The length , radius , tension and density of string`A` are twice the same parameters of string `B`. Find the ratio of fundamental frequency of `B` to the fundamental frequency of `A`.

Explanation

To find the ratio of the fundamental frequency of string B to the fundamental frequency of string A, we can follow these steps:

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

### Step 2: Express mass per unit length
The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the string. The mass can also be expressed in terms of density \( \rho \) and volume \( V \):
\[ m = \rho \cdot V \]
For a cylindrical string, the volume \( V \) is given by the cross-sectional area \( A \) multiplied by the length \( L \):
\[ V = A \cdot L = \pi r^2 \cdot L \]
Thus, we can write:
\[ m = \rho \cdot (\pi r^2 \cdot L) \]
Substituting this into the equation for \( \mu \):
\[ \mu = \frac{\rho \cdot (\pi r^2 \cdot L)}{L} = \rho \cdot \pi r^2 \]

### Step 3: Substitute \( \mu \) into the frequency formula
Now substituting \( \mu \) back into the frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot \pi r^2}} \]
This can be simplified to:
\[ f \propto \frac{1}{L} \sqrt{\frac{T}{\rho \cdot r^2}} \]

### Step 4: Write the frequencies for strings A and B
Let’s denote the parameters for string A as \( L_A, T_A, \rho_A, r_A \) and for string B as \( L_B, T_B, \rho_B, r_B \).

From the problem, we know:
- \( L_A = 2L_B \)
- \( T_A = 2T_B \)
- \( \rho_A = 2\rho_B \)
- \( r_A = 2r_B \)

Now, we can express the frequencies:
\[ f_A \propto \frac{1}{L_A} \sqrt{\frac{T_A}{\rho_A \cdot r_A^2}} \]
\[ f_B \propto \frac{1}{L_B} \sqrt{\frac{T_B}{\rho_B \cdot r_B^2}} \]

### Step 5: Calculate the ratio of frequencies
Now we can find the ratio \( \frac{f_B}{f_A} \):
\[ \frac{f_B}{f_A} = \frac{\frac{1}{L_B} \sqrt{\frac{T_B}{\rho_B \cdot r_B^2}}}{\frac{1}{L_A} \sqrt{\frac{T_A}{\rho_A \cdot r_A^2}}} \]

Substituting the values:
\[ = \frac{L_A}{L_B} \cdot \sqrt{\frac{T_B \cdot \rho_A \cdot r_A^2}{T_A \cdot \rho_B \cdot r_B^2}} \]

### Step 6: Substitute the known ratios
Substituting the known ratios:
- \( L_A = 2L_B \) gives \( \frac{L_A}{L_B} = 2 \)
- \( T_A = 2T_B \) gives \( \frac{T_B}{T_A} = \frac{1}{2} \)
- \( \rho_A = 2\rho_B \) gives \( \frac{\rho_A}{\rho_B} = 2 \)
- \( r_A = 2r_B \) gives \( \frac{r_A^2}{r_B^2} = 4 \)

Putting it all together:
\[ \frac{f_B}{f_A} = 2 \cdot \sqrt{\frac{1/2 \cdot 2 \cdot 4}{1}} = 2 \cdot \sqrt{4} = 2 \cdot 2 = 4 \]

### Final Result
Thus, the ratio of the fundamental frequency of string B to that of string A is:
\[ \frac{f_B}{f_A} = 4 \]