A) \[2\pi \sqrt{\frac{M}{k}}\]
B) \[2\pi \sqrt{\frac{2M}{k}}\]
C) \[2\pi \sqrt{\frac{M}{2k}}\]
D) \[2\pi \sqrt{\frac{3M}{2k}}\]
Correct Answer: D
Solution :
When cylinder is slightly displaced and released, let at an instant, v be the velocity of the cylinder, \[\omega \] be its angular velocity and x be the extension in the spring. Total energy of the system is \[E=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}k{{x}^{2}}\] \[=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}\left( \frac{1}{2}M{{R}^{2}} \right){{\omega }^{2}}+\frac{1}{2}k{{x}^{2}}\] \[\left[ \because \,\,\,I=\frac{1}{2}M{{R}^{2}} \right]\] \[=\frac{1}{2}m{{v}^{2}}+\frac{1}{4}m{{v}^{2}}+\frac{1}{2}k{{x}^{2}}=\frac{3}{4}m{{v}^{2}}+\frac{1}{2}k{{x}^{2}}\] As energy is conserved, so \[\frac{dE}{dt}=0=\frac{3}{2}Mv\frac{dv}{dt}+kx\frac{dx}{dt}\] or \[\frac{dv}{dt}=-\frac{2k}{3M}x\] or Acceleration \[a=\frac{dv}{dt}=-\frac{2k}{3M}x\] ..... (i) Comparing (i) with \[a=-{{\omega }^{2}}x,\] we have \[\omega =\sqrt{\frac{2k}{3M}}\]\[\therefore \] \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{3M}{2k}}\]
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