A uniform cylinder of length `(L)` and mass `(M)` having cross sectional area `(A)` is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half - submerged in a liquid of density `(rho)` at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with small amplitude. If the force constant of the spring is `(k)`, the frequency of oscillation of the cylinder is.

Explanation

To find the frequency of oscillation of the uniform cylinder suspended from a spring and partially submerged in a liquid, we can follow these steps:

### Step 1: Identify the Forces Acting on the Cylinder
When the cylinder is in equilibrium, the forces acting on it include:
- The buoyant force (upward)
- The spring force (upward)
- The weight of the cylinder (downward)

### Step 2: Write the Expression for the Buoyant Force
The buoyant force \( F_b \) can be expressed as:
\[ F_b = \rho V g \]
where \( V \) is the volume of the liquid displaced by the submerged part of the cylinder. Since the cylinder is half-submerged, the volume displaced can be calculated as:
\[ V = A \cdot \frac{L}{2} \]
Thus, the buoyant force becomes:
\[ F_b = \rho \left( A \cdot \frac{L}{2} \right) g \]

### Step 3: Write the Expression for the Spring Force
The spring force \( F_s \) when the spring is extended by a distance \( x \) is given by Hooke's law:
\[ F_s = kx \]

### Step 4: Set Up the Equation of Motion
At equilibrium, the sum of the forces acting on the cylinder is zero. When displaced by a small distance \( x \) and released, the net force \( F_{net} \) acting on the cylinder is:
\[ F_{net} = F_b + F_s - Mg \]
Substituting the expressions for \( F_b \) and \( F_s \):
\[ F_{net} = \rho \left( A \cdot \frac{L}{2} \right) g + kx - Mg \]

### Step 5: Apply Newton's Second Law
According to Newton's second law, the net force is also equal to mass times acceleration:
\[ F_{net} = M a \]
where \( a \) is the acceleration of the cylinder. Since acceleration can be expressed as \( a = \frac{d^2x}{dt^2} \), we can write:
\[ M \frac{d^2x}{dt^2} = \rho \left( A \cdot \frac{L}{2} \right) g + kx - Mg \]

### Step 6: Rearrange the Equation
Rearranging gives us:
\[ M \frac{d^2x}{dt^2} + kx + \rho \left( A \cdot \frac{L}{2} \right) g - Mg = 0 \]
This can be simplified to:
\[ \frac{d^2x}{dt^2} + \frac{k}{M} x = -\frac{\rho A g L}{2M} + \frac{g}{M} \]

### Step 7: Identify the Angular Frequency
The equation of motion is in the form of simple harmonic motion (SHM):
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
where:
\[ \omega^2 = \frac{k}{M} + \frac{\rho A g}{2M} \]

### Step 8: Calculate the Frequency
The frequency \( f \) of the oscillation is given by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{M} + \frac{\rho A g}{2M}} \]

### Final Answer
Thus, the frequency of oscillation of the cylinder is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{M} + \frac{\rho A g}{2M}} \]