AP EAMCET · PHYSICS · Oscillations
A particle is executing simple harmonic motion with an amplitude of \(2 \mathrm{~m}\). The difference in the magnitudes of its maximum acceleration and maximum velocity is 4 . The time-period of its oscillation and its velocity when it is 1 \(\mathrm{m}\) away from the mean position are respectively
- A \(2 \mathrm{~s}, 2 \sqrt{3} \mathrm{~ms}^{-1}\)
- B \(\frac{7}{22} \mathrm{~s}, 4 \sqrt{3} \mathrm{~ms}^{-1}\)
- C \(\frac{22}{7} \mathrm{~s}, 2 \sqrt{3} \mathrm{~ms}^{-1}\)
- D \(\frac{44}{7} \mathrm{~s}, 4 \sqrt{3} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(\frac{22}{7} \mathrm{~s}, 2 \sqrt{3} \mathrm{~ms}^{-1}\)
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