A string 1 has twice the length, twice the radius, twice the tension and twice the density of another string 2. The relation between the fundamental frequency of 1 and 2 is

Options

A. `f_(1)=2f_(2)`
B. `f_(1)=4f_(2)`
C. `f_(2)=4f_(1)`
D. `f_(1)=f_(2)`

Answer

To find the relationship between the fundamental frequencies of two strings (string 1 and string 2) given their respective properties, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Parameters:**
- For string 1:
- Length \(L_1 = 2L_2\)
- Radius \(R_1 = 2R_2\)
- Tension \(T_1 = 2T_2\)
- Density \(\rho_1 = 2\rho_2\)

2. **Fundamental Frequency Formula:**
The fundamental frequency \(f\) of a string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \(\mu\) (linear mass density) is defined as:
\[ \mu = \rho \cdot A \]
and \(A\) (cross-sectional area) for a circular string is:
\[ A = \pi R^2 \]
Thus, we can express \(\mu\) as:
\[ \mu = \rho \cdot \pi R^2 \]

3. **Express the Frequencies:**
- For string 1:
\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{\mu_1}} = \frac{1}{2(2L_2)} \sqrt{\frac{2T_2}{2\rho_2 \cdot \pi (2R_2)^2}} \]
Simplifying this:
\[ f_1 = \frac{1}{4L_2} \sqrt{\frac{2T_2}{2\rho_2 \cdot \pi \cdot 4R_2^2}} = \frac{1}{4L_2} \sqrt{\frac{T_2}{\rho_2 \cdot \pi R_2^2}} \]

- For string 2:
\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{\mu_2}} = \frac{1}{2L_2} \sqrt{\frac{T_2}{\rho_2 \cdot \pi R_2^2}} \]

4. **Find the Ratio of Frequencies:**
Now, we can find the ratio of the fundamental frequencies:
\[ \frac{f_1}{f_2} = \frac{\frac{1}{4L_2} \sqrt{\frac{T_2}{\rho_2 \cdot \pi R_2^2}}}{\frac{1}{2L_2} \sqrt{\frac{T_2}{\rho_2 \cdot \pi R_2^2}}} \]
This simplifies to:
\[ \frac{f_1}{f_2} = \frac{1}{4} \cdot \frac{2L_2}{L_2} = \frac{1}{2} \]

5. **Final Relationship:**
Thus, we can conclude that:
\[ f_1 = \frac{1}{2} f_2 \]
or equivalently:
\[ f_2 = 2f_1 \]

### Conclusion:
The fundamental frequency of string 2 is twice that of string 1.