Question

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.

Answer 1

The piston acts as the closed end of the organ pipe. At the piston, there will be pressure antinode while at the open end there will be a pressure node. Since in both the cases the frequency of the vibration (or the wavelength of the sound) is same = 512 Hz, the next resonance will occur when the length of the air column is increased to λ/2 because one more node will be induced in the air column. From the given problem,

λ/2 = 32 cm = 0.32 m

→ λ= 0.64 m

ν = 512 Hz

Hence the speed of the sound in air V = νλ

→V = 5120.64 m/ *≈ 328 m/.**