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 \(\alpha=\)____________.
- A \(11\)
- B \(22\)
- C \(12\)
- D \(15\)
Answer & Solution
Correct Answer
(C) \(12\)
Step-by-step Solution
Detailed explanation
\( \mathrm{V}_{\mathrm{at} \text { mann position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \) \( \omega=\frac{5}{2} \) \( \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \) \( 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \)…
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