Two organ pipes of lengths 50cm and 50.5 cm long are sounded together, 3 beats per second are heard. Find their frequencies ?

Answer

To find the frequencies of the two organ pipes, we can follow these steps:

### Step 1: Identify the given values
- Length of the first pipe (L1) = 50 cm = 0.50 m
- Length of the second pipe (L2) = 50.5 cm = 0.505 m
- Beat frequency (f_beat) = 3 beats per second

### Step 2: Use the formula for frequency of an organ pipe
The frequency of a pipe closed at one end is given by the formula:
\[ f = \frac{V}{4L} \]
where:
- \( V \) is the speed of sound in air (approximately 343 m/s at room temperature).
- \( L \) is the length of the pipe.

### Step 3: Calculate the frequencies of both pipes
For the first pipe (L1):
\[ f_1 = \frac{V}{4L_1} = \frac{343}{4 \times 0.50} = \frac{343}{2} = 171.5 \text{ Hz} \]

For the second pipe (L2):
\[ f_2 = \frac{V}{4L_2} = \frac{343}{4 \times 0.505} = \frac{343}{2.02} \approx 169.3 \text{ Hz} \]

### Step 4: Determine the difference in frequencies
The difference in frequencies (which causes the beats) is given by:
\[ |f_1 - f_2| = f_{beat} \]
Given that the beat frequency is 3 Hz, we can set up the equation:
\[ |171.5 - 169.3| = 2.2 \text{ Hz} \]
However, since we know the beat frequency is 3 Hz, we can adjust our calculations to find the correct frequencies.

### Step 5: Adjust the frequencies based on the beat frequency
Since we have a beat frequency of 3 Hz, we can express the relationship as:
\[ f_1 - f_2 = 3 \text{ Hz} \]
Assuming \( f_1 > f_2 \):
\[ f_1 = f_2 + 3 \]

### Step 6: Solve for the frequencies
From the ratio of the lengths of the pipes:
\[ \frac{f_1}{f_2} = \frac{L_2}{L_1} = \frac{50.5}{50} = 1.01 \]
This gives us:
\[ f_1 = 1.01 f_2 \]

Substituting \( f_1 \) in the beat frequency equation:
\[ 1.01 f_2 - f_2 = 3 \]
\[ 0.01 f_2 = 3 \]
\[ f_2 = 300 \text{ Hz} \]

Now substituting back to find \( f_1 \):
\[ f_1 = 1.01 \times 300 \approx 303 \text{ Hz} \]

### Final Frequencies
- Frequency of the first pipe \( f_1 \approx 303 \text{ Hz} \)
- Frequency of the second pipe \( f_2 \approx 300 \text{ Hz} \)