A piston is fitted in a cylindrical tube of small cross section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the speed of sound in the air of the tube.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Resonance Condition
The tube is open at one end and has a piston at the other end. When the piston is pulled out, the air column in the tube can resonate at specific frequencies. The first resonance occurs at a certain length of the air column, and the second resonance occurs when the piston is pulled out by 32 cm.

### Step 2: Relate the Change in Length to Wavelength
When the piston is pulled out by 32 cm, it corresponds to a change in the length of the air column. In an open pipe, the resonances occur at odd multiples of half wavelengths. The distance pulled out (32 cm) represents a half wavelength (λ/2) for the second resonance.

### Step 3: Calculate the Wavelength
Since the distance pulled out corresponds to λ/2:
\[ \frac{\lambda}{2} = 32 \text{ cm} = 0.32 \text{ m} \]
Thus, the wavelength (λ) can be calculated as:
\[ \lambda = 2 \times 0.32 \text{ m} = 0.64 \text{ m} \]

### Step 4: Use the Speed of Sound Formula
The speed of sound (v) can be calculated using the formula:
\[ v = f \times \lambda \]
where \( f \) is the frequency of the tuning fork, which is given as 512 Hz.

### Step 5: Substitute the Values
Now, substituting the values into the formula:
\[ v = 512 \text{ Hz} \times 0.64 \text{ m} = 327.68 \text{ m/s} \]

### Step 6: Round the Answer
Rounding off the speed of sound to three significant figures gives:
\[ v \approx 328 \text{ m/s} \]

### Final Answer
The speed of sound in the air of the tube is approximately **328 m/s**.

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