Physics · Oscillations

A particle performs simple harmonic motion with amplitude A. Its...

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 2A3\dfrac{2A}{3} from equilibrium position. The new amplitude of the motion is:

Options

  1. AA3A\sqrt{3}
  2. B7A3\dfrac{7A}{3}
  3. CA341\dfrac{A}{3}\sqrt{41}
  4. D3A3A

Correct answer

B. 7A3\dfrac{7A}{3}

Step-by-step solution

For Original SHM,

using v=ωA2−X2v = \omega\sqrt{{A}^{2} - {X}^{2}}

v=ωA2−4A29v = \omega\sqrt{{A}^{2} - \dfrac{4{A}^{2}}{9}}

=ω5A3= \dfrac{\omega\sqrt{5}A}{3}

New SHM will be,

3v=ωAN2−XN23v =\omega\sqrt{{A}_{N}^{2} - {X}_{N}^{2}}

3ω5A3=ωAN2−4A29\dfrac{3\omega\sqrt{5}A}{3} = \omega\sqrt{{A}_{N}^{2} - \dfrac{4{A}^{2}}{9}}

5A2=AN2−4A295{A}^{2} = {A}_{N}^{2} - \dfrac{4{A}^{2}}{9}

AN2=49A29{A}_{N}^{2} = \dfrac{49{A}^{2}}{9}

AN=7A3{A}_{N} = \dfrac{7A}{3}
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