A solid uniform cylinder of mass `m` performs small oscillations due to the action of two springs of stiffness `k` each (figure). Find the period of these oscillation in the absence of sliding.
`x`

Answer

To find the period of oscillation of a solid uniform cylinder of mass \( m \) performing small oscillations due to the action of two springs with stiffness \( k \), we can follow these steps:

### Step 1: Understanding the System
The cylinder is subjected to the restoring force from two springs when it is displaced. The displacement causes a torque about the center of the cylinder.

### Step 2: Define the Displacement
Let the angle of displacement of the cylinder be \( \theta \). The horizontal displacement of the ends of the springs due to this angle is \( 2\theta r \), where \( r \) is the radius of the cylinder.

### Step 3: Calculate the Restoring Force
The restoring force from each spring when displaced by \( 2\theta r \) is:
\[ F = k \cdot (2\theta r) \]
Since there are two springs, the total restoring force is:
\[ F_{\text{total}} = 2k(2\theta r) = 4k\theta r \]

### Step 4: Calculate the Torque
The torque \( \tau \) due to this force about the center of the cylinder is given by:
\[ \tau = F_{\text{total}} \cdot r = 4k\theta r \cdot r = 4kr^2\theta \]

### Step 5: Relate Torque to Angular Acceleration
Using Newton's second law for rotation, we have:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the cylinder about its axis and \( \alpha \) is the angular acceleration. The moment of inertia \( I \) for a solid cylinder is:
\[ I = \frac{1}{2} m r^2 \]

### Step 6: Substitute and Rearrange
Substituting the expressions for torque and moment of inertia, we get:
\[ 4kr^2\theta = \left(\frac{1}{2} m r^2\right) \alpha \]
Since \( \alpha = \frac{d^2\theta}{dt^2} \), we can rewrite this as:
\[ 4kr^2\theta = \frac{1}{2} m r^2 \frac{d^2\theta}{dt^2} \]

### Step 7: Simplify the Equation
Dividing both sides by \( r^2 \) (assuming \( r \neq 0 \)):
\[ 4k\theta = \frac{1}{2} m \frac{d^2\theta}{dt^2} \]
Rearranging gives:
\[ \frac{d^2\theta}{dt^2} + \frac{8k}{m}\theta = 0 \]

### Step 8: Identify the Angular Frequency
This is a standard form of the simple harmonic motion equation:
\[ \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \]
where \( \omega^2 = \frac{8k}{m} \). Thus, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{8k}{m}} \]

### Step 9: Calculate the Period
The period \( T \) of oscillation is given by:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{8k}} = \frac{\pi}{4} \sqrt{\frac{m}{k}} \]

### Final Answer
The period of oscillation of the cylinder is:
\[ T = \frac{\pi}{4} \sqrt{\frac{m}{k}} \]

---