An open end organ pipe is 0.50 m long. If the velocity of sound is `320 ms^(-1)`, find the frequency of the fundamental note.

Answer

To find the frequency of the fundamental note of an open-end organ pipe, we can follow these steps:

### Step 1: Understand the formula for the fundamental frequency
For an open-end organ pipe, the fundamental frequency (f) is given by the formula:
\[ f = \frac{v}{2L} \]
where:
- \( f \) = fundamental frequency
- \( v \) = velocity of sound in air
- \( L \) = length of the pipe

### Step 2: Identify the given values
From the question, we have:
- Length of the pipe, \( L = 0.50 \, m \)
- Velocity of sound, \( v = 320 \, m/s \)

### Step 3: Substitute the values into the formula
Now, we substitute the values into the formula:
\[ f = \frac{320 \, m/s}{2 \times 0.50 \, m} \]

### Step 4: Calculate the denominator
Calculate \( 2L \):
\[ 2L = 2 \times 0.50 \, m = 1.00 \, m \]

### Step 5: Calculate the frequency
Now, substitute \( 2L \) back into the frequency formula:
\[ f = \frac{320 \, m/s}{1.00 \, m} = 320 \, Hz \]

### Conclusion
Thus, the frequency of the fundamental note of the open-end organ pipe is:
\[ f = 320 \, Hz \]

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