Hi could someone see if I have answered this correctly 1)-Explain how travelling standing waves can be formed by travelling waves, and hence how the freq's. of a standing waves in an open tube are related to the speed at which sound travels along the tube? standing travelling waves formed when to waves travelling in opposite directions with same freq. and amplitude cancel each other out-but nodes and antinodes exist coausing regions of high amp. and regions of no sound. but I'm not sure on how they are related to the speed?? Thanks physics_06er
Think about the boundary conditions on the standing wave. You have to have a certain number of waves in the tube to form a standing wave, hence given that sound will travel at the same speed, only certain frequencies will give a standing wave.
I've thought about this for a while since reading your post, and it's a surprisingly difficult question to answer. I think the place to start might be two kids swinging a large skipping rope - a single antinode between two nodes. Why does the rope swing at the speed that it does? What happens as it's swung faster or slower? Now extend the scenario by adding another skipping rope between one of the children and a third child. Think of the child in the middle. What is the natural way to swing the second rope compared to the first? Why? What's different about swinging the two ropes in different ways? What forces act on the child in each case? I don't know if this line of thought will be productive... I'm still thinking. It might be productive to think of each antinode as a pendulum of some sort. O
On a purely mathematical level, it's rather nice - a travelling wave described by A=sin(wt-kx) is completely reflected at a boundary, to give a second wave, A'=sin(wt+kx). with a little rearrangement, the sum of the two waves can be shown to be 2*sin(wt)*cos(kx), which demonstrates the oscillations of amplitude with time, and the fixed positions of the nodes.
hmm...still somewhat confused-have an exam tomorrow!-pretty sure a question like this will come up (similar have come up in the past)-hopefully will understand by then! Please Register or Log in to view the hidden image! ...anyways thanks for the replies...with the rope analogy if it was swung faster/slower the freq. would change right? cos the children would jump less/more (freq.) Please Register or Log in to view the hidden image! thanks
That's right... where I was going with it is that to swing the rope faster, then children must pull on it harder. That means that the tension in the rope is getting higher, and the speed of waves in the rope is increasing. For a child in between two ropes (a inode), the easiest thing is to have the ropes always opposite each other (one arm down while the other is up, one arm forward while the other is back) so that the pull of the ropes is balanced. This means that if the ropes are tied together at this point, the child wouldn't need to be there - the ropes would swing each other around, instead of the child swinging them. I don't know if this helps. If you can't get an intuitive grasp on it, don't worry - focus on being able to do the maths. When you grok the maths, the intuitive understanding will follow.
yeah kinda does-but if i were to explain in an open tube-how could i talk bout "increasing the speed" in a tube-its "easier" it talk about it in a string-but how could i say that tightening the tube would increase speed etc-:bugeye:
