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. independent of (a)C. proportional to `sqrt(a)`D. proportional to `a^(3//2)`. |
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Answer» Correct Answer - A `U (x) = k|x|^3` :. `[k] = ([U])/([x^3]) = (M L^2 T^-2)/(L^3) = ML ^-1 T^-2` now time period may depend on ` T prop (mass)^x (amplitude)^y (k)^z` :. `[M^0 L^0T] = [M]^x [L]^y [M L^-1 T^-2]^z` =`[M^x + z L^y -z T^(-2z)]` Equating the powers, we get `-2z = 1 or -1//2` `y - z = 0 or y = z = - 1//2` Hence `T prop (amplitude)^(-1//2) prop a^(1//2)`. |
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