For a certain organ pipe three successive resonance frequencies are observed at 425Hz, 595 Hz and 765Hz respectively. If the speed of sound air is 340m/s, then the length of the pipe is

Options

A. 2.0m
B. 0.4m
C. 1.0m
D. 0.2m

Answer

To find the length of the organ pipe based on the given resonance frequencies, we can follow these steps:

### Step 1: Identify the Frequencies
The three successive resonance frequencies given are:
- \( f_1 = 425 \, \text{Hz} \)
- \( f_2 = 595 \, \text{Hz} \)
- \( f_3 = 765 \, \text{Hz} \)

### Step 2: Determine the Frequency Ratio
We can find the ratios of these frequencies:
- The ratio of \( f_1 : f_2 : f_3 \) can be simplified.
- Dividing each frequency by 425 gives:
- \( f_1 = 425 \, \text{Hz} \) (5th harmonic)
- \( f_2 = 595 \, \text{Hz} \) (7th harmonic)
- \( f_3 = 765 \, \text{Hz} \) (9th harmonic)

Thus, the ratio can be expressed as \( 5 : 7 : 9 \).

### Step 3: Identify the Type of Pipe
Since the ratios of the frequencies are odd integers, we conclude that the organ pipe is a closed pipe. In a closed organ pipe, the harmonics are given by odd multiples of the fundamental frequency.

### Step 4: Use the Formula for Resonant Frequencies
For a closed organ pipe, the frequency of the nth harmonic is given by:
\[ f_n = \frac{nV}{4L} \]
Where:
- \( f_n \) = frequency of the nth harmonic
- \( V \) = speed of sound in air (given as 340 m/s)
- \( L \) = length of the pipe

### Step 5: Set Up the Equation for the 5th Harmonic
Using \( f_1 = 425 \, \text{Hz} \) which corresponds to the 5th harmonic:
\[ 425 = \frac{5 \times 340}{4L} \]

### Step 6: Solve for Length \( L \)
Rearranging the equation to solve for \( L \):
\[ L = \frac{5 \times 340}{4 \times 425} \]

Calculating the values:
\[ L = \frac{1700}{1700} = 1 \, \text{m} \]

### Final Answer
The length of the pipe is \( 1 \, \text{meter} \).

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