A pipe closed at one end, produces a fundamental note 412 Hz. It is now cut into two pieces of equal length. What will be the frequency of the fundamental note produced by each piece?

Answer

To solve the problem, we need to find the fundamental frequency of each piece of the pipe after it is cut into two equal lengths.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- The initial frequency of the closed pipe is given as \( f_0 = 412 \, \text{Hz} \).
- Let the initial length of the pipe be \( L \).

2. **Determine the speed of sound in the pipe:**
- For a pipe closed at one end, the fundamental frequency is given by the formula:
\[ f_0 = \frac{V}{4L} \]
- Rearranging this gives us the speed of sound \( V \):
\[ V = 4L f_0 \]
- Substituting \( f_0 = 412 \, \text{Hz} \):
\[ V = 4L \times 412 \]

3. **Cut the pipe into two equal lengths:**
- After cutting, the length of each piece becomes:
\[ L_f = \frac{L}{2} \]

4. **Calculate the frequency of the open pipe:**
- The first piece (open at both ends) will have its fundamental frequency given by:
\[ f' = \frac{V}{2L_f} \]
- Substituting \( L_f = \frac{L}{2} \):
\[ f' = \frac{V}{2 \times \frac{L}{2}} = \frac{V}{L} \]
- Now substituting \( V = 4L f_0 \):
\[ f' = \frac{4L f_0}{L} = 4 f_0 \]
- Therefore:
\[ f' = 4 \times 412 = 1648 \, \text{Hz} \]

5. **Calculate the frequency of the closed pipe:**
- The second piece (closed at one end) will have its fundamental frequency given by:
\[ f'' = \frac{V}{4L_f} \]
- Substituting \( L_f = \frac{L}{2} \):
\[ f'' = \frac{V}{4 \times \frac{L}{2}} = \frac{V}{2L} \]
- Again substituting \( V = 4L f_0 \):
\[ f'' = \frac{4L f_0}{2L} = 2 f_0 \]
- Therefore:
\[ f'' = 2 \times 412 = 824 \, \text{Hz} \]

### Final Results:
- The frequency of the fundamental note produced by the piece that is open at both ends is **1648 Hz**.
- The frequency of the fundamental note produced by the piece that is closed at one end is **824 Hz**.