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/(2pi)sqrt((k)/(m))`
B. `(1)/(2pi)sqrt((karhog)/(m))`
C. `(1)/(2pi)sqrt((m+arhog)/(k))`
D. `(1)/(2pi)sqrt((k+arhog)/(m))`

Explanation

To solve the problem, we need to find the frequency of oscillation of a uniform cylinder that is half-submerged in a liquid and suspended from a spring. Here’s a step-by-step solution:

### Step 1: Identify the Forces at Equilibrium
At the equilibrium position, the forces acting on the cylinder must balance out. The forces include:
- The upward force from the spring, which is given by Hooke's law: \( F_{\text{spring}} = k \cdot x \)
- The upward buoyant force, which can be calculated as: \( F_{\text{buoyant}} = \rho \cdot A \cdot g \cdot \frac{L}{2} \) (since half of the cylinder is submerged)
- The downward gravitational force: \( F_{\text{gravity}} = M \cdot g \)

At equilibrium, the net force is zero:
\[ F_{\text{spring}} + F_{\text{buoyant}} - F_{\text{gravity}} = 0 \]
This implies:
\[ k \cdot x + \rho \cdot A \cdot g \cdot \frac{L}{2} - M \cdot g = 0 \]

### Step 2: Displace the Cylinder
When the cylinder is displaced downward by a small distance \( x \) and released, it will oscillate. The forces acting on it when displaced are:
- The spring force: \( F_{\text{spring}} = k \cdot x \)
- The change in buoyant force due to the additional submerged volume: \( F_{\text{buoyant}} = \rho \cdot A \cdot g \cdot x \)

The net force acting on the cylinder when displaced is:
\[ F_{\text{net}} = -k \cdot x - \rho \cdot A \cdot g \cdot x \]
This can be expressed as:
\[ F_{\text{net}} = -(k + \rho \cdot A \cdot g) \cdot x \]

### Step 3: Relate Force to Acceleration
According to Newton's second law, the net force is also equal to mass times acceleration:
\[ F_{\text{net}} = M \cdot a \]
Substituting the expression for net force gives:
\[ -(k + \rho \cdot A \cdot g) \cdot x = M \cdot a \]
Since acceleration \( a \) can be expressed as \( \frac{d^2x}{dt^2} \), we have:
\[ -(k + \rho \cdot A \cdot g) \cdot x = M \cdot \frac{d^2x}{dt^2} \]

### Step 4: Formulate the Equation of Motion
This leads to the equation of motion for simple harmonic motion:
\[ \frac{d^2x}{dt^2} + \frac{(k + \rho \cdot A \cdot g)}{M} \cdot x = 0 \]
This is of the form:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
where \( \omega^2 = \frac{k + \rho \cdot A \cdot g}{M} \).

### Step 5: Find the Frequency of Oscillation
The frequency \( f \) of oscillation is related to \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
Substituting for \( \omega \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{k + \rho \cdot A \cdot g}{M}} \]

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

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