An open pipe is 85 cm long. If the velocity of sound is 340 `ms^(-1)`, find the frequency of the fundamental note of the pipe. What would be the length of a closed pipe which produces a fundamental note of the same frequency?

Answer

To solve the problem step by step, we will first find the frequency of the fundamental note of the open pipe and then determine the length of the closed pipe that produces the same frequency.

### Step 1: Find the frequency of the fundamental note of the open pipe.

**Formula:**
The frequency (ν) of the fundamental note for an open pipe is given by the formula:
\[ \nu = \frac{v}{2L} \]
where:
- \( v \) is the velocity of sound,
- \( L \) is the length of the pipe.

**Given:**
- Length of the open pipe, \( L = 85 \, \text{cm} = 0.85 \, \text{m} \) (convert cm to m)
- Velocity of sound, \( v = 340 \, \text{m/s} \)

**Substituting the values:**
\[ \nu = \frac{340 \, \text{m/s}}{2 \times 0.85 \, \text{m}} = \frac{340}{1.7} = 200 \, \text{Hz} \]

### Step 2: Find the length of the closed pipe that produces the same frequency.

**Formula:**
The frequency (ν) of the fundamental note for a closed pipe is given by the formula:
\[ \nu = \frac{v}{4L'} \]
where:
- \( L' \) is the length of the closed pipe.

**Since both pipes have the same frequency, we can set the frequencies equal:**
\[ 200 \, \text{Hz} = \frac{340 \, \text{m/s}}{4L'} \]

**Rearranging to find \( L' \):**
\[ L' = \frac{340 \, \text{m/s}}{4 \times 200 \, \text{Hz}} = \frac{340}{800} = 0.425 \, \text{m} = 42.5 \, \text{cm} \]

### Final Answers:
- The frequency of the fundamental note of the open pipe is \( 200 \, \text{Hz} \).
- The length of the closed pipe that produces the same frequency is \( 42.5 \, \text{cm} \).

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