A solid cube of side l is made to oscillate about a horizontal axis passing through one of its edges. Its time period will be

Options

A. `2pisqrt((2sqrt2)/(3)(l)/(g))`
B. `2pisqrt((2)/(3)(l)/(g))`
C. `2pisqrt((sqrt3)/(2)(l)/(g))`
D. `2pisqrt((2)/(sqrt3)(l)/(g))`

Answer

To find the time period of oscillation of a solid cube of side \( l \) about a horizontal axis passing through one of its edges, we can follow these steps:

### Step 1: Understand the Setup
We have a solid cube with side length \( l \) oscillating about a horizontal axis that passes through one of its edges. The center of mass of the cube is located at its geometric center.

### Step 2: Identify Forces and Torque
When the cube is displaced by a small angle \( \theta \), the weight \( mg \) acts vertically downward. The component of the weight acting parallel to the direction of displacement is \( mg \sin \theta \), which provides the torque necessary for oscillation.

### Step 3: Calculate the Torque
The torque \( \tau \) about the axis of rotation (which is at the edge of the cube) due to the weight of the cube is given by:
\[ \tau = mg \sin \theta \cdot d \]
where \( d \) is the perpendicular distance from the line of action of the weight to the axis of rotation. For a cube, this distance can be calculated as:
\[ d = \frac{l}{\sqrt{2}} \]
Thus, the torque becomes:
\[ \tau = mg \sin \theta \cdot \frac{l}{\sqrt{2}} \]

### Step 4: Moment of Inertia
Next, we need to calculate the moment of inertia \( I \) of the cube about the axis of rotation. The moment of inertia of a solid cube about an axis through its center of mass (perpendicular to one of its faces) is:
\[ I_{cm} = \frac{ml^2}{6} \]
Using the parallel axis theorem, the moment of inertia about the edge is:
\[ I = I_{cm} + m \left(\frac{l}{\sqrt{2}}\right)^2 = \frac{ml^2}{6} + m \cdot \frac{l^2}{2} = \frac{ml^2}{6} + \frac{3ml^2}{6} = \frac{2ml^2}{3} \]

### Step 5: Relate Torque and Angular Acceleration
According to Newton's second law for rotation:
\[ \tau = I \alpha \]
Substituting the expressions for torque and moment of inertia:
\[ mg \sin \theta \cdot \frac{l}{\sqrt{2}} = \frac{2ml^2}{3} \alpha \]

### Step 6: Small Angle Approximation
For small angles, we can use the approximation \( \sin \theta \approx \theta \):
\[ mg \theta \cdot \frac{l}{\sqrt{2}} = \frac{2ml^2}{3} \alpha \]

### Step 7: Express Angular Acceleration
The angular acceleration \( \alpha \) can be expressed as:
\[ \alpha = \frac{d^2\theta}{dt^2} \]
Substituting this into the equation gives:
\[ mg \theta \cdot \frac{l}{\sqrt{2}} = \frac{2ml^2}{3} \frac{d^2\theta}{dt^2} \]

### Step 8: Rearranging the Equation
Rearranging the equation leads to:
\[ \frac{d^2\theta}{dt^2} + \frac{3g}{2l/\sqrt{2}} \theta = 0 \]
This is a standard form of a simple harmonic motion equation, where the term \( \frac{3g}{2l/\sqrt{2}} \) is the angular frequency squared \( \omega^2 \).

### Step 9: Finding the Time Period
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Thus,
\[ \frac{2\pi}{T} = \sqrt{\frac{3g}{2l/\sqrt{2}}} \]
Squaring both sides gives:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{3g\sqrt{2}}{2l} \]
Rearranging for \( T \):
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]

### Final Answer
The time period of oscillation \( T \) is:
\[ T = 2\pi \sqrt{\frac{2l}{3g}} \]