8
$\begingroup$

About two weeks ago there was a mock test in Korea, and a physics question asked if a plucked guitar (it was actually a gayageum, a traditional instrument, but I'll just call it a guitar for convenience) string creates a standing wave.

I've learned in school that this is true, and the answer was true as well. But today my physics teacher said that this is actually false. Because a standing wave is caused by two identical waves traveling in opposite directions, a guitar string cannot create a standing wave. So a plucked guitar string only makes a vibration, not a standing wave.

But this is also mentioned in school textbooks. On the page explaining standing waves, there's a picture of a vibrating string and the caption says, "A string tied at both ends makes a standing wave, causing resonance."

I am confused. Does plucking a guitar string make a standing wave on the string? Or is this just a vibration?

$\endgroup$
  • 2
    $\begingroup$ This is a discussion of nomenclature, which is kind of pointless. It constructs some artificial difference between a "standing wave" and a "vibration". To defend the name standing wave, with the definition your teacher gave, with respect to the guitar: Actually, there are counter-propagating waves on a guitar string (and they are reflected at the fixed ends, so the wave propagates towards the fixed end, is reflected and interferes with its reflection to form a standing wave). $\endgroup$ – Sebastian Riese Jun 20 '18 at 15:20
  • $\begingroup$ That's what I learned in school, but if that forms a standing wave, shouldn't it be possible to make antinodes as well? (I don't know how else to phrase this since I learned in Korean) A string tied at only one end also has counter-propagating waves. You can make standing waves with antinodes on these strings, but you can't with guitar strings. $\endgroup$ – Alvin Kim Jun 20 '18 at 15:41
  • $\begingroup$ Related: physics.stackexchange.com/q/111780/44126 $\endgroup$ – rob Jun 20 '18 at 15:48
  • 4
    $\begingroup$ Actually, there are both nodes and antinodes on a vibrating guitar string vibrating at a single frequency. The nodes are the places where the string can be touched without changing the vibration frequency, and the antinodes are halfway between the nodes. Touch the string there, and the vibration will be damped most quickly. $\endgroup$ – S. McGrew Jun 20 '18 at 16:04
  • $\begingroup$ Yes, but can it make more than one antinode? I don't think it's possible to do so on a guitar string $\endgroup$ – Alvin Kim Jun 20 '18 at 22:33
13
$\begingroup$

Yes, plucking a guitar string does create standing waves, but...
No, plucking a guitar string does not create a standing wave, as the sum of standing waves is in general not a standing wave (thanks for Ben Crowell for pointing this out), since a standing wave must have a stationary spatial dependence and a well-defined frequency:

$$ y(x,t) \propto \sin(2\pi x/\lambda)\cos(\omega t).$$

The initial perturbation is not sinusoidal, but instead contains a plethora of frequencies, of which only remain, after a transient, the resonant ones - which correspond to some of the possible standing waves. It's the sum of those that compose the vibration you'll observe.

The counter-propagating waves, if you want to model each of the standing waves this way, you get from the reflections at the cord's ends.

For more details see this answer and, especially, the answers to the question Why do harmonics occur when you pluck a string?.

$\endgroup$
  • 2
    $\begingroup$ As a matter of terminology, I think a standing wave is defined as something that has a well-defined frequency. It follows that the sum of standing waves is not necessarily a standing wave, and that the waves on a guitar string, in particular, are not. If you define any sum of standing waves to be a standing wave, then you get conclusions that don't make sense, because you can represent a traveling wave as a sum of standing waves. $\endgroup$ – Ben Crowell Jun 20 '18 at 18:00
  • $\begingroup$ @BenCrowell Thanks a lot! Despite the previous upvotes the answer was actually wrong. Now it's fixed. $\endgroup$ – stafusa Jun 20 '18 at 19:24
  • $\begingroup$ @BenCrowell because of the inharmonicity in a real stretched strings, the vibration of a guitar string is not even periodic, so it can't be a standing wave by any reasonable definition of the term. There is also the nonlinear effect of finite amplitude vibrations - but that is probably second order compared with the inharmonicity. $\endgroup$ – alephzero Jun 20 '18 at 22:12
  • $\begingroup$ @alephzero That's true. Even more obvious an effect than inharmonicity is damping, which also makes it aperiodic, but I think the OP asked about the ideal case. $\endgroup$ – stafusa Jun 20 '18 at 22:24
  • $\begingroup$ I'm not sure I understood everything, but do you mean that plucking a guitar string makes standing waves, but the end result is not one? $\endgroup$ – Alvin Kim Jun 20 '18 at 22:40
3
$\begingroup$

The wave created will obey the boundary conditions for all time and in initial conditions at the moment of "pluck". To satisfy the initial conditions one needs to Expand (Fourier transform) the initial shape and initial velocity profile (derivative of shape with respect to time) in an infinite series of the "standing wave" solutions. The time evolution of the profile will cause energy to move back and forth from different modes, if there is a damping mechanism higher frequencies will decay faster eventually leaving the fundamental as the only noticeable mode vibrating.

