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

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is \(\frac{2 \mathrm{~A}}{3}\). The new amplitude of motion is \(\frac{\mathrm{nA}}{3}\). The value of \(\mathrm{n}\) is____.

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Answer & Solution

Correct Answer

(A) \(7\)

Step-by-step Solution

Detailed explanation

\(v=\omega \sqrt{A^2-x^2}\)

\(\text { at } x=\frac{2 A}{3}\)

\(v=\omega \sqrt{A^2-\left(\frac{2 A}{3}\right)^2}=\frac{\sqrt{5} A \omega}{3}\)

New amplitude \(=\mathrm{A}^{\prime}\)

\(\mathrm{v}^{\prime}=3 \mathrm{v}=\sqrt{5} \mathrm{~A} \omega=\omega \sqrt{\left(\mathrm{A}^{\prime}\right)^2-\left(\frac{2 \mathrm{~A}}{3}\right)^2}\)

\(\mathrm{A}^{\prime}=\frac{7 \mathrm{~A}}{3}\)