If a simple pendulum of length l has maximum angular displacement `theta`, then the maximum velocity of the bob is

Options

A. `sqrt((L)/(g))`
B. `2 sqrtgl sin ((theta)/(2))`
C. `sqrt(2gl) sin theta`
D. `sqrt(2g l)`

Answer

To find the maximum velocity of the bob of a simple pendulum with length \( l \) and maximum angular displacement \( \theta \), we can use the principles of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
In a simple pendulum, the total mechanical energy is conserved. This means that the sum of kinetic energy (KE) and potential energy (PE) at any point in the swing is constant.

### Step 2: Identify Points of Interest
Let’s denote:
- Point A: The highest point of the swing (maximum angular displacement \( \theta \)).
- Point B: The lowest point of the swing (mean position).

At point A, the bob has maximum potential energy and zero kinetic energy. At point B, the bob has maximum kinetic energy and minimum potential energy.

### Step 3: Write the Energy Equations
At point A (maximum height):
- Kinetic Energy \( KE_A = 0 \)
- Potential Energy \( PE_A = mgh \) where \( h \) is the height of the bob above the lowest point.

At point B (lowest point):
- Kinetic Energy \( KE_B = \frac{1}{2} mv^2 \)
- Potential Energy \( PE_B = 0 \) (taking this point as the reference level).

### Step 4: Calculate the Height \( h \)
The height \( h \) can be expressed in terms of the length of the pendulum \( l \) and the angle \( \theta \):
\[ h = l - l \cos \theta = l(1 - \cos \theta) \]

### Step 5: Set Up the Conservation of Energy Equation
Using the conservation of energy:
\[ PE_A = KE_B \]
\[ mgh = \frac{1}{2} mv^2 \]
Substituting \( h \):
\[ mg(l(1 - \cos \theta)) = \frac{1}{2} mv^2 \]

### Step 6: Simplify the Equation
Cancel out the mass \( m \) from both sides:
\[ g(l(1 - \cos \theta)) = \frac{1}{2} v^2 \]

### Step 7: Solve for Maximum Velocity \( v \)
Rearranging gives:
\[ v^2 = 2g(l(1 - \cos \theta)) \]
Taking the square root:
\[ v = \sqrt{2g(l(1 - \cos \theta))} \]

### Step 8: Use the Identity for \( 1 - \cos \theta \)
Using the trigonometric identity:
\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \]
Substituting this into the equation for \( v \):
\[ v = \sqrt{2g(l \cdot 2 \sin^2\left(\frac{\theta}{2}\right))} = \sqrt{4g l \sin^2\left(\frac{\theta}{2}\right)} \]
Thus, the maximum velocity \( v \) of the bob is:
\[ v = 2 \sqrt{g l} \sin\left(\frac{\theta}{2}\right) \]

### Final Answer
The maximum velocity of the bob at the mean position is:
\[ v = 2 \sqrt{g l} \sin\left(\frac{\theta}{2}\right) \]

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