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The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in meters. The period of small oscillation (in sec) isA. `2pi`B. `pi`C. `(pi)/(2)`D. `sqrt2pi` |
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Answer» Correct Answer - C `F=-(dU)/(dx)=-8sin2x` For small oscillation, `sin2x=2x` i.e., Since `amu-x`, the oscillation are simple harmonic in nature. `T=pisqrt(|(x)/(a)|)=2pisqrt((1)/(16))=(pi)/(2)s` |
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