A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the…
Physics · JEE Advanced · NTA Exams — Oscillations and Waves
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
- \(\frac{1}{2 \pi \sqrt{3}}\)
- \(2 \pi \sqrt{3}\)
- \(\frac{2 \pi}{\sqrt{3}}\)
- \(\frac{\sqrt{3}}{2 \pi}\)
Answer
(C) 2 3
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