I learnt standing waves to be a combination of a forward travelling and a backward travelling wave: $$ y_1 = A\sin(kx-\omega t) \\ y_2 = A\sin(kx+\omega t) \\ y = y_1 + y_2 = 2A\sin(kx)\cdot\cos(\omega t) $$ The second wave $y_2$ is the reflected wave travelling in the opposite direction. All the conditions for the length of the tube and the positions of node and antinode can be found from this equation, for a tube closed at one end. However, in the qualitative discussion, I read that "the tuning fork sends forth a compression, which reflects from the sealed end of the tube as a compression, returns back and again reflects from the open end of the tube (becoming a rarefaction). If at this instant, the tuning fork produces a rarefaction, then they constructively interfere and produce an antinode at that position." They use this theory to conclude that (for minimum frequency), $$2L/v = T/2 \implies T = 4L/v \quad\text{ or, }\quad f=v/4L$$ My point is, first they say two travelling waves going in opposite directions are involved, and next they say that a compression which has reflected as a rarefaction, interferes with a nwely produced rarefaction travelling in the same direction, interfere to produce the antinode.

Exactly which waves interfere to produce the standing wave, those going in the same direction, or those going in opposite directions?

And even though I know the answer is the latter, I need an explanation of what is going on here really!

  • $\begingroup$ There are two waves in opposite directions. The tuning fork produces a wave that goes down the tube. The wave reflects at the far end and comes back up the tube. $\endgroup$ – mmesser314 Feb 4 '17 at 14:24
  • $\begingroup$ @mmesser314 Yes, but the theoretical explanation says, the new rarefaction interferes with the old rarefaction, which is obtained after two reflections! This implies that both the said waves are travelling in the same direction (the original wave after 2 reflections is in the same direction as the new one). $\endgroup$ – FreezingFire Feb 4 '17 at 18:07
  • $\begingroup$ It depends on the length of the pipe. The tuning fork vibrates at a fixed rate. That is, the time between one rarefaction and the next is fixed. Likewise the length between one compression and the next rarefaction. The time for the wave to travel the length of the pipe, be reflected, and travel back depends on the length. If you choose the length right, you can arrange for the tuning for to produce the next rarefaction just as a reflected rarefaction returns, or just as a reflected compression returns. You can choose constructive or destructive interference by choosing the length of the pipe. $\endgroup$ – mmesser314 Feb 5 '17 at 0:09
  • $\begingroup$ That describes interference between waves from the tuning fork and waves reflected from the far end. The returning wave also reflects from the open end. It goes down the pipe again and interferes with the wave headed up the pipe. Interference just means waves add. The wave you measure is the sum of all the waves traveling up and down the pipe. Waves going the same direction just sit on top of each other and make a bigger or smaller wave. Waves going in opposite directions make standing waves, which look different than a wave (or sum of waves) traveling together in the same direcction. $\endgroup$ – mmesser314 Feb 5 '17 at 0:14

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