A) Proportional to \[\frac{1}{\sqrt{a}}\]
B) Independent of a
C) Proportional to \[\sqrt{a}\]
D) Proportional to \[{{a}^{3/2}}\]
Correct Answer: A
Solution :
[a] \[U(x)=K|x{{|}^{3}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,[K]=\frac{[U]}{{{[x]}^{3}}}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[{{L}^{3}}]}=[M{{L}^{-1}}{{T}^{-2}}]\] Now time period may depend on \[T\propto {{(mass)}^{x}}{{(amplitude)}^{y}}{{(K)}^{z}}\] or \[[{{M}^{0}}{{L}^{0}}{{T}^{1}}]={{[M]}^{x}}{{[L]}^{y}}{{[M{{L}^{-1}}{{T}^{-2}}]}^{z}}\] \[=[{{M}^{x+z}}{{L}^{y-z}}{{T}^{-2z}}]\] Equating the powers, we get \[y-z=0\] or \[y=z=-\frac{1}{2}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,T\propto {{(amplitude)}^{-1/2}}\] or \[T\propto {{(a)}^{-1/2}}\] or \[T\propto \frac{1}{\sqrt{a}}\]
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