The fundamental frequency for a closed (organ) pipe can be expressed as:

$ f = \frac{v}{4\ell} $

For the first air column, with length $ \ell_1 $, the frequency $ f_1 $ is:

$ f_1 = \frac{v}{4\ell_1} $

For the second air column, with length $ \ell_2 $, the frequency $ f_2 $ is:

$ f_2 = \frac{v}{4\ell_2} $

The beat frequency, which is the difference in these two frequencies ($ f_1 - f_2 $), is given as 15 beats per second:

$ \text{Beat} = f_1 - f_2 = \frac{v}{4} \left( \frac{1}{\ell_1} - \frac{1}{\ell_2} \right) $

Substitute the given lengths into the formula:

$ 15 = \frac{v}{4} \left( \frac{1}{1} - \frac{1}{1.2} \right) $

Simplify the equation:

$ 15 = \frac{v}{4} \left( \frac{0.2}{1.2} \right) $

Solve for $ v $:

$ v = \frac{15 \times 4 \times 1.2}{0.2} = 60 \times 6 = 360 \, \text{m/s} $

Thus, the velocity of sound in the air column is 360 m/s.