A piece of cardboard tubing, closed at one end, is just the right length so that when it is cut into two pieces,
the lowest resonant frequency is 256 Hz for the piece with the closed end and 440 Hz for the other.
a) What is the lowest resonant frequency that would have been produced by the original cardboard tubing?
b) How long was the original tubing?
The figure is below.The speed of sound in dry air at 20 degree Celsius is `v =343 m/s` .
In the portion of tube closed at one end there is only one quarter of wavelength (the standing wave in this tube need to have a minimum at the closed end and a maximum at the open end).
`L1 =lambda/4 = v*T/4 = v/(4F) =343/(4*256) =0.335 m`
In the portion of the tube open at both ends there is half of wavelength present (the standing wave in this tube need to have maxima of oscillation at both ends).
`L2 = lambda/2 =(v*T)/2 =V/(2F) =343/(2*440) =0.390m`
Thus the original tube length is
`L = L1+L2 =0.335+0.390 =0.725 m`
This original tube is also closed at one end and open at the other end, and therefore inside it at the lowest resonant frequency there need to be only one quarter of wavelength.
`L = lambda/4`
`lambda = 4*L`
`F = v/(4L) =343/(4*0.725) =118.28 Hz =118 Hz`
Answer: The lowest resonant frequency produced by the original tube is 118 Hz and the length of tube is 0.725 m.