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.
Solution
Given:
Frequency of tuning fork f = 512 Hz
Let the speed of sound in the tube be v.
Let l1 be the length at which the piston resonates for the first time and l2 be the length at which the piston resonates for the second time.
We have:
l2 =2l1 = 2 \[\times\]32 = 64 cm =0.64 m
Velocity v = f \[\times\] l2
\[\Rightarrow\] v = 512 × 0.64 = 328 m/s
Hence, the speed of the sound in the tube is 328 m/s.
