Final answer:
Using the properties of air columns in tubes with open and closed ends, the original tube, when open at both ends, should have double the frequency of the cut piece open at one end. The frequency should be 1362 Hz (which is twice 681 Hz), but this option is not listed in the choices provided. There could be an error in the question or the options given.
Step-by-step explanation:
The fundamental frequency of the original tube can be determined by understanding the behavior of air columns in tubes with open and closed ends. A tube that is open at one end and closed at the other has a fundamental frequency that is half that of a tube of the same length open at both ends because the latter has symmetrical boundary conditions and can support a half-wavelength standing wave. Given that the piece open at both ends has a fundamental frequency of 412 Hz and the piece open at one end has a fundamental frequency of 681 Hz, we can deduce that the original tube was open at both ends. This is because the piece with a fundamental frequency of 412 Hz, which is open at both ends, would have had a frequency of 206 Hz (half of 412 Hz) if it were closed at one end, just like the other piece.
Therefore, the original tube, when it was open at both ends, should have had a fundamental frequency that is twice that of the 681 Hz piece (when it was still attached and acting as a closed tube), which makes it 1362 Hz (2 x 681 Hz). However, none of the options provided matches 1362 Hz, so it seems there might be an error in the question or the options given. If considering the question strictly as is, without additional context or possibility of an error, no correct option is listed.