Masses `M_(A) "and" M_(B)` hanging from the ends of strings of lengths `L_(A) "and" L_(B)` are executing simple harmonic motions. If their frequencies are `f_(A) = 2f_(B)`, then

Options

A. `l_(A)=4l_(B)` regardles of masses
B. `l_(B)=4l_(A)` reagardless of masses
C. `M_(A)=2M_(B),l_(A) = 2l_(B)`
D. `M_(B)=2M_(A),l_(B) = 2l_(A)`

Answer

To solve the problem, we need to analyze the relationship between the frequencies of the simple harmonic motions of two masses \( M_A \) and \( M_B \) hanging from strings of lengths \( L_A \) and \( L_B \). Given that the frequency \( f_A \) is twice that of \( f_B \), we can derive the relationship between the lengths of the strings.

### Step-by-Step Solution:

1. **Understand the Frequency Formula**:
The frequency of a mass-spring system (or a pendulum) is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \]
where \( g \) is the acceleration due to gravity and \( L \) is the length of the string.

2. **Write the Frequencies for Both Masses**:
For mass \( M_A \):
\[ f_A = \frac{1}{2\pi} \sqrt{\frac{g}{L_A}} \]
For mass \( M_B \):
\[ f_B = \frac{1}{2\pi} \sqrt{\frac{g}{L_B}} \]

3. **Establish the Given Relationship**:
We are given that:
\[ f_A = 2f_B \]

4. **Substitute the Frequencies**:
Substitute the expressions for \( f_A \) and \( f_B \):
\[ \frac{1}{2\pi} \sqrt{\frac{g}{L_A}} = 2 \left( \frac{1}{2\pi} \sqrt{\frac{g}{L_B}} \right) \]

5. **Simplify the Equation**:
Cancel \( \frac{1}{2\pi} \) from both sides:
\[ \sqrt{\frac{g}{L_A}} = 2 \sqrt{\frac{g}{L_B}} \]

6. **Square Both Sides**:
Squaring both sides gives:
\[ \frac{g}{L_A} = 4 \frac{g}{L_B} \]

7. **Cancel \( g \)**:
Since \( g \) is a constant and non-zero, we can cancel it:
\[ \frac{1}{L_A} = \frac{4}{L_B} \]

8. **Rearrange to Find the Relationship**:
Rearranging gives:
\[ L_B = 4L_A \]

### Conclusion:
The relationship derived indicates that the length of string \( L_B \) is four times the length of string \( L_A \):
\[ L_B = 4L_A \]