I cant conceptually understand standing waves!

[Meredith92]

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I understand all the rules about standing waves but I'm having understand conceptually getting the motion of it. so you have a wave traveling that "reflects" of a stationary end and inverts 180 degrees so that the incident and reflected wave constructively interfere to create a larger wave as seen with the pic on this website (http://hyperphysics.phy-astr.gsu.edu/‌hbase/waves/standw.html)

Where I'm confused is with the picture I've attached from a diagram in TPR's chapter on waves. How do the antinodes move up and down like that? I've been trying to draw it out but I cant understand what is going on with the incident/ reflected waves to cause them to summate somewhat destructively at these points, while the nodes stay in the same spot. Can anyone help explain this to me?

Thanks so much!

 

[Francium87]

I understand all the rules about standing waves but I'm having understand conceptually getting the motion of it. so you have a wave traveling that "reflects" of a stationary end and inverts 180 degrees so that the incident and reflected wave constructively interfere to create a larger wave as seen with the pic on this website (http://hyperphysics.phy-astr.gsu.edu/‌hbase/waves/standw.html)

Where I'm confused is with the picture I've attached from a diagram in TPR's chapter on waves. How do the antinodes move up and down like that? I've been trying to draw it out but I cant understand what is going on with the incident/ reflected waves to cause them to summate somewhat destructively at these points, while the nodes stay in the same spot. Can anyone help explain this to me?

Thanks so much!

the reason why they r moving up and down because when a wave travelling at some medium and hit a denser medium it reflects back (the dotted lines) ofcourse if no energy is dissipated then the wave would reflect the same energy (same amplitude)

 

[Meredith92]

right, but why do the antinodes move up and down? are they overlapping at lower parts of their crests creating a smaller constructive amplitude?

 

[ScienceNerdUofU]

It looks to me like destructive interference and that is why you get the smaller amplitude. The amplitude of the incident wave is depending on the energy of the reflected wave because it is destructively interfering.

 

[milski]

Illustrating standing waves with a static pictures is rather hard. Of all the possible snapshots in time that could have been used, the one on hyperphysics is probably the worst, or at least the most confusing. Here is something slightly better:

Let's see if I can explain what's going on. First we have two waves, f1 and f2. f1 is moving from left to right and f2 is moving from right to left and they are 180 degress out of phase with each other. At a nodal point, f1(t)=-f2(t) at any time. The resultant wave is f1(t)+f2(t)=0 which is what we expect.

Now let's consider a point slightly further to the right from the node. Let's say that the node is at the upslope of f1 at this moment. This means that f1 to the right is slightly higher than f1 at the node. Since f2 propagates in the opposite direction and is 180 degree out of phase with f1, at a point slightly to the right of the node it will be exactly the opposite of what f1 is to the left of the node, or slightly higher than what it was at the node.

In the graph below, the node is at the vertical line around 3.1 and the point that we discuss later at the vertical line around 3.3. f1 is blue, f2 is red and their sum green. The most important point is that slightly to the right of the node both f1 and f2 are larger than what they were at the node.

 

[ScienceNerdUofU]

I think I misread your question but too answer it, I would say that the anti-nodes are increasing in amplitude because of the energy of the reflected wave. It shows multiple possibilities depending on the amount of constructive interference.

 

[Meredith92]

AH omg Milski, I could marry you. This is what I've been looking for! Its been so hard to find this explanation, but it makes so much more sense now! Thanks for your help as well, ScienceNerdUofU!

I hate to ask another question, but I've been struggling on this for a bit now and would love someone help. I've attached a picture of an open pipe. Why is the first reflected pulse low pressure, when the second reflected pulse is high pressure? How is this possible when they are both exposed to the same open air interface?

Thanks again!

Edit: does this have to do with the wave inverting when it reaches the open interface since its going from a dense to a less dense area?

 

[milski]

Nice to know that my physics major is useful for something. 😉

Yes, you've got it already. It's pretty much the equivalent of the inversion of the wave. I don't want to go in details since I'll probably butcher it but when the high pressure reaches the end of the pipe a bit of extra air from the pipe escapes with it. That creates a slightly lower pressure in the pipe which propagates back along it. At the other end you have the opposite situation and the result is another high pressure front moving the opposite way and so on and so on...

 

[Meredith92]

Thanks again! my only question is... what makes the opposite happen at the other end? When the wave goes to the open end it loses a bit of pressure, and then it makes its way to the other open end, which i would have thought would result in the same loss of pressure.

However, when I think of it more in terms of a wave inversion from a higher to lower pressured area, it makes sense that the wave would invert from a high pressure wave to a low pressure wave (like the crest of wave flipping over to become the trough).


