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. porprotional to `(1)/(sqrt(a))`B. independent of aC. 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)|) = ([ML^(2)T^(-2)])/([L^(3)]) = [ML^(-1)T^(-2)]` Now, time period may depend on `T prop ("mass")^(x)("amplitude")^(y) (k)^(z)` `rArr [M^(0)L^(0)T] = [M]^(x)[L]^(y)[ML^(-1)T^(-2)]^(2) = [M^(x+2)L^(y-2)T^(-2x)]` Equating powers, we get `2z = 1` or `Z = -1//2` `y -z = 0` or `y = z = -1//2` Hence, `Tprop("amplitude")^(-1//2)` `rArr T prop (a)^(-1//2) rArr T prop (1)/(sqrt(a))` |
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