Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A particle of mass 200 gm executes S.H.M. The restoring force is provided by a spring of force constant 80 N / m. The time period of oscillations is
Text Solution
Verified by ExpertsThe correct answer is:
A
$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.2}{80}} = 0.31 sec$
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