A closed tube has a frequency n. If its length is doubled and radius is halved its frequency will become

Options

A. n/3
B. n/2
C. not in a gas but in a metal
D. 2 n

Explanation

To solve the problem, we need to analyze how the frequency of a closed tube changes when its length is doubled and its radius is halved.

### Step-by-Step Solution:

1. **Understanding the Closed Tube:**
A closed tube (or pipe) has one closed end and one open end. The fundamental frequency (first harmonic) of a closed tube is determined by its length. The relationship between the length of the tube (L) and the wavelength (λ) is given by:
\[ L = \frac{\lambda}{4} \]
Thus, the wavelength can be expressed as:
\[ \lambda = 4L \]

2. **Frequency and Wavelength Relationship:**
The frequency (f) of a wave is related to its speed (v) and wavelength (λ) by the formula:
\[ f = \frac{v}{\lambda} \]
Substituting the expression for wavelength, we have:
\[ f = \frac{v}{4L} \]
This shows that frequency is inversely proportional to the length of the tube:
\[ f \propto \frac{1}{L} \]

3. **Initial Frequency:**
Let the initial frequency of the tube be \( f_1 = n \) when the length is \( L_1 = L \).

4. **Changing the Length:**
If the length of the tube is doubled, the new length \( L_2 \) becomes:
\[ L_2 = 2L \]

5. **Calculating New Frequency:**
Using the inverse relationship between frequency and length, we can express the new frequency \( f_2 \):
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} \]
Substituting the known values:
\[ \frac{n}{f_2} = \frac{2L}{L} \]
Simplifying gives:
\[ \frac{n}{f_2} = 2 \implies f_2 = \frac{n}{2} \]

6. **Effect of Radius:**
The radius of the tube does not affect the fundamental frequency in a closed tube, so halving the radius does not change the frequency.

7. **Final Result:**
Therefore, the new frequency after doubling the length and halving the radius is:
\[ f_2 = \frac{n}{2} \]

### Conclusion:
The frequency of the closed tube after the changes will be \( \frac{n}{2} \).