A small ball of density `4rho_(0)` is released from rest just below the surface of a liquid. The density of liquid increases with depth as `rho=rho_(0)(1+ay)` where `a=2m^(-1)` is a constant. Find the time period of its oscillation. (Neglect the viscosity effects).

Options

A. `(2pi)/(sqrt(5))sec`
B. `(pi)/(sqrt(5))sec`
C. `(pi)/(2sqrt(5))sec`
D. `(3pi)/(2sqrt(5))sec`

Answer

To find the time period of oscillation of a small ball of density \(4\rho_0\) released just below the surface of a liquid with varying density, we can follow these steps:

### Step 1: Understand the problem
The ball has a density of \(4\rho_0\) and is submerged in a liquid whose density increases with depth according to the formula:
\[ \rho = \rho_0(1 + ay) \]
where \(a = 2 \, \text{m}^{-1}\) and \(y\) is the depth below the surface.

### Step 2: Determine the equilibrium position
At the equilibrium position, the buoyant force acting on the ball equals its weight. The buoyant force \(F_b\) can be expressed as:
\[ F_b = \rho \cdot V \cdot g \]
where \(V\) is the volume of the ball and \(g\) is the acceleration due to gravity. The weight \(W\) of the ball is:
\[ W = 4\rho_0 \cdot V \cdot g \]

Setting the buoyant force equal to the weight:
\[ \rho_0(1 + ay) \cdot V \cdot g = 4\rho_0 \cdot V \cdot g \]

### Step 3: Simplify the equation
Dividing both sides by \(Vg\) (assuming \(V \neq 0\) and \(g \neq 0\)):
\[ \rho_0(1 + ay) = 4\rho_0 \]
This simplifies to:
\[ 1 + ay = 4 \]
Thus:
\[ ay = 3 \quad \Rightarrow \quad y = \frac{3}{a} = \frac{3}{2} \, \text{m} \]

### Step 4: Analyze the motion
When the ball is displaced from its equilibrium position by a small distance \(y\), the net force acting on it can be derived from the change in buoyancy:
\[ F = -\Delta \rho \cdot V \cdot g \]
where \(\Delta \rho\) is the change in density of the liquid at the new depth \(y\):
\[ \Delta \rho = \rho_0(1 + a(y + \frac{3}{2})) - 4\rho_0 \]
This simplifies to:
\[ \Delta \rho = \rho_0(1 + ay + \frac{3a}{2}) - 4\rho_0 \]
Substituting \(y = 0\) (for equilibrium):
\[ \Delta \rho = \rho_0(1 + 3) - 4\rho_0 = 0 \]

### Step 5: Calculate the restoring force
The restoring force can be expressed as:
\[ F = -\left(\rho_0(1 + a(y + \frac{3}{2})) - 4\rho_0\right)gV \]
For small oscillations, we can approximate:
\[ F \approx -\left(2\rho_0 a y\right) gV \]
This leads to:
\[ F = -k y \quad \text{where} \quad k = 2\rho_0 a g V \]

### Step 6: Write the equation of motion
The equation of motion for simple harmonic motion is given by:
\[ m \frac{d^2y}{dt^2} = -k y \]
where \(m = 4\rho_0 V\). Thus:
\[ 4\rho_0 V \frac{d^2y}{dt^2} = -2\rho_0 a g V y \]
This simplifies to:
\[ \frac{d^2y}{dt^2} + \frac{2a g}{4} y = 0 \]

### Step 7: Identify the angular frequency
The angular frequency \(\omega\) is given by:
\[ \omega^2 = \frac{2ag}{4} = \frac{ag}{2} \]

### Step 8: Calculate the time period
The time period \(T\) is given by:
\[ T = 2\pi \sqrt{\frac{1}{\omega^2}} = 2\pi \sqrt{\frac{2}{ag}} \]
Substituting \(a = 2 \, \text{m}^{-1}\) and \(g \approx 10 \, \text{m/s}^2\):
\[ T = 2\pi \sqrt{\frac{2}{2 \cdot 10}} = 2\pi \sqrt{\frac{1}{10}} = 2\pi \cdot \frac{1}{\sqrt{10}} = \frac{2\pi}{\sqrt{10}} \]

### Final Answer
Thus, the time period of oscillation is:
\[ T = \frac{2\pi}{\sqrt{10}} \, \text{s} \]