MHT CET · Physics · Oscillations
Starting from mean position, a body oscillates simple harmonically with a period 'T'. After what time will its kinetic energy be \(75 \%\) of the total energy? \(\left(\sin 30^{\circ}=0.5\right)\)
- A \(\frac{\mathrm{T}}{8}\)
- B \(\frac{\mathrm{T}}{12}\)
- C \(\frac{\mathrm{T}}{16}\)
- D \(\frac{\mathrm{T}}{24}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{T}}{12}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{KE}=75 \% \text { of TE } \\ \therefore & \frac{1}{2} \mathrm{ma}^2 \omega^2 \cos ^2 \omega \mathrm{t}=\frac{75}{100} \times \frac{1}{2} \mathrm{ma}^2 \omega^2 \\ \therefore & \cos ^2 \omega \mathrm{t}=\frac{3}{4} \\ \therefore \quad & \cos \omega \mathrm{t}=\frac{\sqrt{3}}{2} \\ \therefore \quad & \omega \mathrm{t}=\frac{\pi}{6} \\ \therefore \quad & \frac{2 \pi}{\mathrm{~T}} \mathrm{t}=\frac{\pi}{6} \quad \therefore \mathrm{t}=\frac{\mathrm{T}}{12}\end{array}\)
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