AP EAMCET · PHYSICS · Oscillations
A particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is 1.5 s, then the minimum time at which the ratio of the kinetic and total energies of the particle becomes \(3: 4\) is
- A \(\frac{1}{4} \mathrm{~s}\)
- B \(\frac{1}{12} \mathrm{~s}\)
- C \(\frac{1}{8} \mathrm{~s}\)
- D \(\frac{1}{6} \mathrm{~s}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{8} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(\frac{KE}{TE} = \cos^2(\omega t)\) \(\cos^2(\omega t) = \frac{3}{4}\) \(\cos(\omega t) = \frac{\sqrt{3}}{2}\) \(\omega t = \frac{\pi}{6}\) \(\frac{2\pi}{T} t = \frac{\pi}{6}\) \(\frac{2\pi}{1.5} t = \frac{\pi}{6}\) \(t = \frac{1.5}{2 \times 6}\)…
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