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InterviewSolution
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A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.A. proportional to `(1)/(sqrta)`B. independent of aC. proportional to `sqrta`D. proportaional to `a^(3//2)` |
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Answer» Correct Answer - A `u=k|x|^(3)` `U_("max")=ka^(3)` The oscillation energy is given by `(1)/(2)ma^(2)omega^(2)`. `therefore" "(1)/(2)ma^(2)omega^(2)=ka^(2)a` or `omega^(2)=(2k)/(m)a` or `omega=sqrt([((2k)/(m))a])` `therefore" "T=(2pi)/(omega)=2pisqrt((m)/(2ka))` `therefore" "Tprop(1)/(sqrta)` |
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