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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2024
  • Hard
  • Numerical

A particle executes simple harmonic motion with an amplitude of \(4 \mathrm{~cm}\). At the mean position, velocity of the particle is \(10 \mathrm{~cm} / \mathrm{s}\). The distance of the particle from the mean position when its speed becomes \(5 \mathrm{~cm} / \mathrm{s}\) is \(\sqrt{\alpha} \mathrm{cm}\), where \(\alpha=\)____________.

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Answer & Solution

Correct Answer

(C) \(12\)

Step-by-step Solution

Detailed explanation

\( \mathrm{V}_{\mathrm{at} \text { mann position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \)

\( \omega=\frac{5}{2} \)

\( \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \)

\( 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \)

\( \mathrm{x}=\sqrt{12} \mathrm{~cm}\)

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