Question

In an open organ pipe \( \nu_3 \) and \( \nu_6 \) are 3rd and 6th harmonic frequencies, respectively. If \( \nu_6 - \nu_3 = 2200\ \text{Hz} \), then the length of the pipe is _________ mm. (Take velocity of sound in air as \(330\ \text{m s}^{-1}\).)

• A. 200
• B. 225
• C. 250
• D. 275

Hint:

In an open pipe, all harmonics are present and the frequency difference between harmonics depends only on the pipe length.

Answer 1

The Correct Option is B

Step 1: Write the expression for harmonic frequencies of an open pipe.
For an open organ pipe, \[ \nu_n = \frac{n v}{2L} \]
Step 2: Write expressions for \( \nu_3 \) and \( \nu_6 \).
\[ \nu_3 = \frac{3v}{2L}, \quad \nu_6 = \frac{6v}{2L} \]
Step 3: Use the given frequency difference.
\[ \nu_6 - \nu_3 = \frac{6v}{2L} - \frac{3v}{2L} = \frac{3v}{2L} \] \[ \frac{3v}{2L} = 2200 \]
Step 4: Substitute the value of velocity and solve for \(L\).
\[ \frac{3 \times 330}{2L} = 2200 \Rightarrow 2L = \frac{990}{2200} \] \[ L = 0.225\ \text{m} = 225\ \text{mm} \]
Final Answer: \[ \boxed{225\ \text{mm}} \]