Note about open$-$open or closed$-$closed $($symmetric$)$ tubes:

For a tube open at both ends, resonance occurs when an integer number of half$-$wavelengths

fit inside the tube. This can be stated as L $=$ n$($$/2),$ where n is a positive integer $(1,2,3,4,$

$...).$ With a little algebraic shuffling, and using v $=$ f to substitute for $,$ we note that the

resonant frequencies occur at integer multiples of the fundamental: f $=$ n$($v$/2$L$).$ Thus for a

tube that is open at both ends, the difference between two successive resonant frequencies is

the same as the fundamental frequency $($not including end effects$).$

Note about closed$-$open $($anti$-$symmetric$)$ tubes:

For a tube closed at one end, resonance occurs when an odd number of quarter$-$wavelengths

fit inside the tube, or L $=$ n$($$/4),$ where n is an odd integer $(1,3,5,7,...).$

e$.$ Use the notes above and v $=$ f to determine the frequency difference between two

consecutive resonances for a closed$-$open tube. How does this difference compare to the

tube$$s fundamental frequency?