the linear density of a wire under tension t varies linearly from `mu_(1)` to `mu_(20`.Calculate the time that a pulse would need to pass one to the other.The length of the wire is l

Options

A. `2/3(varphi)/sqrtT((mu_2-mu_1^(3//2))/((mu_2-mu_1)))`
B. `2/3(sqrtT(mu_2-mu_1))/(varphi(mu_2^(3//2)-mu_1^(3//2)))`
C. `2/3(varphi(mu_2-mu_1))/(sqrtT(mu_2^(3//2)-mu_1^(3//2)))`
D. `2/3(varphi(mu_2^(3//2)-mu_1^(3//2)))/(sqrtT(mu_2-mu_1))`

Answer

To solve the problem of calculating the time it takes for a pulse to travel from one end of a wire to the other when the linear density varies linearly from \( \mu_1 \) to \( \mu_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between velocity, tension, and linear density**:
The velocity \( v \) of a wave in a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the linear mass density.

2. **Express the linear density as a function of position**:
Since the linear density varies linearly from \( \mu_1 \) to \( \mu_2 \) over a length \( L \), we can express \( \mu \) as:
\[ \mu(x) = \mu_1 + \left(\frac{\mu_2 - \mu_1}{L}\right)x \]
where \( x \) is the position along the wire.

3. **Substitute \( \mu(x) \) into the velocity formula**:
The velocity as a function of position \( x \) becomes:
\[ v(x) = \sqrt{\frac{T}{\mu(x)}} = \sqrt{\frac{T}{\mu_1 + \left(\frac{\mu_2 - \mu_1}{L}\right)x}} \]

4. **Relate the time \( dt \) to the distance \( dx \)**:
The relationship between time and distance can be expressed as:
\[ dt = \frac{dx}{v(x)} = \frac{dx}{\sqrt{\frac{T}{\mu_1 + \left(\frac{\mu_2 - \mu_1}{L}\right)x}}} \]

5. **Integrate to find the total time**:
We need to integrate from \( x = 0 \) to \( x = L \):
\[ t = \int_0^L \frac{dx}{\sqrt{\frac{T}{\mu_1 + \left(\frac{\mu_2 - \mu_1}{L}\right)x}}} \]
This simplifies to:
\[ t = \frac{1}{\sqrt{T}} \int_0^L \sqrt{\mu_1 + \left(\frac{\mu_2 - \mu_1}{L}\right)x} \, dx \]

6. **Evaluate the integral**:
The integral can be solved using substitution or standard integral formulas. After evaluating, we find:
\[ t = \frac{2}{3} \cdot \frac{L}{\sqrt{T}} \cdot \frac{\mu_2^{3/2} - \mu_1^{3/2}}{\mu_2 - \mu_1} \]

### Final Answer:
Thus, the time \( t \) that a pulse would need to pass from one end of the wire to the other is given by:
\[ t = \frac{2}{3} \cdot \frac{L}{\sqrt{T}} \cdot \frac{\mu_2^{3/2} - \mu_1^{3/2}}{\mu_2 - \mu_1} \]