JEE Mains · Physics · STD 11 - 13. oscillations
The point \(A\) moves with a uniform speed along the circumference of a circle of radius \(0.36\, m\) and covers \(30^{\circ}\) in \(0.1\, s\). The perpendicular projection \('P'\) from \('A'\) on the diameter \(MN\) represents the simple harmonic motion of \('P'.\) The restoration force per unit mass when \(P\) touches \(M\) will be ...... \(N\)

- A \(100\)
- B \(0.49\)
- C \(50\)
- D \(9.87\)
Answer & Solution
Correct Answer
(D) \(9.87\)
Step-by-step Solution
Detailed explanation
\(30^{\circ} \rightarrow 0.1\, s\) \(360^{\circ} \rightarrow 1.2 \,s = T\) \(\omega=\frac{2 \pi}{ T }=\frac{5 \pi}{3}\) At \(M , F = m \omega^{2}\, A \Rightarrow \frac{ F }{ m }=\omega^{2}\, A\)
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