InterviewSolution
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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 `1sqrt(a)`B. independent of `a`C. proportional to `sqrt(a)`D. proportional to `a^(3//2)` |
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Answer» `V=k|x|^(3)` `F=(dv)/(dx)=-3k|x|^(2)`…….`(1)` The equation of simple harmonic motion is given as `x=a sinomega t` `rArrm(d^(2)x)/(dt^(2))=m(-aomega^(2)sinomegat)=-momega^(2).x`……`(2)` Using `(1)` and `(2)`, we obtain `3k|x|^(2)=momega^(2)xrArromega=sqrt(3kx//m)` `rArr T=2pisqrt((m)/(3kx))rArrT=2pisqrt((m)/(3kasinomegat))` `T prop (1)/(sqrt(a))` |
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