The frequency of the fundamental note of a tube closed at one end is 200 Hz. What will be the frequency of the fundamental note of a similar tube of the same length but open at both ends ?

Answer

To solve the problem, we need to find the frequency of the fundamental note of a tube that is open at both ends, given that the frequency of a similar tube closed at one end is 200 Hz.

### Step-by-Step Solution:

1. **Understand the relationship between frequency and length of the tube**:
- For a tube closed at one end, the fundamental frequency (f1) is given by the formula:
\[ f_1 = \frac{v}{4L_1} \]
where \(v\) is the speed of sound in air and \(L_1\) is the length of the tube.

2. **Express the length of the closed tube in terms of frequency**:
- Rearranging the formula gives us:
\[ L_1 = \frac{v}{4f_1} \]
- Substituting \(f_1 = 200 \, \text{Hz}\):
\[ L_1 = \frac{v}{4 \times 200} \]

3. **Write the formula for the fundamental frequency of a tube open at both ends**:
- The fundamental frequency (f2) for a tube open at both ends is given by:
\[ f_2 = \frac{v}{2L_2} \]
where \(L_2\) is the length of the tube.

4. **Since the lengths of the tubes are the same**:
- We have \(L_1 = L_2\). Therefore, we can write:
\[ L_1 = L_2 \]

5. **Set the two equations for lengths equal**:
- From the earlier expressions, we have:
\[ \frac{v}{4f_1} = \frac{v}{2f_2} \]

6. **Cancel out the speed of sound (v)**:
- This simplifies to:
\[ \frac{1}{4f_1} = \frac{1}{2f_2} \]

7. **Cross-multiply to solve for f2**:
- Cross-multiplying gives:
\[ 2f_2 = 4f_1 \]
- Dividing both sides by 2:
\[ f_2 = 2f_1 \]

8. **Substitute the known frequency**:
- Since \(f_1 = 200 \, \text{Hz}\):
\[ f_2 = 2 \times 200 = 400 \, \text{Hz} \]

### Final Answer:
The frequency of the fundamental note of the tube open at both ends is **400 Hz**.