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. independent of aB. proportional to `sqrt(a)`C. proportional to `a^(3//2)`D. proportional to `(1)/(sqrt(a))` |
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Answer» Correct Answer - D Potential energy `U = k|x|^(2)` Hence force `SHM F = (dU)/(dx) = -3k k|x|^(2)`….(i) Also for `SHM x= a sin omega t` and `d^(2)x)/(dt^(2)) + omega^(2) x= 0` `rArr` Acceleration `omega = ((d^(2)x)/(dt^(2)) = - omega^(2)x`…(ii) From (i) and (ii) we get `omega = sqrt((3ks)/(m))` `rArr T = (2pi)/(omega) = 2pi sqrt((m)/(3ks)) = 2pi sqrt((m)/(3k(a sin omega t)))` `rArr T prop (1)/(sqrt(a))` |
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