A cylinder of mass $\mathrm{M}$ and radius $\mathrm{R}$ lies on a plank of mass $\mathrm{M}$ as shown. The surface between plank and ground is smooth, and between cylinder and plank is rough. Assuming no slipping between cylinder and plank, the time period of oscillations (when displaced from equilibrium) of the system is

• A

  $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{3\mathrm{k}}}$
• B

  $4\mathrm{\pi }\sqrt{\frac{2\mathrm{M}}{3\mathrm{K}}}$
• Correct answer
  C

  $4\mathrm{\pi }\sqrt{\frac{\mathrm{M}}{3\mathrm{k}}}$
• D

  $\mathrm{\pi }\sqrt{\frac{3\mathrm{M}}{2\mathrm{K}}}$

Solution

Using no Slipping condition:



${\mathrm{a}}_{\mathrm{c}}+\mathrm{\alpha }\mathrm{R}={\mathrm{a}}_{\mathrm{p}}$

$\frac{\mathrm{f}}{\mathrm{M}}+\frac{2\mathrm{f}\mathrm{R}}{\mathrm{M}{\mathrm{R}}^2}\mathrm{R}=\frac{\mathrm{k}\mathrm{x}-\mathrm{f}}{\mathrm{M}}$

$\Rightarrow 4\mathrm{f}=\mathrm{k}\mathrm{x}$

$\therefore \mathrm{k}\mathrm{x}-\frac{\mathrm{k}\mathrm{x}}{4}=\mathrm{M}{\mathrm{a}}_{\mathrm{p}}\Rightarrow {\mathrm{a}}_{\mathrm{p}}=\left(\frac{3\mathrm{k}}{4\mathrm{M}}\right)\mathrm{x}$

$\therefore \mathrm{\omega }=\sqrt{\frac{3\mathrm{k}}{4\mathrm{M}}}$

$\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{4\mathrm{M}}{3\mathrm{k}}}=4\mathrm{\pi }\sqrt{\frac{\mathrm{M}}{3\mathrm{k}}}$

Hence, the correct answer is option: 'C'.