Length of string in a standing wave

  • #1
laser
81
11
Homework Statement
see description
Relevant Equations
v = fY, where Y is lambda
Screenshot_6.png

In part (c), I have no problem if the string could stretch, but consider an inextensible string. This could still form a standing wave, according to google.
Screenshot_7.png

But then how is the string (the wave) equal to L? So can we actually equate lambda/4 with L? i.e. it is curved and varies with amplitude.

In addition, I don't understand how if you have an inextensible string fixed between two points (a different scenario), how you can even pluck it. Because surely this will stretch the string? But the string is inextensible!
 
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  • #2
laser said:
if you have an inextensible string fixed between two points (a different scenario), how you can even pluck it
Quite. All strings are extensible. An inextensible string is an idealisation. The reliable way to solve problems with idealisations is to analyse a realistic version then take the limit as the idealisation is approached.
Please post the link to the claim that an inextensible string can form a standing wave.
 
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  • #3
That said ...

The wave equation for transversal waves on a string is already an approximation in itself building upon linearisation of the system in the deviation from rest ##u(x,t)##. This is only ever valid for small ##u## and for small ##u##, the length of the string receives corrections which are of second order in ##u_x(x,t)##. Hence, to the approximation order (i.e., linear in ##u##) the string is indeed well described as inextensible because the first correction to this enters at quadratic order.
 
  • #4
laser said:
But then how is the string (the wave) equal to L?
The string is an object. Length is a property. It doesn't make sense to say they are equal.

Perhaps you mean wavelength? That is not equal to ##L##. The wavelength is equal to ##4L##.
laser said:
So can we actually equate lambda/4 with L? i.e. it is curved and varies with amplitude.
That figure greatly exaggerates the curvature of the string. We are using an approximation where the amount of curvature is negligible. (See below for further explanation).

laser said:
In addition, I don't understand how if you have an inextensible string fixed between two points (a different scenario), how you can even pluck it. Because surely this will stretch the string? But the string is inextensible!
There is no such thing as a string that can't be stretched. The term inextensible refers to an approximation where the amount the string is stretched is negligible compared to the length of the string.

Note that frictionless is another example of this type of jargon. It just means that friction is negligible. In others words, the effects of friction are so small that they cannot be detected in a particular circumstance.
 
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  • #5
laser said:
In part (c), I have no problem if the string could stretch, but consider an inextensible string. This could still form a standing wave, according to google.
Screenshot_7.png
I didn't notice this at first, but the standing wave you have drawn is not possible for a vibrating string. Both ends of the string have to be fixed, that is, they must be displacement nodes. Thus you cannot have a quarter of a wave.

What you have drawn is possible for other wave-carrying media, such as a sound wave in air (the so-called open pipe) but not for a string.

I know that the problem states that one end of the string is fixed, but it doesn't say only one end is fixed. It's kind of a trick question. Either that, or it's nonsense.
 
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  • #6
Mister T said:
I know that the problem states that one end of the string is fixed, but it doesn't say only one end is fixed. It's kind of a trick question. Either that, or it's nonsense.
When waves are being sent down from the other end, e.g. by moving your hands up and down, is that not an antinode? I kind of see your point though - by barely moving your hands it would still oscillate, and you could consider both ends a node. But what happens if you move your hands a noticeable amount, such that it is equal to the amplitude? Or is this possible?!
 
  • #7
Mister T said:
I didn't notice this at first, but the standing wave you have drawn is not possible for a vibrating string. Both ends of the string have to be fixed, that is, they must be displacement nodes. Thus you cannot have a quarter of a wave.
This is incorrect. It depends on the type of boundary condition you impose. It is certainly possible to have Neumann boundary conditions on either end.

Edit: If you have access to my book, Example 3.14 discusses how Neumann boundary conditions arise for the case of a vibrating string.

Mister T said:
What you have drawn is possible for other wave-carrying media, such as a sound wave in air (the so-called open pipe) but not for a string.
See above.
 
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  • #8
laser said:
When waves are being sent down from the other end, e.g. by moving your hands up and down, is that not an antinode? I kind of see your point though - by barely moving your hands it would still oscillate, and you could consider both ends a node. But what happens if you move your hands a noticeable amount, such that it is equal to the amplitude? Or is this possible?!
The above being said, the forcing motion of one end will fix the position of that end. This will lead to excitations of the modes where the end points are indeed fixed so in this particular case, the minimum wavelength is L/2 and not L/4.
 
  • #9
Orodruin said:
the forcing motion of one end will fix the position of that end.
Take a piece of string and move your hands up and down. Your hands are moving, so the position isn't fixed, right?
 
  • #10
laser said:
Take a piece of string and move your hands up and down. Your hands are moving, so the position isn't fixed, right?
It is fixed to be where your hand is. This is what is relevant. It will not be a position of max amplitude. You can easily check this as you will be able to achieve quite large amplitudes using small hand movements (if you use the appropriate frequency).
 
  • #11
Orodruin said:
You can easily check this as you will be able to achieve quite large amplitudes using small hand movements
Yes but if you use large hand movements the size is comparable to the amplitude? Or perhaps a standing wave would not form...?
 
  • #12
laser said:
Yes but if you use large hand movements the size is comparable to the amplitude? Or perhaps a standing wave would not form...?
The formation of a standing wave depends on the frequency of your hand movement. To create a standing wave you need to (approximately) match a resonant frequency of the string. If you do not, then the string motion will not be much larger than that of your hand and the string shape will not be restricted to the shape of one of the standing wave modes.
 
  • #13
Orodruin said:
The formation of a standing wave depends on the frequency of your hand movement. To create a standing wave you need to (approximately) match a resonant frequency of the string. If you do not, then the string motion will not be much larger than that of your hand and the string shape will not be restricted to the shape of one of the standing wave modes.
So, you could just adjust the frequency of your hand movement... I don't understand what you are trying to say.
 
  • #14
If you match the resonant frequency, then you will have a significantly larger amplitude of oscillations than the amplitude of your hand movement.
 
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  • #15
Orodruin said:
This is incorrect. It depends on the type of boundary condition you impose. It is certainly possible to have Neumann boundary conditions on either end.
You're right! I'd forgotten about this possibility. In practice it can be achieved by attaching a ring to the end of the string, with the ring free to slide transversely on a rod. I'll go back and fix my post.
 
  • #16
laser said:
When waves are being sent down from the other end, e.g. by moving your hands up and down, is that not an antinode? I kind of see your point though - by barely moving your hands it would still oscillate, and you could consider both ends a node. But what happens if you move your hands a noticeable amount, such that it is equal to the amplitude? Or is this possible?!
Yes, it's possible. I was wrong. It's also possible using a ring and a rod. (See Post #15).
 
  • #17
Orodruin said:
If you have access to my book, Example 3.14 discusses how Neumann boundary conditions arise for the case of a vibrating string.
How can I get access? Is it available online? If not, what's the title, etc.?
 
  • #19
Mister T said:
How can I get access? Is it available online? If not, what's the title, etc.?
Do you still have access to the school library where you taught? If they don't have it on the shelves already, they should be able to get it via inter-library loan, I would think. Worst case I can mail you my copy for a few weeks to look through to see if you might be interested in purchasing a copy for yourself.

https://www.amazon.com/Mathematical...attias-ebook/dp/B086H3LMZF/?tag=pfamazon01-20
 
  • #20
berkeman said:
Worst case I can mail you my copy for a few weeks to look through to see if you might be interested in purchasing a copy for yourself.
Thanks. But I should be able to get it from the library.
 
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