Explanation

For an open organ pipe, the frequency of the $n^{th}$ harmonic is given by

$$ \nu_n = \frac{n v}{2L} $$

where $v$ is the velocity of sound in air and $L$ is the length of the pipe.

Given that $\nu_3$ is the 3rd harmonic frequency,

$$ \nu_3 = \frac{3v}{2L} $$

Similarly, $\nu_6$ is the 6th harmonic frequency,

$$ \nu_6 = \frac{6v}{2L} $$

Now find the difference between these two frequencies,

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

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

According to the question,

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

Substitute the value of velocity of sound $v = 330\ \text{m/s}$,

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

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

$$ 990 = 4400L $$

$$ L = \frac{990}{4400} = 0.225\ \text{m} $$

Convert length into millimetres,

$$ L = 0.225 \times 1000 = 225\ \text{mm} $$

Hence, the length of the pipe is

225 mm

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