A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. Speed of sound in air `= 340 m s^-1`.

Explanation

To find the length of a closed organ pipe that vibrates at a minimum frequency of 500 Hz, we can use the relationship between the speed of sound, frequency, and wavelength. Here’s the step-by-step solution:

### Step 1: Understand the relationship
For a closed organ pipe, the fundamental frequency (minimum frequency) is given by the formula:
\[ f = \frac{V}{4L} \]
where:
- \( f \) is the frequency (in Hz),
- \( V \) is the speed of sound in air (in m/s),
- \( L \) is the length of the pipe (in meters).

### Step 2: Rearrange the formula
We need to find the length \( L \) of the pipe, so we can rearrange the formula:
\[ L = \frac{V}{4f} \]

### Step 3: Substitute the known values
We know:
- \( f = 500 \, \text{Hz} \)
- \( V = 340 \, \text{m/s} \)

Substituting these values into the rearranged formula:
\[ L = \frac{340 \, \text{m/s}}{4 \times 500 \, \text{Hz}} \]

### Step 4: Calculate the length
Now, calculate the denominator:
\[ 4 \times 500 = 2000 \]
So we have:
\[ L = \frac{340}{2000} \]

Calculating this gives:
\[ L = 0.17 \, \text{m} \]

### Final Answer
The length of the closed organ pipe is \( 0.17 \, \text{m} \) or \( 17 \, \text{cm} \).

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