JEE Mains · Physics · STD 11 - 13. oscillations
A simple harmonic oscillator has an amplitude \(A\) and time period \(6 \pi\) second. Assuming the oscillation starts from its mean position, the time required by it to travel from \(x=A\) to \(x=\frac{\sqrt{3}}{2} A\) will be \(\frac{\pi}{x}\) s, where \(x=\) _______.
- A \(2\)
- B \(12\)
- C \(4\)
- D \(9\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
From phasor diagram particle has to move from \(\mathrm{P}\) to \(Q\) in a circle of radius equal to amplitude of \(SHM\) \( \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} \) \( \phi=\frac{\pi}{6}\) Now, \(\frac{\pi}{6}=\omega t\)…
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