[Q. Nos. 3-4] Difference in frequencies between 3rd overtone of closed pipe and 5th harmonic of the same pipe is 400 Hz. Further 3rd hormonic of this closed pipe is equal to 6th hormonic of another open pipe. If speed sound is 330 m/s. Then lengths of closed pipe and open pipe are a) 0.4125 m, 0.825 m b) 3.3m, 1.65 m c) 0.825 m, 0.825 m d) 1.65 m, 0.825 m

Options

A. 0.4125 m, 0.825 m
B. 3.3m, 1.65 m
C. 0.825 m, 0.825 m
D. 1.65 m, 0.825 m

Answer

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Understand the Frequencies of the Closed Pipe
The difference in frequencies between the 3rd overtone of a closed pipe and the 5th harmonic of the same pipe is given as 400 Hz.

- The frequency of the nth overtone (or harmonic) in a closed pipe is given by:
\[ f_n = \frac{(2n - 1) v}{4L} \]
where \( n \) is the harmonic number, \( v \) is the speed of sound, and \( L \) is the length of the pipe.

### Step 2: Calculate the Frequencies
For the 3rd overtone (which is the 4th harmonic, \( n = 4 \)):
\[ f_4 = \frac{(2 \cdot 4 - 1) v}{4L} = \frac{7v}{4L} \]

For the 5th harmonic (\( n = 5 \)):
\[ f_5 = \frac{(2 \cdot 5 - 1) v}{4L} = \frac{9v}{4L} \]

### Step 3: Set Up the Equation for Frequency Difference
The difference in frequencies is given as:
\[ f_4 - f_5 = 400 \text{ Hz} \]
Substituting the expressions for \( f_4 \) and \( f_5 \):
\[ \frac{7v}{4L} - \frac{9v}{4L} = 400 \]
This simplifies to:
\[ \frac{-2v}{4L} = 400 \]
\[ \frac{-v}{2L} = 400 \]
\[ v = -800L \]

### Step 4: Substitute the Speed of Sound
Given that the speed of sound \( v = 330 \text{ m/s} \):
\[ 330 = 800L \]
Solving for \( L \):
\[ L = \frac{330}{800} = 0.4125 \text{ m} \]

### Step 5: Relate Closed and Open Pipe Frequencies
It is given that the 3rd harmonic of the closed pipe is equal to the 6th harmonic of another open pipe.

For the closed pipe (3rd harmonic):
\[ f_3 = \frac{(2 \cdot 3 - 1)v}{4L} = \frac{5v}{4L} \]

For the open pipe (6th harmonic):
\[ f_6 = \frac{6v}{2L_o} = \frac{3v}{L_o} \]

Setting these equal:
\[ \frac{5v}{4L} = \frac{3v}{L_o} \]

### Step 6: Cancel \( v \) and Rearrange
Cancel \( v \) from both sides:
\[ \frac{5}{4L} = \frac{3}{L_o} \]
Cross-multiplying gives:
\[ 5L_o = 12L \]
Substituting \( L = 0.4125 \text{ m} \):
\[ L_o = \frac{12 \cdot 0.4125}{5} = 0.825 \text{ m} \]

### Final Answer
The lengths of the closed pipe and the open pipe are:
- Length of closed pipe \( L = 0.4125 \text{ m} \)
- Length of open pipe \( L_o = 0.825 \text{ m} \)

Thus, the correct option is:
**a) 0.4125 m, 0.825 m**