An open organ pipe produces a note of frequency 512 Hz at `15^@ C`, calculate the length of the pipe.Velocity of sound at `0^@C` is `335 ms^(-1)`.

Answer

To solve the problem of finding the length of an open organ pipe that produces a note of frequency 512 Hz at 15°C, we can follow these steps:

### Step 1: Calculate the velocity of sound at 15°C
The velocity of sound at a temperature \( T \) can be calculated using the formula:
\[ v = v_0 + 0.61 \times T \]
where \( v_0 \) is the velocity of sound at 0°C (335 m/s) and \( T \) is the temperature in Celsius.

Substituting the values:
\[ v = 335 \, \text{m/s} + 0.61 \times 15 \]
Calculating \( 0.61 \times 15 \):
\[ 0.61 \times 15 = 9.15 \]
Now, add this to 335:
\[ v = 335 + 9.15 = 344.15 \, \text{m/s} \]

### Step 2: Use the formula for the fundamental frequency of an open organ pipe
The fundamental frequency \( f \) of an open organ pipe is given by the formula:
\[ f = \frac{4v}{L} \]
where \( L \) is the length of the pipe.

### Step 3: Rearrange the formula to find the length of the pipe
We can rearrange the formula to solve for \( L \):
\[ L = \frac{4v}{f} \]

### Step 4: Substitute the values into the formula
Now, substituting the values we have:
\[ L = \frac{4 \times 344.15}{512} \]

Calculating the numerator:
\[ 4 \times 344.15 = 1376.6 \]
Now, divide by the frequency:
\[ L = \frac{1376.6}{512} \approx 2.69 \, \text{m} \]

### Step 5: Final calculation
Calculating the final value:
\[ L \approx 0.269 \, \text{m} \text{ (or 26.9 cm)} \]

Thus, the length of the open organ pipe is approximately **0.269 m**.