Explain how the bridges should be placed in order to divide a wire 1.10 m long into three segments whose fundamental frequencies are in the ratio 2:3:4.

Answer

To solve the problem of dividing a wire of length 1.10 m into three segments with fundamental frequencies in the ratio of 2:3:4, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to divide a wire of length 1.10 m into three segments (L1, L2, L3) such that their fundamental frequencies are in the ratio 2:3:4.

2. **Set Up the Length Equation**: Since the total length of the wire is 1.10 m, we can express this as:
\[ L_1 + L_2 + L_3 = 1.10 \text{ m} \]
Converting this to centimeters for easier calculations, we have:
\[ L_1 + L_2 + L_3 = 110 \text{ cm} \]

3. **Relate Frequencies to Lengths**: The fundamental frequency \( n \) for a segment of wire is given by:
\[ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension and \( \mu \) is the linear mass density. For segments L1, L2, and L3, we can write:
\[ n_1 = \frac{1}{2L_1} \sqrt{\frac{T}{\mu}}, \quad n_2 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}}, \quad n_3 = \frac{1}{2L_3} \sqrt{\frac{T}{\mu}} \]

4. **Express Frequencies in Terms of Lengths**: From the ratio of frequencies \( n_1:n_2:n_3 = 2:3:4 \), we can express:
\[ \frac{n_1}{n_2} = \frac{2}{3} \quad \text{and} \quad \frac{n_1}{n_3} = \frac{2}{4} = \frac{1}{2} \]
This gives us the relationships:
\[ \frac{L_2}{L_1} = \frac{3}{2} \quad \text{and} \quad \frac{L_3}{L_1} = \frac{1}{2} \]

5. **Express L2 and L3 in Terms of L1**:
- From \( \frac{L_2}{L_1} = \frac{3}{2} \), we have:
\[ L_2 = \frac{3}{2} L_1 \]
- From \( \frac{L_3}{L_1} = \frac{1}{2} \), we have:
\[ L_3 = \frac{1}{2} L_1 \]

6. **Substitute into the Length Equation**: Substitute \( L_2 \) and \( L_3 \) into the total length equation:
\[ L_1 + \frac{3}{2}L_1 + \frac{1}{2}L_1 = 110 \]
This simplifies to:
\[ L_1 + 1.5L_1 + 0.5L_1 = 110 \implies 3L_1 = 110 \]

7. **Solve for L1**:
\[ L_1 = \frac{110}{3} \approx 36.67 \text{ cm} \]

8. **Calculate L2 and L3**:
- For \( L_2 \):
\[ L_2 = \frac{3}{2} \times 36.67 \approx 55.00 \text{ cm} \]
- For \( L_3 \):
\[ L_3 = \frac{1}{2} \times 36.67 \approx 18.33 \text{ cm} \]

9. **Final Lengths**: The lengths of the segments are approximately:
- \( L_1 \approx 36.67 \text{ cm} \)
- \( L_2 \approx 55.00 \text{ cm} \)
- \( L_3 \approx 18.33 \text{ cm} \)

### Summary of Segment Lengths:
- \( L_1 \approx 36.67 \text{ cm} \)
- \( L_2 \approx 55.00 \text{ cm} \)
- \( L_3 \approx 18.33 \text{ cm} \)