A solid cylinder is attached to a horizontal massless spring so that it can roll without slipping on a horizontal surface. The spring constant k is 3N/m. If the system is released from rest, find the time period of small oscillation of centre of mass of the cylinder

2 π \sqrt{\frac{m}{k}}

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2 π \sqrt{\frac{3 m}{2 k}}

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2 π \sqrt{\frac{2 m}{3 k}}

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2 π \sqrt{\frac{m}{2 k}}

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Solution

The correct option is B $2 π \sqrt{\frac{3 m}{2 k}}$
$\begin{matrix} E = \frac{1}{2} k x^{2} + \frac{1}{2} I_{c m} ω^{2} + \frac{1}{2} m v_{c m}^{2} E = \frac{1}{2} k x^{2} + \frac{1}{2} (\frac{1}{2} m R^{2}) (v_{c m} R^{2}) + \frac{1}{2} m v_{c m}^{2} [∴ c_{c m} = R ω] E = \frac{1}{2} k x^{2} + \frac{1}{4} m v_{c m}^{2} + \frac{1}{2} m v_{c m}^{2} w e t a k e t h a t t h e e n e r g y c o n s tan t s o t h e d e r i v a t i v e o f t h e e n e r g y w i t h r e s p e c t t o t i m e i s z e r o \frac{d E}{d t} = \frac{1}{2} 2 k x \frac{d x}{d t} + \frac{3}{2} 3 m v_{c m} \frac{d v_{c m}}{d t} \frac{d E}{d t} = \frac{1}{2} 2 k x v_{c m} + \frac{3}{2} 2 m v_{c m} \frac{d v_{c m}}{d t} 0 = v_{c m} (k x + \frac{3}{2} m \frac{d^{2} x}{d t^{2}}) 0 = k x + \frac{3}{2} m \frac{d^{2} x}{d t^{2}} \frac{3}{2} m \frac{d^{2} x}{d t^{2}} + k x = 0 \frac{d^{2} x}{d t^{2}} + \frac{2 k}{3 m} x = 0 L e t t h e a n g u l a r f r e q u e n c y b e ω = \sqrt{\frac{2 k}{3 m}} t h e a n g u l a r f r e q u e n c y i s e q u a l t o ω = 2 π f a n d t i m e p e r i o d T = \frac{1}{f} T h e n ω = 2 π \frac{1}{T} T = 2 π \frac{1}{ω} T = 2 π \frac{1}{\sqrt{\frac{2 k}{3 m}}} T = 2 π \sqrt{\frac{3 m}{2 k}} . \end{matrix}$
Hence, the option $B$ is the correct answer.

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