I'm having trouble with the explanation that a standing wave in a string is the superposition of traveling waves.

standing wave
(source: physicsclassroom.com)

The nodes in the diagram above are points where the particles of the string's medium undergo zero displacement, i.e. they do not move at all. But if they do not move, how is the disturbance of (any internal traveling wave) propagated past the node?

The usual explanation for how a wave is propagated is that when one particle is disturbed (say, moved up), it exerts a pull on another, which in turn exerts a pull on the next one, and so on. In other words, to exert a pull or push on the next particle there must be some movement/disturbance of the previous one. But the particle(s) at the node point do not move at all, so how does the disturbance of a traveling wave propagation pass through them? (I'm trying to understand the picture in terms of the mechanical forces between particles).


The nodes do not change position, but the forces on them change. The forces are the cause of displacement.

It may help to use a slinky instead of a string. The slinky stretches into a sinusoidal shape and shrinks to a line. As the point at the node is pulled up by one by one traveling wave and equally down by the other, it stretches. As the slinky shrinks, so does the stretching at the node.

The pattern of stretching does pass through the stationary node.

  • $\begingroup$ Are you implying that if, for some reason, the string after a node were still, energy from the oscillating mode would propagate and make the still part oscillate? $\endgroup$ – auxsvr May 5 '14 at 14:01
  • $\begingroup$ @auxsvr: that's the case. You can test it with a guitar string: (1) Play a string holding it at the 12th (octave) fret. (2) Gently release the string from the fret, so that the sound continues, as if you were playing a harmonic. (3) Touch the string with two fingers anywhere along the previously non-vibrating half. You'll feel the vibrations for a moment before you damp the sound away. (4) Damp the string with one finger at the 5th or 7th fret and you'll hear that you are exciting higher overtones in the still part of the string as well. $\endgroup$ – rob May 7 '14 at 16:20
  • $\begingroup$ @rob That's not what my question was about. A standing wave must have its endpoints fixed. $\endgroup$ – auxsvr May 8 '14 at 17:10
  • $\begingroup$ @auxsvr Then I don't understand your previous comment. A shortened guitar string does have its endpoints fixed. Release the fret and you have a new fixed endpoint, at the nut, and a string which is half oscillating and half at rest. Very rapidly the entire string begins to oscillate. mmesser314 has explained why. $\endgroup$ – rob May 8 '14 at 17:39
  • $\begingroup$ @rob mmesser314 didn't explain what you had in mind, because he mentions nodes that "do not change position". $\endgroup$ – auxsvr May 8 '14 at 18:38

The answer is that it doesn't, a standing wave does not have energy propagating. Between the nodes you observe the effect of the energy the string already had when the wave formed, because a standing wave is formed by two waves with equal energy density traveling in opposite directions.


It's not that there is no movement of the node particle, it's that the particle rotates in the same place. When the string is pulled to one side, the node particle rotates in one direction to minimize the stress. When the string is pulled to the other side, the node particle is rotated the other way.

  • $\begingroup$ So is it something like this : i.imgur.com/Hk5v3x3.png ? I.e. the particle at the node can be viewed as a ball/disc rotating around the center axis, with 'handles' that attach to the string (the handles representing interatomic/intermolecular forces that keep the string together) $\endgroup$ – user45848 May 5 '14 at 6:36
  • 1
    $\begingroup$ Also, how does this explain nodes in longitudinal waves, like for example shown here, where there is no rotation at the nodes? $\endgroup$ – user45848 May 5 '14 at 6:54
  • $\begingroup$ @user45848 It's obvious that I was not referring to longitudinal waves, so you should delete your comment. $\endgroup$ – LDC3 May 6 '14 at 2:01

As @auxsvr says.There is a contradiction between "traveling wave", which does not have stationary nodes, and the diagram you are displaying, which is of a standing wave.

Here is a standing wave:

standing wave

The energy is stationary in the x direction, that is why nodes are formed .

Here is an animation of a traveling wave:

traveling wave


Let's consider a simpler oscillator than a continuous string: two masses with a sprint between them, both connected to two walls:

enter image description here

(image source)

When two masses are equal, and then oscillate in opposite phases, the wave you can see is the single-frequency standing wave. Now, as they have opposite phases, this means that there's a node between the masses. Note that the node is not on the mass — here it's on the spring. Now how does the wave pass the node? Simple: half of the time its energy stored in the spring deformation, another half it's in the mass motion.

When the masses come close to each other, they deform the string too much to continue motion. They stop, then go backwards. The wave now is returning back from the spring to the masses — until the spring is deformed too much, now by stretching, when the masses are again stopped and then go backwards.

If you put a third mass between these two, then, if that mass appears at the node, the wave again will pass the node because its energy stored not in the masses all the time. It'll push-pull the side masses (the moving ones) because half of the time it's mostly stored in the springs, which in this case surround the stationary mass.

Now as you add more and more masses and springs between them, you'll get in the limit your continuous string.


This is a great paradox! I've wondered this myself as a trumpet player, and it took me a while to figure out where the confusion comes from.

I think people hear that standing waves are the sum of a traveling wave and its reflection, and get irritated because +1 + -1 is still zero. It just seems like a math trick.

But there's a big difference between an object with no force on it, and an object with equal and opposite forces on it, even though they both have zero acceleration. The difference is that the object with equal and opposite forces is probably (assuming it's elastic, blah blah blah) transferring those forces across itself.

Force detour

Consider an aluminum block (negligible mass) sitting on a concrete floor. It has no(-ish) force on it. Now stand on the block. The position of the block hasn't changed, but your weight is now transferred through the block to the floor, and the floor's 3rd Law response is transferred through the block to you. You can feel this, in your feet (sitting certainly feels better!). It's not just a math trick, it's real.

Applied to waves

Likewise, standing waves show up when a traveling wave interferes with its own reflection. Meaning nodes aren't points of zero force, they're points where the (non-zero) forces for each wave sum to zero acceleration. But they still transfer the force of each wave to neighboring particles. And for those neighboring particles, the accelerations don't cancel out (at least not most of the time).

Transferring force without deformation

user45848 pointed out that the aluminum block mentioned ought to deform a bit, and that was bugging me too. But I don't think making the block incompressible changes anything: the top layer should be able to transfer force to lower layers even if it doesn't move at all.

That still sounds a bit nonsensical, but it helped when I imagined this scenario:

Imagine three adjacent particles in a line with no force on them, then replace the outer particles with oppositely charged particles. None of the particles have moved, but now force is being transferred through the center particle.

  • $\begingroup$ >But they still transfer the force of each wave to >neighboring particles. $\endgroup$ – user45848 May 5 '14 at 23:45
  • $\begingroup$ But when you stand on the aluminum block, the particles under your feet do move (though a little bit), because the block slightly deforms. It is only because it deformed a little that it is able to transmit that force to the ground (the particles in the block push on each other in turn until the bottom most layer pushes on the ground). If the layer right under your feet had not deformed (i.e. moved) at all, then the configuration of the block's particles would be identical to if you had not stepped on it - and then where would any new force come from? $\endgroup$ – user45848 May 5 '14 at 23:56
  • $\begingroup$ @user45848: Made an edit to try and resolve that stuff. $\endgroup$ – Vanessa Phipps May 7 '14 at 16:03

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