Formula for harmonics in an open organ pipe

For an open organ pipe of length $L$, the frequency of the $n^{\text{th}}$ harmonic is $f_n = \frac{n v}{2L}$ where $v$ is the speed of sound.

Frequency of the 6th harmonic for the first pipe

Given: $L_1 = 60\text{ cm} = 0.60\text{ m}, \quad n_1 = 6$

So, $f_1 = \frac{6 \times 333}{2 \times 0.60}$

$f_1 = \frac{1998}{1.2} = 1665\text{ Hz}$

Frequency of the 5th harmonic for the second pipe

Given: $L_2 = 90\text{ cm} = 0.90\text{ m}, \quad n_2 = 5$

So, $f_2 = \frac{5 \times 333}{2 \times 0.90}$

$f_2 = \frac{1665}{1.8} = 925\text{ Hz}$

Difference of frequencies

$\Delta f = f_1 - f_2 = 1665 - 925 = 740\text{ Hz}$

Final answer

The required difference of frequencies is $\boxed{740}$