velocity of sound in air is 320 m/s. A pipe closed at one end has a length of 1 m. Neglecting end corrections, the air column in the pipe can resonates for sound of frequency

Options

A. 80 Hz
B. 240 Hz
C. 320 Hz
D. 400 Hz

Explanation

To find the frequencies at which the air column in a closed-end pipe can resonate, we will follow these steps:

### Step 1: Understand the fundamental frequency formula
The fundamental frequency (first harmonic) for a closed-end pipe is given by the formula:
\[ f_1 = \frac{v}{4L} \]
where:
- \( f_1 \) = fundamental frequency
- \( v \) = velocity of sound in air
- \( L \) = length of the pipe

### Step 2: Substitute the given values
From the question, we know:
- The velocity of sound \( v = 320 \, \text{m/s} \)
- The length of the pipe \( L = 1 \, \text{m} \)

Now, substituting these values into the formula:
\[ f_1 = \frac{320 \, \text{m/s}}{4 \times 1 \, \text{m}} \]

### Step 3: Calculate the fundamental frequency
Calculating the above expression:
\[ f_1 = \frac{320}{4} = 80 \, \text{Hz} \]

### Step 4: Determine higher harmonics
For a closed-end pipe, the higher harmonics are odd integral multiples of the fundamental frequency. The formula for the \( n \)-th harmonic is:
\[ f_n = (2n - 1) f_1 \]
where \( n \) is an integer (1, 2, 3,...).

### Step 5: Calculate the first few harmonics
- For \( n = 1 \):
\[ f_1 = 80 \, \text{Hz} \]
- For \( n = 2 \):
\[ f_2 = (2 \times 2 - 1) \times 80 = 3 \times 80 = 240 \, \text{Hz} \]
- For \( n = 3 \):
\[ f_3 = (2 \times 3 - 1) \times 80 = 5 \times 80 = 400 \, \text{Hz} \]
- For \( n = 4 \):
\[ f_4 = (2 \times 4 - 1) \times 80 = 7 \times 80 = 560 \, \text{Hz} \]

### Step 6: List the possible frequencies
The possible resonant frequencies for the air column in the pipe are:
- \( f_1 = 80 \, \text{Hz} \)
- \( f_2 = 240 \, \text{Hz} \)
- \( f_3 = 400 \, \text{Hz} \)
- \( f_4 = 560 \, \text{Hz} \)

### Conclusion
The air column in the pipe can resonate at frequencies of 80 Hz, 240 Hz, and 400 Hz.