AP EAMCET · PHYSICS · Oscillations
The potential energy of a simple harmonic oscillator of mass \(2 \mathrm{~kg}\) at its mean position is \(5 \mathrm{~J}\). If its total energy is \(9 \mathrm{~J}\) and amplitude is \(1 \mathrm{~cm}\), then its time period is
- A \(\frac{\pi}{100} \mathrm{~s}\)
- B \(\frac{\pi}{50} \mathrm{~s}\)
- C \(\frac{\pi}{20} \mathrm{~s}\)
- D \(\frac{\pi}{10} \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{100} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Given, total energy \(=9 \mathrm{~J}\) \(\mathrm{PE}\) at mean position \(=5 \mathrm{~J}\) So, maximum \(\mathrm{KE}=9 \mathrm{~J}-5 \mathrm{~J}=4 \mathrm{~J}\) Now, in SHM Maximum (at mean) \(\mathrm{KE}=\) Maximum PE (at extremes)…
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