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 :
\[U=k|x{{|}^{3}}\Rightarrow F==-\frac{dU}{dx}=-3k|x{{|}^{2}}\] ...(i) Also, for SHM \[x=a\sin \omega \,t\] and \[\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{\omega }^{2}}x=0\] \[\Rightarrow \]acceleration \[=\frac{{{d}^{2}}x}{d{{t}^{2}}}=-{{\omega }^{2}}x\Rightarrow F=ma\] \[=m\frac{{{d}^{2}}x}{d{{t}^{2}}}=-m{{\omega }^{2}}x\] ...(ii) From equation (i) & (ii) we get \[\omega =\sqrt{\frac{3kx}{m}}\] \[\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3kx}}=2\pi \sqrt{\frac{m}{3k(a\sin \omega \,t)}}\]\[\Rightarrow T\propto \frac{1}{\sqrt{a}}\].
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