For example, the displacement of the air molecules in a closed straight tube must be zero at the end plates. Thus if the length of the tube is an integer multiple of the half- wavelength, there will be nodes, i.e. zero-amplitude points, at the two ends of the tube and a standing wave is maintained easily, in spite of losses during propagation and reflection. The figure above shows such a case with four half-waves. But if you change the length slightly, the bouncing waves will not have a stable phase relationship and no standing wave develops. You can still see some oscillations, but they are complicated and the amplitude is very low. Thus standing waves form in a closed tube only if 12/2 = n vT/2 = L, (7) where n is an integer. This equation restricts the possible wavelengths and - through the wave velocity, v, the possible periods and frequencies of a standing wave that can happen for a given tube length. QUESTION 5 Using equation 7 for n = 5, how many wavelengths (A) would fit in a closed tube? Assume the frequency and speed are just right produce a standing wave. O 2.5A O 5A O 7.5X O 10A