A non-uniform wire of length `l` and mass `M` has a variable linear mass density given by `mu = kx`, where `x` is distance from one end of wire and `k` is a constant. Find the time taken by a pulse starting at one end to reach the other end when the tension in the wire is `T`.

Explanation

To find the time taken by a pulse starting at one end of a non-uniform wire to reach the other end, we will follow these steps:

### Step 1: Understand the problem
We have a wire of length \( l \) and mass \( M \) with a variable linear mass density given by \( \mu = kx \), where \( x \) is the distance from one end of the wire and \( k \) is a constant. We need to find the time taken for a pulse to travel the length of the wire under a tension \( T \).

### Step 2: Determine the velocity of the pulse
The velocity \( v \) of a wave in a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Since \( \mu = kx \), we can express the velocity as:
\[ v = \sqrt{\frac{T}{kx}} \]

### Step 3: Relate velocity to distance and time
The relationship between velocity, distance, and time is given by:
\[ v = \frac{dx}{dt} \]
Thus, we can write:
\[ \frac{dx}{dt} = \sqrt{\frac{T}{kx}} \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ dt = \frac{dx}{\sqrt{\frac{T}{kx}}} \]
This simplifies to:
\[ dt = \sqrt{\frac{k}{T}} \cdot \frac{\sqrt{x}}{dx} \]

### Step 5: Integrate to find total time
Now we need to integrate both sides. The total time \( t \) taken for the pulse to travel from \( x = 0 \) to \( x = l \) is:
\[ t = \int_0^l \sqrt{\frac{k}{T}} \cdot \frac{\sqrt{x}}{dx} \]
This can be expressed as:
\[ t = \sqrt{\frac{k}{T}} \int_0^l \sqrt{x} \, dx \]

### Step 6: Calculate the integral
The integral \( \int_0^l \sqrt{x} \, dx \) can be calculated as follows:
\[ \int_0^l \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^l = \frac{2}{3} l^{3/2} \]

### Step 7: Substitute back into the equation for time
Substituting the result of the integral back into the equation for time gives:
\[ t = \sqrt{\frac{k}{T}} \cdot \frac{2}{3} l^{3/2} \]

### Step 8: Find the value of \( k \)
We know that the total mass \( M \) of the wire can be expressed as:
\[ M = \int_0^l \mu \, dx = \int_0^l kx \, dx = k \cdot \frac{l^2}{2} \]
From this, we can solve for \( k \):
\[ k = \frac{2M}{l^2} \]

### Step 9: Substitute \( k \) back into the time equation
Substituting \( k \) into the time equation gives:
\[ t = \sqrt{\frac{2M}{l^2 T}} \cdot \frac{2}{3} l^{3/2} \]
This simplifies to:
\[ t = \frac{2}{3} \cdot \frac{2M}{\sqrt{T}} \cdot \frac{l^{3/2}}{l} \]
Thus, the final expression for the time \( t \) is:
\[ t = \frac{4M}{3\sqrt{T}} \cdot l^{1/2} \]

### Final Result
The time taken by the pulse to travel from one end of the wire to the other is:
\[ t = \frac{4M}{3\sqrt{T}} \cdot l^{1/2} \]