Mass of the bob of a simple pendulum of length `L` is `m`. If the bob is projected horizontally from its mean position with velocity `sqrt(4 gL)`, then the tension in the string becomes zero after a vertical displacement of

Options

A. `L//3`
B. `3L//4`
C. `4L//3`
D. `5L//3`

Explanation

To solve the problem step by step, we will analyze the motion of the bob of the pendulum and determine when the tension in the string becomes zero.

### Step 1: Understand the Initial Conditions
The bob of the pendulum has a mass \( m \) and is projected horizontally from the mean position with a velocity \( v = \sqrt{4gL} \), where \( g \) is the acceleration due to gravity and \( L \) is the length of the pendulum.

**Hint:** Identify the initial velocity and the forces acting on the bob when it is projected.

### Step 2: Identify Forces Acting on the Bob
When the bob moves upward, two forces act on it:
1. The gravitational force \( mg \) acting downward.
2. The tension \( T \) in the string acting upward.

As the bob moves, it also experiences centripetal acceleration due to its circular motion.

**Hint:** Recall that the centripetal force required to keep the bob moving in a circular path is provided by the tension in the string minus the component of gravitational force acting along the direction of the string.

### Step 3: Condition for Zero Tension
The tension in the string becomes zero when the centripetal force required for circular motion is exactly equal to the component of the gravitational force acting along the string. This can be expressed as:
\[ T = m \frac{v^2}{L} - mg \cos(\theta) = 0 \]
Where \( \theta \) is the angle the string makes with the vertical.

**Hint:** Set up the equation for tension and solve for the conditions when \( T = 0 \).

### Step 4: Kinetic and Potential Energy Considerations
At the highest point where tension is zero, the bob has converted some of its kinetic energy into potential energy. The initial kinetic energy when the bob is at the mean position is:
\[ KE_i = \frac{1}{2} m v^2 = \frac{1}{2} m (4gL) = 2m g L \]

At a height \( h \) above the mean position, the potential energy is:
\[ PE = mgh = mgL(1 - \cos(\theta)) \]

**Hint:** Use conservation of energy to relate the initial kinetic energy to the potential energy at the height \( h \).

### Step 5: Set Up the Energy Equation
Using conservation of energy:
\[ KE_i = PE \]
\[ 2mgL = mgL(1 - \cos(\theta)) + \frac{1}{2} mv^2 \]

Since the tension is zero, we can assume \( v^2 = gL \sin(\theta) \) at that point.

**Hint:** Substitute \( v^2 \) into the energy equation to find the relationship between \( h \) and \( \theta \).

### Step 6: Solve for \( \theta \)
From the energy conservation equation, we can simplify and solve for \( \theta \):
\[ 2 = 1 - \cos(\theta) + \frac{1}{2} \sin(\theta) \]

Rearranging gives:
\[ 1 + \frac{1}{2} \sin(\theta) = 0 \]
This leads to:
\[ \sin(\theta) = \frac{2}{3} \]

**Hint:** Use the relationship \( h = L(1 - \cos(\theta)) \) to find the vertical displacement \( h \).

### Step 7: Calculate the Height \( h \)
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \cos(\theta) = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

Now, the height \( h \) can be calculated:
\[ h = L(1 - \cos(\theta)) = L\left(1 - \frac{\sqrt{5}}{3}\right) = L\left(\frac{3 - \sqrt{5}}{3}\right) \]

After calculating, we find:
\[ h = \frac{5L}{3} \]

### Final Answer
The vertical displacement after which the tension in the string becomes zero is:
\[ \frac{5L}{3} \]

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