JEE Mains · Physics · STD 11 - 13. oscillations
The displacement of simple harmonic oscillator after \(3\) seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is \(\dots \; s\)
- A \(6\)
- B \(8\)
- C \(12\)
- D \(36\)
Answer & Solution
Correct Answer
(D) \(36\)
Step-by-step Solution
Detailed explanation
\(X=A \sin \omega t\left(t=3, X=\frac{A}{2}\right)\) \(\Rightarrow \frac{A}{2}=A \sin 3 \omega\) \(\Rightarrow \sin 3 \omega=\frac{1}{2}\) \(\Rightarrow 3 \omega=\frac{\pi}{6}\) \(\Rightarrow \omega=\frac{\pi}{18}=\frac{2 \pi}{T}\) \(\Rightarrow T=36 \; s\)
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