On closing one end of an open pipe, the frequency of fifth harmonic of closed pipe becomes larger by 200 Hz than the fundamental frequency of open pipe. Calcuate the fundamental frequency of open pipe.

Answer

To solve the problem, we need to analyze the frequencies of the open pipe and the closed pipe.

### Step 1: Define the fundamental frequency of the open pipe
The fundamental frequency (ν₀) of an open pipe is given by the formula:
\[ ν₀ = \frac{V}{2L} \]
where \( V \) is the speed of sound in air and \( L \) is the length of the pipe.

### Step 2: Define the frequency of the closed pipe
When one end of the open pipe is closed, it becomes a closed pipe. The fundamental frequency (ν₁) of a closed pipe is given by:
\[ ν₁ = \frac{V}{4L} \]
The frequency of the 5th harmonic (ν₅) of the closed pipe is:
\[ ν₅ = 5 \cdot ν₁ = 5 \cdot \frac{V}{4L} = \frac{5V}{4L} \]

### Step 3: Establish the relationship between the frequencies
According to the problem, the frequency of the 5th harmonic of the closed pipe is larger by 200 Hz than the fundamental frequency of the open pipe. Therefore, we can write:
\[ ν₅ = ν₀ + 200 \text{ Hz} \]
Substituting the expressions for \( ν₅ \) and \( ν₀ \):
\[ \frac{5V}{4L} = \frac{V}{2L} + 200 \]

### Step 4: Solve for the speed of sound
To solve for \( V \), we first find a common denominator and rearrange the equation:
\[ \frac{5V}{4L} - \frac{2V}{4L} = 200 \]
This simplifies to:
\[ \frac{3V}{4L} = 200 \]
Now, multiply both sides by \( 4L \):
\[ 3V = 800L \]
Thus, we have:
\[ V = \frac{800L}{3} \]

### Step 5: Substitute back to find the fundamental frequency of the open pipe
Now, we substitute \( V \) back into the formula for the fundamental frequency of the open pipe:
\[ ν₀ = \frac{V}{2L} = \frac{800L/3}{2L} = \frac{800}{6} = \frac{400}{3} \text{ Hz} \]
Calculating this gives:
\[ ν₀ \approx 133.33 \text{ Hz} \]

### Final Answer
The fundamental frequency of the open pipe is approximately **133.33 Hz**.

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