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Physics Question on Oscillations

A particle of mass m is executing oscillation about the origin on the x-axis. Its potential energy is U(x)=k[x]3, U (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 1/2 1/ \sqrt 2

B

independent of a

C

proportional to a \sqrt a

D

proportional to a3/2 a^{3/2}

Answer

proportional to 1/2 1/ \sqrt 2

Explanation

Solution

U(x)=k∣x∣3U(x) = k |x|^3
∴   [k]=[U][x3]=[ML2T−2][L3]=[ML−1T−2]\therefore \, \, \, [k] = \frac{ {[ U]}}{ [x^3]} = \frac{[ ML^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]
Now, time period may depend on
T∝(mass)x(amplitude)y(k)zT \propto (mass )^x (amplitude)^y (k)^z
[M0L0T]=[M]x[L]y[ML−1T−2]z[M^0 L ^0 T ] = [M ]^x [L]^y [ML^{-1} T^{-2}]^z
       =[Mx+zLy−zT−2z]\, \, \, \, \, \, \, = [ M^{x+z} L^{y-z} T^{-2z }]
Equating the powers, we get
powers, we get
-2z = 1 or z = -l/2

y - z = 0 or y = z = - 1/2

Hence, T∝(amplitude)−1/2∝(a)−1/2T \propto (amplitude)^{-1/2} \propto (a)^{-1/2}
or       T∝1a\, \, \, \, \, \, T \propto \frac{1}{\sqrt a}