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 the tension in two identical strings (x and y) with fundamental frequencies of 450 Hz and 300 Hz respectively, we can follow these steps:

### Step-by-Step Solution:

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

2. **Set up the equations for both strings**:
For string x (with frequency \( f_x = 450 \, \text{Hz} \)):
\[ f_x = \frac{1}{2L} \sqrt{\frac{T_x}{\mu}} \]
For string y (with frequency \( f_y = 300 \, \text{Hz} \)):
\[ f_y = \frac{1}{2L} \sqrt{\frac{T_y}{\mu}} \]

3. **Express the tensions in terms of frequencies**:
Rearranging the equations gives us:
\[ T_x = (2Lf_x)^2 \mu \]
\[ T_y = (2Lf_y)^2 \mu \]

4. **Find the ratio of tensions**:
Now, we can find the ratio of the tensions \( \frac{T_x}{T_y} \):
\[ \frac{T_x}{T_y} = \frac{(2Lf_x)^2 \mu}{(2Lf_y)^2 \mu} \]
The \( \mu \) and \( (2L)^2 \) cancel out:
\[ \frac{T_x}{T_y} = \frac{f_x^2}{f_y^2} \]

5. **Substitute the frequencies**:
Now substitute \( f_x = 450 \, \text{Hz} \) and \( f_y = 300 \, \text{Hz} \):
\[ \frac{T_x}{T_y} = \left(\frac{450}{300}\right)^2 \]

6. **Calculate the ratio**:
Simplifying \( \frac{450}{300} \):
\[ \frac{450}{300} = \frac{15}{10} = \frac{3}{2} \]
Now square it:
\[ \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]

7. **Final result**:
Thus, the ratio of the tension in string x to the tension in string y is:
\[ \frac{T_x}{T_y} = \frac{9}{4} \]

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