[Q. Nos. 3-4] Difference in frequencies between 3rd overtone of closed pipe and 5th haronic of the same pipe is 400 Hz. Futher 3rd hormonic of this closed pipe os equal to 6th hormonic of another open pipe.
Fundamental frequencies of closed pipe and open pipe are: a) 200 Hz, 400 Hz b) 150 Hz, 75 Hz c) 200 Hz, 100Hzz d) 400 Hz, 300Hz

Options

A. 200 Hz, 400 Hz
B. 150 Hz, 75 Hz
C. 200 Hz, 100 hz
D. 400 Hz, 300 Hz

Answer

To solve the problem step by step, we need to analyze the information given about the closed pipe and the open pipe.

### Step 1: Understand the Terms
- The **3rd overtone** of a closed pipe corresponds to the **7th harmonic** (since the nth overtone is given by \(2n + 1\)).
- The **5th harmonic** of the same closed pipe corresponds to the **5th harmonic**.
- The difference in frequencies between the 3rd overtone and the 5th harmonic is given as 400 Hz.

### Step 2: Write the Frequency Formulas
For a closed pipe, the frequencies can be expressed as:
- Frequency of the 7th harmonic (3rd overtone):
\[ f_7 = \frac{7V_c}{4L_c} \]
- Frequency of the 5th harmonic:
\[ f_5 = \frac{5V_c}{4L_c} \]

### Step 3: Set Up the Equation
The difference in frequencies is given by:
\[ f_7 - f_5 = 400 \text{ Hz} \]
Substituting the formulas:
\[ \frac{7V_c}{4L_c} - \frac{5V_c}{4L_c} = 400 \]
This simplifies to:
\[ \frac{(7 - 5)V_c}{4L_c} = 400 \]
\[ \frac{2V_c}{4L_c} = 400 \]
\[ \frac{V_c}{2L_c} = 400 \]
\[ V_c = 800L_c \quad \text{(Equation 1)} \]

### Step 4: Relate the Closed Pipe and Open Pipe
According to the problem, the **3rd harmonic** of the closed pipe is equal to the **6th harmonic** of the open pipe:
- Frequency of the 3rd harmonic of the closed pipe:
\[ f_3 = \frac{3V_c}{4L_c} \]
- Frequency of the 6th harmonic of the open pipe:
\[ f_6 = \frac{6V_o}{2L_o} \]

Setting these equal gives:
\[ \frac{3V_c}{4L_c} = \frac{6V_o}{2L_o} \]
Simplifying this:
\[ \frac{3V_c}{4L_c} = \frac{3V_o}{L_o} \]
Cancelling \(3\) from both sides:
\[ \frac{V_c}{4L_c} = \frac{V_o}{L_o} \]
Cross-multiplying gives:
\[ V_c L_o = 4V_o L_c \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 1 into Equation 2
From Equation 1, substitute \(V_c = 800L_c\) into Equation 2:
\[ 800L_c L_o = 4V_o L_c \]
Assuming \(L_c = L_o = L\) (length of the pipes are equal):
\[ 800L L = 4V_o L \]
Cancelling \(L\) (assuming \(L \neq 0\)):
\[ 800 = 4V_o \]
\[ V_o = 200 \text{ m/s} \quad \text{(Equation 3)} \]

### Step 6: Find the Fundamental Frequencies
Now we can find the fundamental frequencies:
- For the closed pipe:
\[ f_c = \frac{V_c}{4L_c} = \frac{800L_c}{4L_c} = 200 \text{ Hz} \]
- For the open pipe:
\[ f_o = \frac{V_o}{2L_o} = \frac{200L_o}{2L_o} = 100 \text{ Hz} \]

### Final Answer
The fundamental frequencies of the closed pipe and open pipe are:
- Closed pipe: 200 Hz
- Open pipe: 100 Hz

Thus, the correct option is:
**c) 200 Hz, 100 Hz**