To tackle this problem, we need to analyze the dynamics of the system involving the block and the ball. Let's break down each part of your question step by step.

1. Displacement of the Block When the Ball Reaches Equilibrium

When the ball is displaced by an angle θ and then released, it will swing downwards due to gravity. As it moves, it will exert a force on the string, which in turn will affect the block. The key here is to understand the conservation of momentum and the geometry of the system.

At the moment the ball reaches its lowest point (the equilibrium position), it has converted its potential energy into kinetic energy. The block will also move horizontally due to the tension in the string. The horizontal displacement of the block can be calculated using the geometry of the situation:

• The vertical drop of the ball when it reaches the lowest point is given by \( L(1 - \cos \theta) \).
• Using the principle of conservation of momentum, the horizontal displacement of the block, \( x \), can be derived as \( x = L \sin \theta \).

Thus, the displacement of the block when the ball reaches the equilibrium position is \( L \sin \theta \).

2. Tension in the String When It Is Vertical

When the ball is at the lowest point of its swing, the forces acting on it include the gravitational force and the tension in the string. At this point, the ball is moving with maximum speed, and we can apply Newton's second law to find the tension.

The forces acting on the ball can be expressed as:

• Weight of the ball: \( mg \) (acting downwards)
• Tension in the string: \( T \) (acting upwards)

At the lowest point, the net force acting on the ball is equal to the centripetal force required to keep it moving in a circular path:

Using the equation: \( T - mg = \frac{mv^2}{L} \), where \( v \) is the speed of the ball at the lowest point, we can rearrange it to find the tension:

Thus, the tension in the string when it is vertical is given by:

\( T = mg + \frac{mv^2}{L} \).

3. Maximum Velocity of the Block During Subsequent Motion

To find the maximum velocity of the block during the motion of the system, we can again use the conservation of energy principle. Initially, when the ball is at height \( h = L(1 - \cos \theta) \), it has potential energy given by:

\( PE = mgh = mgL(1 - \cos \theta) \).

As the ball swings down to its lowest point, this potential energy is converted into kinetic energy of both the ball and the block. The total kinetic energy at the lowest point can be expressed as:

\( KE = \frac{1}{2} mv^2 + \frac{1}{2} Mv_B^2 \), where \( v_B \) is the velocity of the block.

Since the system conserves momentum, we can relate the velocities of the ball and the block. The maximum velocity of the block can be derived from the energy conservation equation:

Setting the initial potential energy equal to the total kinetic energy at the lowest point allows us to solve for \( v_B \). The maximum velocity of the block can be expressed as:

\( v_B = \sqrt{2gL(1 - \cos \theta)} \).

In summary, the answers to your questions are:

• Displacement of the block: \( L \sin \theta \)
• Tension in the string when vertical: \( T = mg + \frac{mv^2}{L} \)
• Maximum velocity of the block: \( v_B = \sqrt{2gL(1 - \cos \theta)} \)