A narrow pipe of length 2m is closed at one end. The velocity of sound in air is 320 m/s. Neglecting end corrections , the air column in the pipe will resonate for sound of frequencies :

Options

A. 40Hz
B. 80Hz
C. 120Hz
D. 160Hz

Answer

To solve the problem, we need to determine the frequencies at which the air column in a closed pipe resonates. Here's a step-by-step solution:

### Step 1: Understand the fundamental frequency of a closed pipe
For a pipe that is closed at one end, the fundamental frequency (first harmonic) can be calculated using the formula:
\[ f_1 = \frac{v}{4L} \]
where:
- \(f_1\) is the fundamental frequency,
- \(v\) is the velocity of sound in air,
- \(L\) is the length of the pipe.

### Step 2: Substitute the values into the formula
Given:
- Length of the pipe, \(L = 2 \, \text{m}\)
- Velocity of sound, \(v = 320 \, \text{m/s}\)

Substituting these values into the formula:
\[ f_1 = \frac{320 \, \text{m/s}}{4 \times 2 \, \text{m}} = \frac{320}{8} = 40 \, \text{Hz} \]

### Step 3: Determine the harmonic frequencies
In a closed pipe, the resonant frequencies are given by the formula:
\[ f_n = (2n - 1) f_1 \]
where \(n\) is a positive integer (1, 2, 3, ...).

### Step 4: Calculate the first few harmonics
- For \(n = 1\):
\[ f_1 = (2 \times 1 - 1) \times 40 \, \text{Hz} = 1 \times 40 \, \text{Hz} = 40 \, \text{Hz} \]

- For \(n = 2\):
\[ f_2 = (2 \times 2 - 1) \times 40 \, \text{Hz} = 3 \times 40 \, \text{Hz} = 120 \, \text{Hz} \]

- For \(n = 3\):
\[ f_3 = (2 \times 3 - 1) \times 40 \, \text{Hz} = 5 \times 40 \, \text{Hz} = 200 \, \text{Hz} \]

### Step 5: List the resonant frequencies
The resonant frequencies for the air column in the pipe are:
- \(40 \, \text{Hz}\)
- \(120 \, \text{Hz}\)
- \(200 \, \text{Hz}\)

### Conclusion
The air column in the pipe will resonate at frequencies of \(40 \, \text{Hz}\), \(120 \, \text{Hz}\), and \(200 \, \text{Hz}\).

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