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 `(1)/(sqrt(a))`B. proportional to `a`C. proportional to `sqrt(a)`D. proportional to `a^(1//2)` |
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Answer» Correct Answer - A `U = k(x)^(2) rArr T = - (dU)/(dx) = - 3k(x)^(2)`….(i) Also the `SHM x = a sin omega t amA(A^(2)x)/(dt^(2)) + omega^(2)x = 0` `rArr "acceleration" (d^(2)x)/(dt^(2)) = - omega^(2)x rArr F = ma` `= m(d^(2)x)/(dt^(2)) = - m omega^(2)x` From equation (i) and (ii) we get `omega = sqrt((3kx)/(m))` `rArr T = (2pi)/(omega) = 2pi sqrt((m)/(3kx)) = 2pi sqrt((m)/(3k(a sin omegat)) rArr T prop (1)/(sqrt(a))` |
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