A metre - long tube open at one end, with a movalble piston at the other end, shows resonance with a fixed frequency source (a tunning fork of frequency 340 Hz)when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Explanation

For an air column in a closed pipe (a pipe, closed at one end and open at another end), frequency of `n^(th)` harmonic is given by,
`f _(n) = (2n -1) (v)/(4L) `(where `n = 1,2,3,…)`
Here foe first time resonance,
`fn_(1) = (2n _(1) -1) (v)/(4L_(1)) = 340 Hz`
Similarly for seocnd time resonance,
f `f n_(2) = (2n _(2) -1) (v)/(4L_(2)) = 340 Hz`
`therefore (2n_(1) -1) (v)/(4L _(1)) = (2n _(2) -1) (v)/(4L_(2))`
`therefore (2n _(1) -1)/( 2n _(1) -1) = (L _(1))/( L _(2)) = (25.5)/(79.3) =0.3216 ~~0.33 =1/3`
`implies n _(1) =1 and n _(2) =2`
`implies L_(1) =25.5 cm` corrsponds to fundamental frequency `f _(1)` and so.
`f _(1) = (v)/(4L _(1)) = 340 Hz `
`therefore v = 4 f _(1) L _(1) =4 xx 340 xx 0.255 =346.8 m/.s`
OR
`L _(2) = 79.3` cm corresponds to first overtone or third harmonic `f _(2) =3f_(1)) `and so.
`f _(2) = (3v)/(4L_(2)) = 340 Hz`
`therefore v = 4/3 xx 340 xx L _(2) = 4/3 xx 340 xx 3L _(1)`
`therefore v =4/3 xx 340 xx 3 xx 0.255 =346.8 (m)/(s)`