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____.
Pick an option to check your answer.
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 \(=\mathrm{A}^{\prime}\)
\(\mathrm{v}^{\prime}=3 \mathrm{v}=\sqrt{5} \mathrm{~A} \omega=\omega \sqrt{\left(\mathrm{A}^{\prime}\right)^2-\left(\frac{2 \mathrm{~A}}{3}\right)^2}\)
\(\mathrm{A}^{\prime}=\frac{7 \mathrm{~A}}{3}\)
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