Question:medium
A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is $50\%$ of total energy? \( (\cos 45^\circ = \frac{1}{\sqrt{2}}) \)
A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is $50\%$ of total energy? \( (\cos 45^\circ = \frac{1}{\sqrt{2}}) \)
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- $K/E = \cos^2(\omega t)$ (starting from mean)
- Use $T = \frac{2\pi}{\omega}$
Updated On: May 4, 2026
- 0.25 s
- 0.50 s
- 0.75 s
- 1 s
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The Correct Option is C
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