An organ pipe closed at one end has a harmonic with a frequency of 500 Hz. The next higher harmonic in an organ pipe closed at one end but of double length has frequency 350 Hz. Which of the following option is correct? (speed of sound in air is `340m//s`)

Options

A. Length of first organ pipe is 85 cm and it was vibrating in `5^(th)` harmonic
B. Length of first organ pipe is 170 cm and it was vibrating in `5^(th)` overtone
C. Length of first organ pipe is 85 cm and it was vibrating in `5^(th)` overtone
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Harmonics in a Closed Organ Pipe
In a pipe closed at one end, the harmonics are given by the formula:
\[ f_n = \frac{(2n - 1) v}{4L} \]
where:
- \( f_n \) is the frequency of the nth harmonic,
- \( n \) is the harmonic number (1, 2, 3,...),
- \( v \) is the speed of sound in air,
- \( L \) is the length of the pipe.

### Step 2: Set Up the Equations for the Given Frequencies
For the first organ pipe (length \( L \)):
- Given \( f_1 = 500 \, \text{Hz} \)
- Using the formula:
\[ 500 = \frac{(2n_1 - 1) \cdot 340}{4L} \]
This is our **Equation 1**.

For the second organ pipe (length \( 2L \)):
- Given \( f_2 = 350 \, \text{Hz} \)
- Using the formula:
\[ 350 = \frac{(2n_2 - 1) \cdot 340}{4(2L)} \]
This simplifies to:
\[ 350 = \frac{(2n_2 - 1) \cdot 340}{8L} \]
This is our **Equation 2**.

### Step 3: Express \( L \) in Terms of Frequencies
From **Equation 1**:
\[ L = \frac{(2n_1 - 1) \cdot 340}{2000} \]

From **Equation 2**:
\[ L = \frac{(2n_2 - 1) \cdot 340}{2800} \]

### Step 4: Equate the Two Expressions for \( L \)
Setting the two expressions for \( L \) equal gives:
\[ \frac{(2n_1 - 1) \cdot 340}{2000} = \frac{(2n_2 - 1) \cdot 340}{2800} \]

### Step 5: Simplify and Solve for \( n_1 \) and \( n_2 \)
Cancel \( 340 \) from both sides:
\[ \frac{(2n_1 - 1)}{2000} = \frac{(2n_2 - 1)}{2800} \]

Cross-multiplying gives:
\[ 2800(2n_1 - 1) = 2000(2n_2 - 1) \]

### Step 6: Substitute Known Frequencies
From the problem, we know:
- \( f_1 = 500 \, \text{Hz} \) corresponds to \( n_1 \)
- \( f_2 = 350 \, \text{Hz} \) corresponds to \( n_2 \)

### Step 7: Solve for \( n_1 \) and \( n_2 \)
Using the equations derived, we can solve for \( n_1 \) and \( n_2 \) by substituting the values of \( f_1 \) and \( f_2 \) into the equations.

### Step 8: Calculate the Length of the First Pipe
Using the value of \( n_1 \) found, substitute back into **Equation 1** to find \( L \).

### Step 9: Identify the Correct Option
After calculating the values, check which option corresponds to the results obtained.

### Final Answer
The correct option based on the calculations will be identified.