ExamBro
ExamBro
JEE Mains · Physics · STD 11 - 13. oscillations

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 \(\alpha=\)____________.

  1. A \(11\)
  2. B \(22\)
  3. C \(12\)
  4. D \(15\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(12\)

Step-by-step Solution

Detailed explanation

\( \mathrm{V}_{\mathrm{at} \text { mann position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \) \( \omega=\frac{5}{2} \) \( \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \) \( 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app