In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:

<p>The fundamental frequency for a closed (organ) pipe can be expressed as:</p> <p>$ f = \frac{v}{4\ell} $</p> <p>For the first air column, with length $ \ell_1 $, the frequency $ f_1 $ is:</p> <p>$ f_1 = \frac{v}{4\ell_1} $</p> <p>For the second air column, with length $ \ell_2 $, the frequency $ f_2 $ is:</p> <p>$ f_2 = \frac{v}{4\ell_2} $</p> <p>The beat frequency, which is the difference in these two frequencies ($ f_1 - f_2 $), is given as 15 beats per second:</p> <p>$ \text{Beat} = f_1 - f_2 = \frac{v}{4} \left( \frac{1}{\ell_1} - \frac{1}{\ell_2} \right) $</p> <p>Substitute the given lengths into the formula:</p> <p>$ 15 = \frac{v}{4} \left( \frac{1}{1} - \frac{1}{1.2} \right) $</p> <p>Simplify the equation:</p> <p>$ 15 = \frac{v}{4} \left( \frac{0.2}{1.2} \right) $</p> <p>Solve for $ v $:</p> <p>$ v = \frac{15 \times 4 \times 1.2}{0.2} = 60 \times 6 = 360 \, \text{m/s} $</p> <p>Thus, the velocity of sound in the air column is <strong>360 m/s</strong>.</p>