if the first overtone of a closed pipe of length 50 cm has the same frequency as the first overtone of an open pipe, then the length of the open pipe is

Options

A. 100 cm
B. 200 cm
C. 66.6 cm
D. 33.3 cm

Answer

To solve the problem, we need to find the length of an open pipe given that the first overtone of a closed pipe of length 50 cm has the same frequency as the first overtone of the open pipe.

### Step-by-step Solution:

1. **Identify the formulas for the frequencies:**
- For a closed pipe, the frequency of the first overtone (n=2) is given by:
\[ f_c = \frac{(2n - 1)v}{4L} \]
where \( n = 2 \) for the first overtone, \( v \) is the speed of sound, and \( L \) is the length of the closed pipe.
- For an open pipe, the frequency of the first overtone (n=2) is given by:
\[ f_o = \frac{nv}{2L} \]
where \( n = 2 \) for the first overtone.

2. **Substituting the values:**
- For the closed pipe of length \( L = 50 \) cm = 0.5 m:
\[ f_c = \frac{(2 \cdot 2 - 1)v}{4 \cdot 0.5} = \frac{3v}{4} \]

3. **Frequency of the open pipe:**
- The frequency for the first overtone of the open pipe is:
\[ f_o = \frac{2v}{2L} = \frac{v}{L} \]

4. **Equating the frequencies:**
- Since the first overtone of the closed pipe has the same frequency as the first overtone of the open pipe, we set \( f_c = f_o \):
\[ \frac{3v}{4} = \frac{v}{L} \]

5. **Solving for \( L \):**
- Cancel \( v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{3}{4} = \frac{1}{L} \]
- Rearranging gives:
\[ L = \frac{4}{3} \]
- Converting to meters:
\[ L = \frac{4}{3} \text{ m} \approx 1.333 \text{ m} = 133.3 \text{ cm} \]

6. **Final answer:**
- The length of the open pipe is approximately **133.3 cm**.