In an open organ pipe 3 and 6 are 3rd and 6th harmonic frequencies, respectively. If 6 - 3 = 2200 Hz, then the length of

Question

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

Accepted Answer

Given: In an open organ pipe, harmonic frequencies $\nu_3$ and $\nu_6$ are the $3$ rd and $6$ th harmonics. Also, $\nu_6 - \nu_3 = 2200\, \text{Hz}$ and speed of sound $v = 330\, \text{m s}^{-1}$ .

Find: The length $L$ of the pipe.

For an open organ pipe,

\nu_n = \frac{nv}{2L}

Therefore,

\nu_3 = \frac{3v}{2L}, \qquad \nu_6 = \frac{6v}{2L}

Using the given frequency difference,

\nu_6 - \nu_3 = \frac{6v}{2L} - \frac{3v}{2L} = \frac{3v}{2L}

So,

\frac{3v}{2L} = 2200

Substitute $v = 330\, \text{m s}^{-1}$ :

\frac{3 \times 330}{2L} = 2200

\frac{990}{2L} = 2200

2L = \frac{990}{2200} = 0.45

L = 0.225\, \text{m} = 225\, \text{mm}

Therefore, the length of the pipe is $225\, \text{mm}$ . The correct option is B.

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