What is a simple pendulum? Find an expression for the time period and frequency of a simple pendulum.

Answer

### Step-by-Step Solution

**Step 1: Definition of a Simple Pendulum**
A simple pendulum consists of a mass (called the bob) attached to a string or rod of fixed length (L) that swings back and forth under the influence of gravity. The motion is periodic and can be approximated as simple harmonic motion (SHM) when the angle of displacement is small.

**Step 2: Analyzing Forces and Torque**
When the pendulum is displaced by a small angle (θ), the gravitational force acting on the bob can be resolved into two components: one acting along the direction of the swing (restoring force) and the other acting perpendicular to it. The torque (τ) about the pivot point due to the weight of the bob can be expressed as:
\[ \tau = -MgL \sin(\theta) \]
For small angles, we can use the approximation \(\sin(\theta) \approx \theta\) (in radians), so:
\[ \tau \approx -MgL \theta \]

**Step 3: Moment of Inertia**
The moment of inertia (I) of the pendulum about the pivot point is given by:
\[ I = mL^2 \]
where m is the mass of the bob.

**Step 4: Relating Torque to Angular Acceleration**
According to Newton's second law for rotation, the torque is also related to angular acceleration (α):
\[ \tau = I \alpha \]
Substituting the expressions for torque and moment of inertia, we have:
\[ -MgL \theta = mL^2 \alpha \]

**Step 5: Expressing Angular Acceleration**
Since angular acceleration is the second derivative of angular displacement with respect to time, we can write:
\[ \alpha = \frac{d^2\theta}{dt^2} \]
Thus, we can rewrite the equation as:
\[ -MgL \theta = mL^2 \frac{d^2\theta}{dt^2} \]

**Step 6: Simplifying the Equation**
Dividing through by mL^2 gives:
\[ -\frac{g}{L} \theta = \frac{d^2\theta}{dt^2} \]
This is a standard form of the equation for simple harmonic motion, where:
\[ \frac{d^2\theta}{dt^2} + \frac{g}{L} \theta = 0 \]

**Step 7: Identifying Angular Frequency**
From the standard form of SHM, we can identify:
\[ \omega^2 = \frac{g}{L} \]
Thus, the angular frequency (ω) is:
\[ \omega = \sqrt{\frac{g}{L}} \]

**Step 8: Finding the Time Period (T)**
The time period (T) of a simple pendulum is related to angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for ω gives:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]

**Step 9: Finding the Frequency (f)**
The frequency (f) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \]

### Final Expressions
- Time Period (T):
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
- Frequency (f):
\[ f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \]