A solid cylinder is attached to a horizontal massless spring as shown in the figure. If the cylinder rolls without slipping, the time period of oscillation of the cylinder is

$2 π \sqrt{\frac{x}{g}}$

$2 π \sqrt{\frac{2 M}{3 K}}$

$2 π \sqrt{\frac{3 M}{8 K}}$

$2 π \sqrt{\frac{3 M}{2 K}}$

Solution

$2 π \sqrt{\frac{3 M}{8 K}}$

Let x be the extension in the spring

Then, x = 2Rθ

Therefore, Torque acting on the cylinder about point contact.

T = Fx2R = - Kx(2R) = - 4R²Kθ

Thereore, $T = 2 π \sqrt{\frac{I}{Ke}} = 2 π \sqrt{\frac{\frac{3}{2} {MR}^{2}}{4 R^{2} K}} = 2 π \sqrt{\frac{3 M}{8 K}}$

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