JEE Mains · Physics · STD 11 - 13. oscillations
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____.
- A \(7\)
- B \(8\)
- C \(9\)
- D \(10\)
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 \(=\mathrm{A}^{\prime}\)…
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