So, in a sense the answer to this is that a plucked string contains an infinite number of standing waves, and eventually only one standing wave. Based on the dictionary definition of standing wave this is not truly a standing wave as the bulk amplitude profile changes in both time and location between the nodes.

$\endgroup$
2
$\begingroup$

Because a standing wave is caused by two identical waves traveling in opposite directions, a guitar string cannot create a standing wave. So a plucked guitar string only makes a vibration, not a standing wave.

This is wrong. Suppose you pluck a string in the middle, i.e. you hold up the middle part and release it. All solutions to the wave equation are traveling waves, moving left or right, but this plucked string starts with zero initial velocity. How can you get a string that doesn't move out of solutions that all move? By superposing identical waves moving in opposite directions.

There's an excellent demonstration of this here. As the two opposite moving waves separate, a plateau is formed. It grows until the whole string is horizontal, at which point its momentum makes the pluck reform, flipped over. This repeats indefinitely in the ideal case; it is the textbook example of a standing wave. You start with a superposition of oppositely moving waves, they reflect off the ends of the string, and the process repeats.

As has been pointed out above, there are other ways to define the term "standing wave", but under the definition your teacher was using, plucking a guitar string definitely forms a standing wave. Sadly, high school physics is bad. You have to spend a lot of time memorizing made-up distinctions, like "standing waves" vs. "vibrations", or "interference" vs. "diffraction" that practicing physicists don't care about, and the teachers don't even define consistently. Then you get quizzed on these terms, because exam writers are too lazy to write questions about actual physics. It’s a bad system, but there’s nothing you can do but put up with it until you get to college.

$\endgroup$
1
$\begingroup$

No, a plucked guitar string does not create a standing wave. You can watch a video of what plucked guitar strings do here: https://www.youtube.com/watch?v=INqfM1kdfUc

Note how you can see repeating shapes travel up and down the strings instead of staying put like in a standing wave. Caveat: the camera is not capturing the full motion of the string, what the video shows is distorted by the camera framerate and function. However, it is enough to demonstrate that the motion doesn't have nodes and so is not due to a standing wave.

$\endgroup$
  • 3
    $\begingroup$ The phenomenon in the linked video is actually an effect of the way that iPhone cameras read data from their sensors, rather than a physical effect. For comparison, you wouldn't look at this airplane video and conclude that rapid rotation transforms an aircraft propeller into a set of disconnected parallel pieces. The same frequencies are involved. $\endgroup$ – rob Jun 20 '18 at 20:34
  • $\begingroup$ As I noted, the video is still enough to demonstrate that the waves are not standing waves, as the rolling shutter effect does not make nodes appear to move, so the absence of nodes in the video shows that the waves are not standing waves. $\endgroup$ – Luke Pritchett Jun 20 '18 at 21:03
1
$\begingroup$

The question itself is complicated, and the answer depends on the degree of realism you want to accept in how you model the vibrating string:

  1. Is the model linear?
  2. Is the model lossless?
  3. What is the meaning of "a" in the question "is a standing wave produced?

The complete solution can determined by solving a differential equation based on initial conditions and boundary conditions. For a lossless linear system, the solution will be found, by applying the principal of superposition, to be the sum of several sine waves.

Generally, plucking a string means the initial conditions will produce a combination many sine waves constrained by the motionless endpoints of the string. Each sine wave will produce its own standing wave.

However, that doesn't mean that there will necessarily be nodes of zero displacement along the string (other than at the end points). While the nodes for each sine wave might be located at some integer fractional spacing along the string, there may be nodes for some frequencies where there is displacement for others.

So if one were to ask "will there be nodes with zero displacement, when the string is plucked?" Then the answer is probably not. That's not the same as saying there are no standing waves.

Since most plucking will not produce a single sine wave, then you can't say that a single standing wave is produced, so if a means one then no, a standing wave is not produced. But if a means one or more, then yes, a standing wave is produced.

$\endgroup$

protected by Qmechanic Jun 21 '18 at 1:08

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.