A pipe 17 cm long is closed at one end. Which harmonic mode of the pipe resonates a 1.5 kHz source? (Speed of sound in air = 340 m `s^(-1)`)

Options

A. First
B. Third
C. Fifth
D. Seventh

Explanation

To solve the problem of determining which harmonic mode of a closed pipe resonates with a 1.5 kHz source, we can follow these steps:

### Step 1: Convert the Length of the Pipe
Convert the length of the pipe from centimeters to meters.
- Given length \( L = 17 \, \text{cm} = 0.17 \, \text{m} \)

**Hint:** Remember to convert all measurements to the same unit system, typically meters for length in physics.

### Step 2: Identify the Frequency
Identify the frequency of the sound source.
- Given frequency \( f = 1.5 \, \text{kHz} = 1500 \, \text{Hz} \)

**Hint:** When dealing with frequency, ensure you are using the correct unit (Hz) for calculations.

### Step 3: Calculate the Wavelength
Use the wave equation \( v = f \lambda \) to find the wavelength \( \lambda \).
- Given speed of sound \( v = 340 \, \text{m/s} \)
- Rearranging the equation gives \( \lambda = \frac{v}{f} = \frac{340 \, \text{m/s}}{1500 \, \text{Hz}} \)
- Calculate \( \lambda \):
\[ \lambda = \frac{340}{1500} \approx 0.2267 \, \text{m} \approx 22.67 \, \text{cm} \]

**Hint:** The wavelength is the distance between successive peaks of a wave; ensure your calculations are accurate.

### Step 4: Determine the Harmonic Mode
For a pipe closed at one end, the harmonics are given by the formula:
\[ L = \frac{(2n + 1) \lambda}{4} \]
Where \( n \) is the harmonic number (0, 1, 2, ...).

Rearranging gives:
\[ 2n + 1 = \frac{4L}{\lambda} \]
Substituting the values:
\[ 2n + 1 = \frac{4 \times 0.17 \, \text{m}}{0.2267 \, \text{m}} \approx \frac{0.68}{0.2267} \approx 3 \]

**Hint:** This equation helps to relate the physical length of the pipe to the wavelength and the harmonic number.

### Step 5: Solve for \( n \)
Now, solve for \( n \):
\[ 2n + 1 = 3 \implies 2n = 2 \implies n = 1 \]

**Hint:** The value of \( n \) indicates the number of quarter wavelengths that fit into the length of the pipe.

### Step 6: Determine the Harmonic Number
The harmonic number for a closed pipe is given by \( 2n + 1 \):
- Therefore, the harmonic mode that resonates is:
\[ \text{Harmonic mode} = 2n + 1 = 3 \]

**Hint:** The harmonic number indicates the mode of vibration; for closed pipes, the harmonics are always odd integers.

### Final Answer
The harmonic mode of the pipe that resonates with a 1.5 kHz source is the **3rd harmonic**.

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