A pipe 30 cm long, is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source? (Speed of sound in air = 330 m `s^(-1)`) a) First b) Third c) Second d) Fourth

Options

A. First
B. Third
C. Second
D. Fourth

Explanation

To solve the problem of determining which harmonic mode of a pipe resonates with a 1.1 kHz source, we will follow these steps:

### Step 1: Convert the Length of the Pipe
The length of the pipe is given as 30 cm. We need to convert this into meters for consistency with the speed of sound.

\[ L = 30 \text{ cm} = 30 \times 10^{-2} \text{ m} = 0.30 \text{ m} \]

### Step 2: Identify the Speed of Sound
The speed of sound in air is given as 330 m/s. We will denote this as \( V \).

\[ V = 330 \text{ m/s} \]

### Step 3: Convert Frequency to Hertz
The frequency of the source is given as 1.1 kHz. We need to convert this to Hertz.

\[ f = 1.1 \text{ kHz} = 1.1 \times 10^3 \text{ Hz} = 1100 \text{ Hz} \]

### Step 4: Use the Formula for Harmonics in an Open Pipe
For a pipe open at both ends, the frequency of the nth harmonic is given by the formula:

\[ f_n = \frac{nV}{2L} \]

Where:
- \( f_n \) is the frequency of the nth harmonic,
- \( n \) is the harmonic number (1, 2, 3, ...),
- \( V \) is the speed of sound,
- \( L \) is the length of the pipe.

### Step 5: Rearrange the Formula to Solve for n
We can rearrange the formula to find \( n \):

\[ n = \frac{2Lf_n}{V} \]

### Step 6: Substitute the Known Values
Now we will substitute the known values into the equation:

\[ n = \frac{2 \times (0.30 \text{ m}) \times (1100 \text{ Hz})}{330 \text{ m/s}} \]

### Step 7: Calculate n
Now we will perform the calculation:

\[ n = \frac{2 \times 0.30 \times 1100}{330} \]

Calculating the numerator:

\[ 2 \times 0.30 \times 1100 = 660 \]

Now divide by the speed of sound:

\[ n = \frac{660}{330} = 2 \]

### Step 8: Determine the Harmonic Mode
Since \( n = 2 \), this indicates that the pipe resonates at the second harmonic mode.

### Conclusion
The correct answer is:

**c) Second**

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