Fundamental frequency of two identical strings x and y are 450Hz and 300Hz resp. then find the ratio of tension in string x and y will be

Options

A. (2/3)^1/2
B. 44078
C. (4/3)^1/2
D. (3/2^1/2

Answer

To find the ratio of tension in two identical strings \( x \) and \( y \) with fundamental frequencies \( f_x = 450 \, \text{Hz} \) and \( f_y = 300 \, \text{Hz} \), we can use the relationship between frequency, tension, and mass per unit length.

### Step-by-Step Solution:

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

2. **Set up the equations for both strings**:
For string \( x \):
\[ f_x = \frac{1}{2L} \sqrt{\frac{T_x}{\mu}} \]
For string \( y \):
\[ f_y = \frac{1}{2L} \sqrt{\frac{T_y}{\mu}} \]

3. **Express the tensions in terms of frequencies**:
Rearranging the equations gives:
\[ T_x = 4L^2 \mu f_x^2 \]
\[ T_y = 4L^2 \mu f_y^2 \]

4. **Find the ratio of tensions**:
The ratio of tensions \( \frac{T_x}{T_y} \) can be expressed as:
\[ \frac{T_x}{T_y} = \frac{4L^2 \mu f_x^2}{4L^2 \mu f_y^2} = \frac{f_x^2}{f_y^2} \]

5. **Substitute the given frequencies**:
Now substituting the values of \( f_x \) and \( f_y \):
\[ \frac{T_x}{T_y} = \frac{(450)^2}{(300)^2} \]

6. **Calculate the squares**:
\[ \frac{T_x}{T_y} = \frac{202500}{90000} = \frac{2025}{900} = \frac{9}{4} \]

7. **Final ratio**:
Therefore, the ratio of tensions \( T_x : T_y \) is:
\[ T_x : T_y = 9 : 4 \]

### Conclusion:
The ratio of tension in string \( x \) to tension in string \( y \) is \( 9 : 4 \).