A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency `512 Hz`. If the tube is open at both ands the fundamental frequency that can be excited is (in Hz)

Options

A. (a) `1024`
B. (b) `512`
C. ( c ) `256`
D. ( d) `128`

Explanation

To solve the problem, we will use the relationship between the frequency, wavelength, and the length of the tube for both closed and open-end conditions.

### Step-by-Step Solution:

1. **Understanding the Closed-End Tube:**
- A tube closed at one end supports a fundamental frequency where the length \( L \) of the tube is equal to \( \frac{\lambda}{4} \), where \( \lambda \) is the wavelength.
- Therefore, we can express the wavelength as:
\[ \lambda = 4L \]

2. **Frequency of the Closed-End Tube:**
- The frequency \( f \) is related to the speed of sound \( c \) and the wavelength \( \lambda \) by the formula:
\[ f = \frac{c}{\lambda} \]
- Substituting the expression for \( \lambda \):
\[ f = \frac{c}{4L} \]
- Given that the fundamental frequency \( f \) is \( 512 \, \text{Hz} \):
\[ 512 = \frac{c}{4L} \]

3. **Understanding the Open-End Tube:**
- A tube open at both ends supports a fundamental frequency where the length \( L \) of the tube is equal to \( \frac{\lambda'}{2} \), where \( \lambda' \) is the wavelength for the open tube.
- Therefore, we can express the wavelength for the open tube as:
\[ \lambda' = 2L \]

4. **Frequency of the Open-End Tube:**
- The frequency \( f' \) for the open tube is given by:
\[ f' = \frac{c}{\lambda'} \]
- Substituting the expression for \( \lambda' \):
\[ f' = \frac{c}{2L} \]

5. **Relating Frequencies of Closed and Open Tubes:**
- From the closed-end tube, we have:
\[ c = 4L \cdot 512 \]
- Substituting this into the frequency equation for the open-end tube:
\[ f' = \frac{4L \cdot 512}{2L} = 2 \cdot 512 \]
- Therefore:
\[ f' = 1024 \, \text{Hz} \]

### Final Answer:
The fundamental frequency that can be excited in the tube when it is open at both ends is **1024 Hz**.

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