A piston is inserted into a cylindrical tube of small cross section whose other end is open. The tube resonates with a tuning fork having a frequency of 512 Hz . As the piston is gradually pulled out of the tube, a second resonance is observed when the piston has been displaced by a distance of 32.0 cm . Determine the speed of sound in the air inside the tube.

Step-by-step solution

The distance between two consecutive resonant lengths in a closed organ pipe is equal to half the wavelength of the sound wave. L = 2 Given that the displacement of the piston between the first and second resonance is 32.0 cm , we have: L = 32.0 cm = 0.32 m 2 = 0.32 = 0.64 m The speed of sound v is given by the relation: v = f Substituting the given frequency f = 512 Hz and the calculated wavelength = 0.64 m : v = 512 0.64 = 327.68 m/s Rounding off to the nearest integer, we get v 328 m/s .