A string of length L consists of two distinct sections. The left half has linear mass density `mu_1 = mu_0 // 2` while the right half has linear mass density `mu_2 = 3mu_0` . Tension in the string is `F_0` . The time required for a transverse wave pulse to travel from one end of the string to the other is

Options

A. `L.4 sqrt((mu_0)/(F_0)) (sqrt2 + sqrt6)`
B. `L/2 sqrt((2mu_0)/(F_0)) (1+ sqrt3) `
C. `L/2 sqrt((mu_0)/(2F_0)) (sqrt2 + sqrt 6)`
D. `L/2 sqrt((mu_0)/(2F_0))(1 + sqrt6)`

Answer

To solve the problem, we need to calculate the time required for a transverse wave pulse to travel from one end of the string to the other, considering the two different sections of the string with distinct linear mass densities.

### Step-by-Step Solution:

1. **Identify the Sections of the String:**
- The string is divided into two halves:
- Left half (length = L/2) with linear mass density \( \mu_1 = \frac{\mu_0}{2} \)
- Right half (length = L/2) with linear mass density \( \mu_2 = 3\mu_0 \)

2. **Calculate the Wave Velocity in Each Section:**
- The velocity of a transverse wave in a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- For the left half:
\[ v_L = \sqrt{\frac{F_0}{\mu_1}} = \sqrt{\frac{F_0}{\frac{\mu_0}{2}}} = \sqrt{\frac{2F_0}{\mu_0}} \]
- For the right half:
\[ v_R = \sqrt{\frac{F_0}{\mu_2}} = \sqrt{\frac{F_0}{3\mu_0}} = \sqrt{\frac{F_0}{3\mu_0}} \]

3. **Calculate the Time Taken to Travel Each Section:**
- The time taken for the wave to travel from the left end to the midpoint (left half):
\[ t_L = \frac{\text{Distance}}{\text{Velocity}} = \frac{\frac{L}{2}}{v_L} = \frac{\frac{L}{2}}{\sqrt{\frac{2F_0}{\mu_0}}} = \frac{L}{2} \cdot \sqrt{\frac{\mu_0}{2F_0}} = \frac{L}{2} \cdot \frac{\sqrt{\mu_0}}{\sqrt{2F_0}} \]

- The time taken for the wave to travel from the midpoint to the right end (right half):
\[ t_R = \frac{\text{Distance}}{\text{Velocity}} = \frac{\frac{L}{2}}{v_R} = \frac{\frac{L}{2}}{\sqrt{\frac{F_0}{3\mu_0}}} = \frac{L}{2} \cdot \sqrt{\frac{3\mu_0}{F_0}} = \frac{L}{2} \cdot \frac{\sqrt{3\mu_0}}{\sqrt{F_0}} \]

4. **Total Time Calculation:**
- The total time \( t \) for the wave to travel from one end of the string to the other is the sum of the two times:
\[ t = t_L + t_R \]
- Substituting the values:
\[ t = \frac{L}{2} \cdot \frac{\sqrt{\mu_0}}{\sqrt{2F_0}} + \frac{L}{2} \cdot \frac{\sqrt{3\mu_0}}{\sqrt{F_0}} \]
- Factor out \( \frac{L}{2} \):
\[ t = \frac{L}{2} \left( \frac{\sqrt{\mu_0}}{\sqrt{2F_0}} + \frac{\sqrt{3\mu_0}}{\sqrt{F_0}} \right) \]
- Simplifying further:
\[ t = \frac{L}{2} \left( \frac{\sqrt{\mu_0}}{\sqrt{2F_0}} + \frac{\sqrt{3\mu_0} \cdot \sqrt{2}}{\sqrt{2F_0}} \right) = \frac{L}{2\sqrt{2F_0}} \left( \sqrt{\mu_0} + \sqrt{6\mu_0} \right) \]
- This gives:
\[ t = \frac{L}{2\sqrt{2F_0}} \sqrt{\mu_0} (1 + \sqrt{6}) \]

### Final Answer:
The time required for a transverse wave pulse to travel from one end of the string to the other is:
\[ t = \frac{L}{2\sqrt{2F_0}} \sqrt{\mu_0} (1 + \sqrt{6}) \]