A wire of length L is fixed at both ends such that F is tension in it. Its mass per unit length varies from one end to other end as `mu=mu_0x`, where `mu_0` is constant. Find time taken by a transverse pulse to move from lighter end to its mid point.

Answer

To solve the problem of finding the time taken by a transverse pulse to move from the lighter end of a wire to its midpoint, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Mass per Unit Length**:
The mass per unit length of the wire varies linearly from one end to the other, given by the equation:
\[ \mu(x) = \mu_0 x \]
where \( \mu_0 \) is a constant and \( x \) is the position along the wire.

2. **Wave Velocity**:
The wave velocity \( v \) in a stretched string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire. In this case, \( T = F \) (constant tension).

3. **Substituting for \( \mu \)**:
We substitute the expression for \( \mu \) into the wave velocity formula:
\[ v(x) = \sqrt{\frac{F}{\mu_0 x}} \]

4. **Time Taken to Travel a Small Distance**:
The time \( dt \) taken for the pulse to travel a small distance \( dx \) is given by:
\[ dt = \frac{dx}{v(x)} = \frac{dx}{\sqrt{\frac{F}{\mu_0 x}}} \]
Simplifying this gives:
\[ dt = \frac{\sqrt{\mu_0 x}}{\sqrt{F}} dx \]

5. **Total Time Calculation**:
To find the total time \( T \) taken to travel from the lighter end (at \( x = 0 \)) to the midpoint (at \( x = \frac{L}{2} \)), we integrate \( dt \) from \( 0 \) to \( \frac{L}{2} \):
\[ T = \int_0^{L/2} \frac{\sqrt{\mu_0 x}}{\sqrt{F}} dx \]

6. **Evaluating the Integral**:
The integral can be evaluated as follows:
\[ T = \frac{\sqrt{\mu_0}}{\sqrt{F}} \int_0^{L/2} x^{1/2} dx \]
The integral \( \int x^{1/2} dx \) evaluates to:
\[ \frac{2}{3} x^{3/2} \Big|_0^{L/2} = \frac{2}{3} \left(\frac{L}{2}\right)^{3/2} = \frac{2}{3} \cdot \frac{L^{3/2}}{2^{3/2}} = \frac{L^{3/2}}{3\sqrt{2}} \]

7. **Final Expression for Time**:
Substituting this back into the expression for \( T \):
\[ T = \frac{\sqrt{\mu_0}}{\sqrt{F}} \cdot \frac{L^{3/2}}{3\sqrt{2}} = \frac{L^{3/2} \sqrt{\mu_0}}{3\sqrt{2F}} \]

### Final Answer:
The time taken by the transverse pulse to move from the lighter end to its midpoint is:
\[ T = \frac{L^{3/2} \sqrt{\mu_0}}{3\sqrt{2F}} \]