The second overtone of an open organ pipe beats with the third overtone of the closed organ pipe. The beat frequency is 3.3 Hz. What will be the length of pipes if fundamental frequency of closed organ pipe is 120 Hz ?
Velocity of sound = 340 m/s

Answer

To solve the problem, we need to find the lengths of the open and closed organ pipes given the beat frequency and the fundamental frequency of the closed organ pipe.

### Step-by-Step Solution:

1. **Identify the Frequencies:**
- The fundamental frequency of the closed organ pipe (f_c) is given as 120 Hz.
- The second overtone of the open pipe corresponds to the third harmonic (f_o = 3 * v / (2 * L_o)).
- The third overtone of the closed organ pipe corresponds to the seventh harmonic (f_c7 = 7 * f_c).

2. **Calculate the Third Overtone Frequency of the Closed Pipe:**
- Using the fundamental frequency:
\[ f_{c7} = 7 \times f_c = 7 \times 120 \text{ Hz} = 840 \text{ Hz} \]

3. **Set Up the Beat Frequency Equation:**
- The beat frequency (f_beat) is given as 3.3 Hz. The equation for the beat frequency is:
\[ |f_o - f_{c7}| = 3.3 \text{ Hz} \]
- Substituting the expressions for the frequencies:
\[ \left| \frac{3v}{2L_o} - 840 \right| = 3.3 \]

4. **Substitute the Velocity of Sound:**
- The velocity of sound (v) is given as 340 m/s. Substitute this value into the equation:
\[ \left| \frac{3 \times 340}{2L_o} - 840 \right| = 3.3 \]
- Simplifying this gives:
\[ \left| \frac{510}{L_o} - 840 \right| = 3.3 \]

5. **Solve for L_o:**
- This absolute value equation can be split into two cases:
- Case 1:
\[ \frac{510}{L_o} - 840 = 3.3 \implies \frac{510}{L_o} = 843.3 \implies L_o = \frac{510}{843.3} \approx 0.603 \text{ m} \]
- Case 2:
\[ \frac{510}{L_o} - 840 = -3.3 \implies \frac{510}{L_o} = 836.7 \implies L_o = \frac{510}{836.7} \approx 0.610 \text{ m} \]
- We will consider the first case as it gives a valid length.

6. **Calculate the Length of the Closed Organ Pipe (L_c):**
- The fundamental frequency of the closed organ pipe is related to its length:
\[ f_c = \frac{v}{4L_c} \implies L_c = \frac{v}{4f_c} = \frac{340}{4 \times 120} = \frac{340}{480} \approx 0.708 \text{ m} \]

### Final Answers:
- Length of the open organ pipe (L_o) ≈ 0.603 m
- Length of the closed organ pipe (L_c) ≈ 0.708 m