A uniform rope of mass M length L hangs vertically from the ceiling, with lower end free. A distbance on the rope trvelling upwards starting from the lower end has a velocity v. At a point P at distance x from the lower end.

Options

A. Tension at point P is mg
B. `v = sqrt(xg)`
C. `v = sqrt(2xg)`
D. Tenstion at point P is `(M/L) xg `

Answer

To solve the problem step by step, we will analyze the situation of the hanging rope and the disturbance traveling upwards.

### Step 1: Determine the mass per unit length of the rope
The mass of the rope is given as \( M \) and its length is \( L \). The mass per unit length \( \mu \) can be calculated as:
\[ \mu = \frac{M}{L} \]

**Hint:** Remember that mass per unit length is the total mass divided by the total length.

### Step 2: Calculate the mass of the rope segment of length \( x \)
For a segment of the rope that is \( x \) meters long, the mass \( m_x \) of this segment can be expressed as:
\[ m_x = \mu \cdot x = \frac{M}{L} \cdot x \]

**Hint:** Use the mass per unit length to find the mass of any segment of the rope.

### Step 3: Determine the tension at point P
The tension \( T \) at point \( P \) (which is at a distance \( x \) from the lower end) is due to the weight of the rope segment below point \( P \). The weight of this segment is given by:
\[ T = m_x \cdot g = \left(\frac{M}{L} \cdot x\right) \cdot g = \frac{M \cdot g \cdot x}{L} \]

**Hint:** The tension in the rope is equal to the weight of the rope segment below the point of interest.

### Step 4: Relate the wave velocity to tension and mass per unit length
The velocity \( v \) of the wave traveling through the rope can be expressed using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the expression for tension \( T \) from step 3:
\[ v = \sqrt{\frac{\frac{M \cdot g \cdot x}{L}}{\frac{M}{L}}} \]

**Hint:** The wave speed depends on the tension and mass per unit length; make sure to simplify correctly.

### Step 5: Simplify the expression for wave velocity
Simplifying the equation from step 4:
\[ v = \sqrt{\frac{M \cdot g \cdot x}{M}} = \sqrt{g \cdot x} \]

**Hint:** Cancel out common terms carefully to arrive at a simpler expression.

### Final Results
1. The tension at point \( P \) is:
\[ T = \frac{M \cdot g \cdot x}{L} \]
2. The velocity of the wave at point \( P \) is:
\[ v = \sqrt{g \cdot x} \]