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

To solve this problem, we need to find the length of an open organ pipe based on given harmonic frequencies. In an open organ pipe, the frequencies of the harmonics are given by:

\(\nu_n = n \cdot \frac{v}{2L}\) 

where:

• \(\nu_n\) is the frequency of the \(n^{th}\) harmonic,
• \(v\) is the velocity of sound,
• \(L\) is the length of the pipe.

The problem states that \(\nu_3\) and \(\nu_6\) are the 3rd and 6th harmonic frequencies, respectively. Thus, we can express them as:

\(\nu_3 = 3 \cdot \frac{v}{2L}\)

\(\nu_6 = 6 \cdot \frac{v}{2L}\)

According to the question, the difference between the 6th and 3rd harmonics is 2200 Hz:

\(\nu_6 - \nu_3 = 2200\)

Substituting the values of \(\nu_3\) and \(\nu_6\) from the expressions above, we have:

\(6 \cdot \frac{v}{2L} - 3 \cdot \frac{v}{2L} = 2200\)

Simplifying this gives:

\((6 - 3) \cdot \frac{v}{2L} = 2200\)

\(3 \cdot \frac{v}{2L} = 2200\)

Plugging in the velocity of sound \(v = 330\ \text{m s}^{-1}\), we get:

\(3 \cdot \frac{330}{2L} = 2200\)

Solving for \(L\):

\(\frac{990}{2L} = 2200\)

\(\frac{990}{2 \times 2200} = L\)

\(\frac{990}{4400} = L\)

\(L = \frac{9}{40}\ \text{m}\)

Converting meters to millimeters:

\(L = \frac{9}{40} \times 1000\ \text{mm} = 225\ \text{mm}\)

Thus, the length of the pipe is 225 mm. Therefore, the correct answer is 225.