Two vibrating strings of the same material but lengths `L` and `2L` have radii `2r` and `r` respectively. They are stretched under the same tension. Both the string vibrate in their fundamental nodes, the one of length `L` with freuqency `v_(1)` and the other with frequency `v_(2). the raio `v_(1)//v_(2)` is given by

Options

A. `2`
B. `4`
C. `8`
D. `1`

Explanation

To solve the problem of finding the ratio of the frequencies \( \frac{v_1}{v_2} \) for two vibrating strings of different lengths and radii, we can follow these steps:

### Step 1: Understand the formula for frequency
The frequency of a vibrating string in its fundamental mode is given by the formula:
\[ v = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( v \) is the frequency,
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

### Step 2: Calculate the mass per unit length for each string
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \frac{m}{L} = \frac{\rho \cdot V}{L} = \frac{\rho \cdot A \cdot L}{L} = \rho \cdot A \]
where:
- \( \rho \) is the density of the material,
- \( A \) is the cross-sectional area of the string.

For the first string (length \( L \) and radius \( 2r \)):
\[ A_1 = \pi (2r)^2 = 4\pi r^2 \]
Thus, the mass per unit length for the first string is:
\[ \mu_1 = \rho \cdot 4\pi r^2 \]

For the second string (length \( 2L \) and radius \( r \)):
\[ A_2 = \pi r^2 \]
Thus, the mass per unit length for the second string is:
\[ \mu_2 = \rho \cdot \pi r^2 \]

### Step 3: Write the frequency equations for both strings
Using the formula for frequency:
For the first string:
\[ v_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu_1}} = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot 4\pi r^2}} \]

For the second string:
\[ v_2 = \frac{1}{2(2L)} \sqrt{\frac{T}{\mu_2}} = \frac{1}{4L} \sqrt{\frac{T}{\rho \cdot \pi r^2}} \]

### Step 4: Find the ratio of the frequencies
Now, we can find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1}{v_2} = \frac{\frac{1}{2L} \sqrt{\frac{T}{\rho \cdot 4\pi r^2}}}{\frac{1}{4L} \sqrt{\frac{T}{\rho \cdot \pi r^2}}} \]
Simplifying this, we have:
\[ \frac{v_1}{v_2} = \frac{4L}{2L} \cdot \frac{\sqrt{\frac{T}{\rho \cdot 4\pi r^2}}}{\sqrt{\frac{T}{\rho \cdot \pi r^2}}} \]
The \( T \), \( \rho \), and \( L \) terms cancel out:
\[ \frac{v_1}{v_2} = 2 \cdot \frac{\sqrt{\pi r^2}}{\sqrt{4\pi r^2}} = 2 \cdot \frac{1}{2} = 1 \]

### Conclusion
Thus, the ratio \( \frac{v_1}{v_2} \) is:
\[ \frac{v_1}{v_2} = 1 \]