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 prequency of oscillation of the cylindcer is.

Options

A. `(1)/(2 pi)((k - A rho g)/(M))^(1//2)`
B. `(1)/(2 pi)((k + A rho g)/(M))^(1//2)`
C. `(1)/(2 pi)((k + rho -gL)/(M))^(1//2)`
D. `(1)/(2 pi)((k + A - rho g)/(M))^(1//2)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the cylinder and derive the frequency of oscillation.

### Step 1: Analyze the Forces at Equilibrium
At the equilibrium position, the forces acting on the cylinder are:
- The weight of the cylinder \( W = Mg \) acting downward.
- The buoyant force \( F_b \) acting upward.

According to Archimedes' principle, the buoyant force is equal to the weight of the liquid displaced by the submerged part of the cylinder.

### Step 2: Calculate the Buoyant Force
The volume of the submerged part of the cylinder is given by:
\[ V = \text{cross-sectional area} \times \text{height submerged} = A \times \frac{L}{2} \]
Thus, the buoyant force can be expressed as:
\[ F_b = \text{density of liquid} \times \text{gravity} \times \text{volume} = \rho g \left( A \frac{L}{2} \right) \]

### Step 3: Set Up the Equilibrium Condition
At equilibrium, the buoyant force equals the weight of the cylinder:
\[ F_b = W \implies \rho g \left( A \frac{L}{2} \right) = Mg \]
We can simplify this equation by canceling \( g \) from both sides:
\[ \rho A \frac{L}{2} = M \]

### Step 4: Analyze the Oscillation
When the cylinder is pushed down by a small distance \( x \) and released, the forces acting on it will be:
- The weight \( Mg \) acting downward.
- The buoyant force \( F_b \) acting upward, which now changes because the submerged volume increases.
- The spring force \( F_s = kx \) acting upward.

The new buoyant force when the cylinder is displaced by \( x \) becomes:
\[ F_b' = \rho g \left( A \left( \frac{L}{2} + x \right) \right) = \rho g A \left( \frac{L}{2} + x \right) \]

### Step 5: Write the Equation of Motion
The net force acting on the cylinder when displaced by \( x \) is:
\[ F_{\text{net}} = F_b' - Mg - kx \]
Substituting the expressions we have:
\[ F_{\text{net}} = \rho g A \left( \frac{L}{2} + x \right) - Mg - kx \]
Using the equilibrium condition \( \rho g A \frac{L}{2} = Mg \), we can simplify:
\[ F_{\text{net}} = \rho g A x - kx \]
This can be rearranged as:
\[ F_{\text{net}} = \left( \rho g A - k \right)x \]

### Step 6: Apply Newton's Second Law
According to Newton's second law, \( F = ma \):
\[ M \frac{d^2x}{dt^2} = \left( \rho g A - k \right)x \]
This is a standard form of simple harmonic motion, where:
\[ \frac{d^2x}{dt^2} + \frac{(\rho g A - k)}{M}x = 0 \]
From this, we can identify the angular frequency \( \omega \):
\[ \omega^2 = \frac{k + \rho g A}{M} \]

### Step 7: Calculate the Frequency of Oscillation
The frequency \( f \) is given by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k + \rho g A}{M}} \]

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