Oscillations

JEE Main level, single correct, medium

52 of 194

Question 52 of 194

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:

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Answer: (B) 7A3\dfrac{7A}{3}

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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