An organ pipe, open from both end produces 5 beats per second when vibrated with a source of frequency 200 Hz . The second harmonic of the same pipes produces 10 beats per second with a source of frequency 420 Hz . The frequency of source is

Options

A. 195 Hz
B. 205 Hz
C. 190 Hz
D. 210 Hz

Answer

To solve the problem step by step, we will analyze the information given in the question and apply the concept of beats and harmonics.

### Step-by-Step Solution:

1. **Understanding Beats**:
- The number of beats per second is the absolute difference between the frequency of the source and the frequency of the organ pipe.
- Given that the organ pipe produces 5 beats per second with a source frequency of 200 Hz, we can express this as:
\[ |f_{pipe} - 200| = 5 \]

2. **Finding the Frequency of the Pipe**:
- From the equation above, we can derive two possible frequencies for the organ pipe:
\[ f_{pipe} = 200 + 5 = 205 \text{ Hz} \quad \text{or} \quad f_{pipe} = 200 - 5 = 195 \text{ Hz} \]

3. **Analyzing the Second Harmonic**:
- The second harmonic frequency of an organ pipe open at both ends is given by \( f_{pipe} = 2N \), where \( N \) is the fundamental frequency of the pipe.
- The second harmonic produces 10 beats per second with a source frequency of 420 Hz:
\[ |2N - 420| = 10 \]

4. **Finding the Frequency of the Second Harmonic**:
- From the equation above, we can derive two possible frequencies for the second harmonic:
\[ 2N = 420 + 10 = 430 \text{ Hz} \quad \text{or} \quad 2N = 420 - 10 = 410 \text{ Hz} \]
- This gives us:
\[ N = \frac{430}{2} = 215 \text{ Hz} \quad \text{or} \quad N = \frac{410}{2} = 205 \text{ Hz} \]

5. **Finding Common Frequency**:
- From the first part, we found that the frequency of the pipe could be 205 Hz or 195 Hz.
- From the second part, we found that the fundamental frequency \( N \) could be 215 Hz or 205 Hz.
- The only common frequency from both parts is:
\[ N = 205 \text{ Hz} \]

6. **Conclusion**:
- Therefore, the frequency of the source is:
\[ \boxed{205 \text{ Hz}} \]