A piston is fitted in a cylindrical tube of small cross section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the speed of sound in the air of the tube.

Sol. Let n base 0 = frequency of the turning fork, T = tension of the string L = 40 cm = 0.4 m, m = 4g = 4 × 10^–3 kg So, m = Mass/Unit length = 10^–2 kg/m n base 0 = 1/2l√T/m. So, 2nd harmonic 2n base 0 = (2/ 2l) √T /m As it is unison with fundamental frequency of vibration in the air column ⇒ 2n base 0 = 340/4 * 1 = 85 Hz ⇒ 85 = 2/2 * 0.4 √T/14 ⇒ T = 85^2 * (0.4)^2 * 10^-2 =- 11.6 Newton.