ExamBro
ExamBro
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____.

  1. A \(7\)
  2. B \(8\)
  3. C \(9\)
  4. D \(10\)
Verified Solution

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 \(=\mathrm{A}^{\prime}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app