How will the pitch of a string change when I stretch it? (sound)

[JT Smith]

It’s the tension change per unit length that is reduced. The length change due to fretting remains constant.

Yes, N/m. That's the string constant.

A concrete example: a 0.010 steel string tuned to E4 on a guitar with a 25.5" scale length. If the action is 2mm the increase in string length is about 12 microns. The spring constant of steel "10" on a guitar of that scale length is roughly 14kN/m. So the increase in tension due to fretting is about 0.2N. That corresponds to a pitch increase of around 0.7Hz, which is only about 2 cents. So it's basically impossible in this example for the tension to reduce by "a few cents". If you increase the action to 5mm then the pitch goes up about 11 cents meaning to lose a few cents the spring constant would have to diminish by more than 25%. Does that seem likely? Only if the action is way too high does it start to seem plausible.

On my electric guitars the bridge position (string length adjustment) depends quite a bit on which string. I attribute this to the variation of the tension delta on fretting depending on wire gauge and string height. What’s your explanation?

From what I have read the bridge adjustment is because of the finite diameter of the string. The larger the string the less free it is to move near the nut and bridge which makes the string effectively shorter. So larger strings need to be longer than an infinitely thin string would have to be. The wound strings start over because the core diameter is what matters.

Edit: You're right that fretting the strings and their gauge also has significant bearing on the bridge adjustments. I think this paper does a good job of describing the problem: https://arxiv.org/pdf/0906.0127.pdf

As for why strings go flat with age I don't have an explanation.

 

[sophiecentaur]

There's been confusion about the effect of the 'dead end' of the guitar string (the section between finger and nut). When you stop a guitar string with a finger, you aren't providing a 'hard' boundary. (It's not a node)_We all know about pressing particularly hard to get a better note when needed. This is because your finger allows the string to move underneath it. For a fundamental or low overtones, the string is only around a half (or very few) wavelength long so the termination is shared by the finger and also by the tension in the dead end. The modulus of the dead bit is actually relevant and an old string will have a lower modulus and it will lower the note. So I think it's not down to the change in length (which is tiny).
End effects are very relevant in musical instruments and are responsible for the interesting sounds they produce. Many of the 'imperfections' in a guitar will make it sound nice. Small frequency errors can be subconsciously corrected for by a good player. I imagine that it's easier to cope with nylon strings. Electric guitar techniques use the long sustain and feedback so that the available notes are on an almost continuous scale.

 

[Paul Colby]

You're right that fretting the strings and their gauge also has significant bearing on the bridge adjustments.

I assume I can take this as a partial vindication. Nice paper BTW.

 

[JT Smith]

There's been confusion about the effect of the 'dead end' of the guitar string (the section between finger and nut). When you stop a guitar string with a finger, you aren't providing a 'hard' boundary. (It's not a node)_We all know about pressing particularly hard to get a better note when needed. This is because your finger allows the string to move underneath it.

Yes. Even with a capo the string is not fixed. For that matter any string above a normal nut or below the bridge also has the effect of reducing the spring constant. A floating bridge does the same.

 

[JT Smith]

I assume I can take this as a partial vindication. Nice paper BTW.

Okay then, you're partially vindicated! :-)