Derive an expression for the velocity of pulse in a in stretched in string under a tension `T` and `mu` is the mass per unit length of the sting.

Answer

To derive the expression for the velocity of a pulse in a stretched string under tension \( T \) with mass per unit length \( \mu \), we can follow these steps:

### Step 1: Understand the relationship between tension, mass, and velocity
The velocity of a wave on a string is influenced by the tension in the string and the mass per unit length. The basic relationship we want to derive involves these two quantities.

### Step 2: Consider the forces acting on a small segment of the string
When a pulse travels along the string, consider a small segment of the string. The tension \( T \) acts on both ends of this segment. The net force acting on this segment can be analyzed using Newton's second law.

### Step 3: Set up the equation of motion
For a small segment of the string, the net upward force due to tension can be expressed as:
\[ F = T \sin(\theta_1) - T \sin(\theta_2) \]
where \( \theta_1 \) and \( \theta_2 \) are the angles the string makes with the horizontal at the ends of the segment.

### Step 4: Use the small angle approximation
For small angles, we can use the approximation \( \sin(\theta) \approx \theta \). Thus, we can rewrite the force as:
\[ F \approx T \theta_1 - T \theta_2 \]

### Step 5: Relate the angles to the displacement
Assuming the segment has a length \( \Delta x \) and a small vertical displacement \( y \), we can relate the angles to the displacement:
\[ \theta_1 \approx \frac{y_1}{\Delta x}, \quad \theta_2 \approx \frac{y_2}{\Delta x} \]
where \( y_1 \) and \( y_2 \) are the vertical displacements at the two ends of the segment.

### Step 6: Substitute the angles into the force equation
Substituting the expressions for \( \theta_1 \) and \( \theta_2 \) into the force equation gives:
\[ F \approx T \left( \frac{y_1 - y_2}{\Delta x} \right) \]

### Step 7: Apply Newton’s second law
According to Newton's second law, the net force is also equal to mass times acceleration. The mass of the segment is \( \mu \Delta x \), and the acceleration can be expressed as \( \frac{\partial^2 y}{\partial t^2} \):
\[ F = \mu \Delta x \frac{\partial^2 y}{\partial t^2} \]

### Step 8: Equate the forces
Setting the expressions for force equal gives:
\[ T \left( \frac{y_1 - y_2}{\Delta x} \right) = \mu \Delta x \frac{\partial^2 y}{\partial t^2} \]

### Step 9: Rearranging the equation
Rearranging this equation leads to:
\[ \frac{T}{\mu} = \frac{\partial^2 y}{\partial t^2} \frac{1}{\Delta x} \]

### Step 10: Identify the wave velocity
Recognizing that the wave velocity \( v \) is defined as:
\[ v = \sqrt{\frac{T}{\mu}} \]
This is the expression for the velocity of a pulse in a stretched string under tension \( T \) and mass per unit length \( \mu \).

### Final Expression
Thus, the final expression for the velocity of a pulse in a stretched string is:
\[ v = \sqrt{\frac{T}{\mu}} \]