TS EAMCET · Physics · Oscillations
For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of \(4 \mathrm{~cm}\) from the mean position is \(\frac{1}{3} \mathrm{rd}\) of the maximum kinetic energy. The amplitude of the motion is
- A \(2\sqrt {6}\)
- B \(\frac{2}{\sqrt{6}} \mathrm{~cm}\)
- C \(\sqrt 2\)
- D \(\frac{6}{\sqrt{2}} \mathrm{~cm}\)
Answer & Solution
Correct Answer
(A) \(2\sqrt {6}\)
Step-by-step Solution
Detailed explanation
Kinetic energy of the particle at \(4 \mathrm{~cm}\) \(=\frac{1}{3} \times\) maximum kinetic energy…
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