A pipe 1 m length is closed at one end. Taking the speed of sound in air as `320ms ^(-1)` .the air column in the pipe cannot resonate for the frequency (in Hz )

Options

A. 80
B. 160
C. 240
D. 560

Answer

To solve the problem step by step, we need to determine the frequencies at which a pipe closed at one end can resonate, and then identify which frequency cannot be achieved.

### Step 1: Understand the Resonance in a Closed Pipe
In a pipe closed at one end, the resonant frequencies are determined by the odd harmonics. The fundamental frequency (first harmonic) corresponds to the first mode of resonance, where the length of the pipe \( L \) is equal to \( \frac{1}{4} \lambda \) (where \( \lambda \) is the wavelength). The subsequent harmonics are odd multiples of this fundamental frequency.

### Step 2: Write the Formula for Wavelength
For a closed pipe, the relationship between the length of the pipe \( L \) and the wavelength \( \lambda \) is given by:
\[ L = \frac{(2n - 1) \lambda}{4} \]
where \( n \) is the harmonic number (1, 2, 3,...).

### Step 3: Solve for Wavelength
Rearranging the formula, we can express the wavelength in terms of the length of the pipe:
\[ \lambda = \frac{4L}{2n - 1} \]

### Step 4: Substitute the Length of the Pipe
Given that the length \( L = 1 \, \text{m} \), we substitute this value into the equation:
\[ \lambda = \frac{4 \times 1}{2n - 1} = \frac{4}{2n - 1} \]

### Step 5: Calculate the Frequencies
The frequency \( f \) is related to the speed of sound \( v \) and the wavelength \( \lambda \) by the formula:
\[ f = \frac{v}{\lambda} \]
Substituting \( v = 320 \, \text{m/s} \):
\[ f = \frac{320}{\frac{4}{2n - 1}} = 320 \times \frac{2n - 1}{4} = 80(2n - 1) \]

### Step 6: Determine Possible Frequencies
Now we calculate the frequencies for different values of \( n \):
- For \( n = 1 \):
\[ f_1 = 80(2 \times 1 - 1) = 80 \, \text{Hz} \]
- For \( n = 2 \):
\[ f_2 = 80(2 \times 2 - 1) = 80 \times 3 = 240 \, \text{Hz} \]
- For \( n = 3 \):
\[ f_3 = 80(2 \times 3 - 1) = 80 \times 5 = 400 \, \text{Hz} \]
- For \( n = 4 \):
\[ f_4 = 80(2 \times 4 - 1) = 80 \times 7 = 560 \, \text{Hz} \]

### Step 7: Identify Frequencies
The possible resonant frequencies are:
- 80 Hz
- 240 Hz
- 400 Hz
- 560 Hz

### Step 8: Conclusion
Now, we check the options provided in the question. If the options include frequencies like 160 Hz, we can conclude that this frequency cannot be achieved since it does not correspond to any of the calculated resonant frequencies.

### Final Answer
The air column in the pipe cannot resonate for the frequency **160 Hz**.

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