InterviewSolution
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InterviewSolution
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An object of mass `0.2 kg` executes simple harmonic oscillation along the `x - axis`with a frequency of `(25//pi) Hz`. At the position `x = 0.04`, the object has Kinetic energy of `0.5 J` and potential energy `0.4 J. The amplitude of oscillations is……m. |
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Answer» `x = 0.04 m, K.E. = 0.5 J and P.E. = 0.4 J` `T.E. = (0.5 + 0.4) J = 0.9 J` Now, `T.E. = (1)/(2)m omega^2 a^2 = (1)/(2) m xx 4 pi^2 v^2 a^2` rArr `0.9 = (1)/(2) xx 0.1 xx 4 pi^2 xx (25)/(pi) xx (25)/(pi) xx a^2 rArr a = 0.06 m`. |
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