Question

In an open organ pipe, the 3rd and 6th harmonic frequencies differ by 3200 Hz. Find the length of the organ pipe. (Given: speed of sound = 320 m s\(^{-1}\))

• A. \(5 \text{ cm}\)
• B. \(10 \text{ cm}\)
• C. \(15 \text{ cm}\)
• D. \(20 \text{ cm}\)

Hint:

For open organ pipes:
All harmonics are present
Frequency difference depends only on harmonic numbers and pipe length

Answer 1

The Correct Option is C

Concept: For an open organ pipe, the frequency of the \(n^\text{th}\) harmonic is given by: \[ f_n = \frac{n v}{2L} \] where \(v\) = speed of sound, \(L\) = length of the pipe.
Step 1: Write expressions for the given harmonics. 3rd harmonic: \[ f_3 = \frac{3v}{2L} \] 6th harmonic: \[ f_6 = \frac{6v}{2L} \]
Step 2: Use the given difference in frequencies. \[ f_6 - f_3 = 3200 \] \[ \frac{6v}{2L} - \frac{3v}{2L} = 3200 \] \[ \frac{3v}{2L} = 3200 \]
Step 3: Substitute the value of speed of sound. \[ \frac{3 \times 320}{2L} = 3200 \] \[ \frac{960}{2L} = 3200 \] \[ 960 = 6400L \] \[ L = 0.15 \text{ m} \]
Step 4: Convert into centimeters. \[ L = 0.15 \text{ m} = 15 \text{ cm} \] \[ \boxed{L = 15 \text{ cm}} \]