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CGP EDU Academic Team
Published on: September 12, 2026
A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The magnitude of the velocity in $msec^{-1}$ at the centre of oscillation is
Text Solution
Verified by ExpertsThe correct answer is:
C
At center $\mathbf{V_{max}} = a \omega = a. \frac{2 \pi}{T} = \frac{0.2 \times 2 \pi}{0.01} = 40 \pi$
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