Two sitar strings A and B are slightly out of tune and produce beats of frequency 5 Hz. When the tension in the string B is slightly increased, the beat frequency is found to reduce to 3 Hz. If the frequency of string A is 427 Hz, the original frequency of string B is

Options

A. 422 Hz
B. 424 Hz
C. 430 Hz
D. 432 Hz

Explanation

To solve the problem, we need to determine the original frequency of string B based on the information about the beat frequencies and the frequency of string A.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Frequency of string A, \( f_A = 427 \, \text{Hz} \)
- Initial beat frequency, \( f_{\text{beat}} = 5 \, \text{Hz} \)
- Beat frequency after increasing tension in string B, \( f'_{\text{beat}} = 3 \, \text{Hz} \)

2. **Understanding Beat Frequency:**
- The beat frequency is the absolute difference between the frequencies of the two strings:
\[ f_{\text{beat}} = |f_A - f_B| \]
- For the initial condition:
\[ 5 = |427 - f_B| \]

3. **Setting Up Equations:**
- This gives us two possible equations:
1. \( 427 - f_B = 5 \) (Case 1)
2. \( f_B - 427 = 5 \) (Case 2)

4. **Solving Case 1:**
- From \( 427 - f_B = 5 \):
\[ f_B = 427 - 5 = 422 \, \text{Hz} \]

5. **Solving Case 2:**
- From \( f_B - 427 = 5 \):
\[ f_B = 427 + 5 = 432 \, \text{Hz} \]

6. **Analyzing the Effect of Increasing Tension:**
- When the tension in string B is increased, its frequency increases, and the new beat frequency is given as 3 Hz:
\[ f'_{\text{beat}} = |f_A - f'_B| \]
- Thus, we have:
\[ 3 = |427 - f'_B| \]

7. **Setting Up New Equations:**
- This gives us two new equations:
1. \( 427 - f'_B = 3 \) (New Case 1)
2. \( f'_B - 427 = 3 \) (New Case 2)

8. **Solving New Case 1:**
- From \( 427 - f'_B = 3 \):
\[ f'_B = 427 - 3 = 424 \, \text{Hz} \]

9. **Solving New Case 2:**
- From \( f'_B - 427 = 3 \):
\[ f'_B = 427 + 3 = 430 \, \text{Hz} \]

10. **Determining the Original Frequency of String B:**
- Since increasing the tension in string B should increase its frequency, we compare the original frequencies:
- If \( f_B = 422 \, \text{Hz} \), then \( f'_B = 424 \, \text{Hz} \) (valid).
- If \( f_B = 432 \, \text{Hz} \), then \( f'_B = 430 \, \text{Hz} \) (invalid because frequency decreased).
- Therefore, the only valid solution is:
\[ f_B = 422 \, \text{Hz} \]

### Final Answer:
The original frequency of string B is \( 422 \, \text{Hz} \).