Edit: I'm trying to understand the graphs on this website but I"m getting confused about pressure vs. movement of air http://www.phys.unsw.edu.au/jw/flutes.v.clarinets.html

 

[milski]

Thanks again! my only question is... what makes the opposite happen at the other end? When the wave goes to the open end it loses a bit of pressure, and then it makes its way to the other open end, which i would have thought would result in the same loss of pressure.

However, when I think of it more in terms of a wave inversion from a higher to lower pressured area, it makes sense that the wave would invert from a high pressure wave to a low pressure wave (like the crest of wave flipping over to become the trough).

It's about the difference between the pressure in the pipe and the outside pressure. When the low/high pressure 'front' reaches the end, there is some air flowing in/out of the pipe from/to outside to restore it back to normal pressure. That creates the opposite high/low pressure front moving back.

It's all symmetrical around the baseline - the atmospheric pressure. If you come up with higher pressure, you leave with lower and vice versa.

 

[Meredith92]

It's about the difference between the pressure in the pipe and the outside pressure. When the low/high pressure 'front' reaches the end, there is some air flowing in/out of the pipe from/to outside to restore it back to normal pressure. That creates the opposite high/low pressure front moving back.

It's all symmetrical around the baseline - the atmospheric pressure. If you come up with higher pressure, you leave with lower and vice versa.

OOO that make a lot of sense, thank you!

 

[dudewheresmymd]

Illustrating standing waves with a static pictures is rather hard. Of all the possible snapshots in time that could have been used, the one on hyperphysics is probably the worst, or at least the most confusing. Here is something slightly better:

Let's see if I can explain what's going on. First we have two waves, f1 and f2. f1 is moving from left to right and f2 is moving from right to left and they are 180 degress out of phase with each other. At a nodal point, f1(t)=-f2(t) at any time. The resultant wave is f1(t)+f2(t)=0 which is what we expect.

Now let's consider a point slightly further to the right from the node. Let's say that the node is at the upslope of f1 at this moment. This means that f1 to the right is slightly higher than f1 at the node. Since f2 propagates in the opposite direction and is 180 degree out of phase with f1, at a point slightly to the right of the node it will be exactly the opposite of what f1 is to the left of the node, or slightly higher than what it was at the node.

In the graph below, the node is at the vertical line around 3.1 and the point that we discuss later at the vertical line around 3.3. f1 is blue, f2 is red and their sum green. The most important point is that slightly to the right of the node both f1 and f2 are larger than what they were at the node.

Animated standing waves? You've outdone yourself with this one, milski.

 

[Meredith92]

Haha agreed! We gotta all chip in and buy milski something for all his help!

 

[milski]

It has been quite beneficial to me as well. Nothing clears up concepts like trying to explain them to somebody else. With that said, I will not turn down any offers for candies, chocolate or marriage proposals. 😀

 

[dudewheresmymd]

Haha yeah I love explaining stuff on SDN for exactly the same reason. Everyone wins.

Though sometimes I wish there was a "push to talk button on SDN" so that you wouldn't have to write out long responses. Now wouldn't that be something?

 

[milski]

I can write pretty fast, it's when it comes to drawings or writing formulas when things get "esthetically displeasing." 🙂

 

[mehc012]

I can write pretty fast, it's when it comes to drawings or writing formulas when things get "aesthetically displeasing." 🙂

Yeah, push to talk would be sad for all of us out here trying to passively learn from everyone else's questions!

 

[ldiaz29]

Illustrating standing waves with a static pictures is rather hard. Of all the possible snapshots in time that could have been used, the one on hyperphysics is probably the worst, or at least the most confusing. Here is something slightly better:

Let's see if I can explain what's going on. First we have two waves, f1 and f2. f1 is moving from left to right and f2 is moving from right to left and they are 180 degress out of phase with each other. At a nodal point, f1(t)=-f2(t) at any time. The resultant wave is f1(t)+f2(t)=0 which is what we expect.

Now let's consider a point slightly further to the right from the node. Let's say that the node is at the upslope of f1 at this moment. This means that f1 to the right is slightly higher than f1 at the node. Since f2 propagates in the opposite direction and is 180 degree out of phase with f1, at a point slightly to the right of the node it will be exactly the opposite of what f1 is to the left of the node, or slightly higher than what it was at the node.

In the graph below, the node is at the vertical line around 3.1 and the point that we discuss later at the vertical line around 3.3. f1 is blue, f2 is red and their sum green. The most important point is that slightly to the right of the node both f1 and f2 are larger than what they were at the node.

This was a really helpful explanation for Standing Waves thank you!