Tension (and Frequency) with Change in Temperature

[Xilus]

we could just take the guitar outside and put it on a guitar tuner.
it should say the frequency exactly.

 

[Lotto]

You did a very impressive analysis for a guy who is doing music education. Dazzling in fact.

I have a few comments. It is not necessary to take into account the change in the cross sectional area because that is negligible. So this should simplify things considerably.

The G doesn't belong in the equation for the fundamental frequency (since gravity has nothing to do with this). The equation should read:
f=2LFμ
where L is the clamped length of the string, F is the tension in the string, and μ is the linear density of the string (mass/length).

The equation for the change in tension is:

ΔF=−αEAΔT
where α is the coefficient of thermal expansion, E is the Young's modulus, and T is the temperature.

So if f is the frequency at temperature T₀ and tension F₀ and f + Δf is the frequency at temperature T₀+ΔT and tension F₀+ΔF, then
Δf=2LF0−(αEAΔT)μ−2LF0μ
If we linearize this with respect to the temperature change ΔT, we obtain:
Δf=−EAαLF0μΔT
The fractional change in frequency is:
Δff=−EAα2F0ΔT

According to these equations, when you increase the temperature, the frequency decreases (because the string expands and "slackens").

Chet

I know it is an old thread, but why is that lambda $2L$, the frequency can be $f=\frac{1}{\lambda}\sqrt{\frac {F}{\mu}}$ in general, so why do you consider only the fundamental wavelenght?

 

[Steve4Physics]

I know it is an old thread, but why is that lambda $2L$, the frequency can be $f=\frac{1}{\lambda}\sqrt{\frac {F}{\mu}}$ in general, so why do you consider only the fundamental wavelenght?

The original question basically asks how to tune a string so that, after a future known temperature change, the string will have the required frequency.

When you tune a string (e.g. on a guitar) you are adjusting the fundamental frequency to a required value. The fundamental frequency is that of a standing wave of wavelength λ, where λ=2L. (L is called the 'scale length' and is the distance between the fixed ends of the vibrating string.) That's because the fixed ends are adjacent nodes of the standing wave.

So the specific case of interest here is when λ=2L.

 

[Lotto]

The original question basically asks how to tune a string so that, after a future known temperature change, the string will have the required frequency.

When you tune a string (e.g. on a guitar) you are adjusting the fundamental frequency to a required value. The fundamental frequency is that of a standing wave of wavelength λ, where λ=2L. (L is called the 'scale length' and is the distance between the fixed ends of the vibrating string.) That's because the fixed ends are adjacent nodes of the standing wave.

So the specific case of interest here is when λ=2L.

And when I have a guitar with an equal temperament tuning, what are lambdas for each string? When I have these frequencies 329.63 Hz, 246.94 Hz, 196.00 Hz, 146.83 Hz, 110.00 Hz, 82.41 Hz, is there any way to find out the lambdas for them? Or is it dependent on the technicue I play the guitar, so it can be $2L$ as well as let's say $\frac12 L$?

 

[Steve4Physics]

And when I have a guitar with an equal temperament tuning, what are lambdas for each string? When I have these frequencies 329.63 Hz, 246.94 Hz, 196.00 Hz, 146.83 Hz, 110.00 Hz, 82.41 Hz, is there any way to find out the lambdas for them?

Get a tape measure, measure the length of the vibrating section of the string. Double it. That's the wavelength of the waves travelling along the string.

Or is it dependent on the technicue I play the guitar, so it can be $2L$ as well as let's say $\frac12 L$?

It's always 2L uness you are considering harmonics; then it's more complicated but I don't want to get into that here.

See if this helps:

First note we are only considering fundamental frequencies (not higher harmonics).

The wavelength (of the transverse waves on a string) for a given note is twice the length of the vibrating part of the string (λ=2L). That’s it. Look at this diagram:
https://www.a-levelphysicstutor.com/images/waves/strings-diag-fund.jpg

It may be worth noting that this wavelength is not the same thing as the wavelength of the sound wave in the air.

The frequency associated with a given wavelength depends on the speed of the wave along the string. Each string has a different wave-speed because each string has a different linear density (and probably a different tension) to the others.

When you fret a note on a guitar, you are shortening the vibrating length . This reduces the wavelength and increases the frequency.

For example, say the scale length (nut to bridge distance), for the 1st (top E) string of your guitar is L₀ = 65cm. When you play an open E (frequency 329.6Hz approx.), the wavelength of waves along the string is 2x65cm=130cm.

Since the frets are positioned for equal temperament tuning, for each semitone change in pitch, the frequency changes by a factor $2^{\frac 1{12}} = 1.0595$ approx.

Suppose you now play an F (1st fret on the E string) so the pitch increases by a semitone. You are shortening the length of the vibrating part of the string to a new length L₁, such that L₁ = L₀/1.0595. The wavelength is now 130cm/1.0595. The frequency increases to 329.6 x 1.0595, giving the ‘F’.

 

[Chrono G. Xay]

You did a very impressive analysis for a guy who is doing music education. Dazzling in fact.

I have a few comments. It is not necessary to take into account the change in the cross sectional area because that is negligible. So this should simplify things considerably.

The G doesn't belong in the equation for the fundamental frequency (since gravity has nothing to do with this). The equation should read:
$$f=\frac{1}{2L}\sqrt{\frac{F}{μ}}$$
where L is the clamped length of the string, F is the tension in the string, and μ is the linear density of the string (mass/length).

The equation for the change in tension is:

$$ΔF = -αEAΔT$$
where α is the coefficient of thermal expansion, E is the Young's modulus, and T is the temperature.

So if f is the frequency at temperature T₀ and tension F₀ and f + Δf is the frequency at temperature T₀+ΔT and tension F₀+ΔF, then
$$Δf=\frac{1}{2L}\sqrt{\frac{F_0-(αEAΔT)}{μ}}-\frac{1}{2L}\sqrt{\frac{F_0}{μ}}$$
If we linearize this with respect to the temperature change ΔT, we obtain:
$$Δf=-\frac{EAα}{L\sqrt{F_0μ}}ΔT$$
The fractional change in frequency is:
$$\frac{Δf}{f}=-\frac{EAα}{2F_0}ΔT$$

According to these equations, when you increase the temperature, the frequency decreases (because the string expands and "slackens").

Chet

It’s been at least a few years since I’ve added to this thread. Starting with the linearized equation you gave earlier for $Δf$ and substituting in the unconverted equation for initial tensile force $$F_0 = {μ\left(2Lf_0\right)^2}$$ as well as $$Δf = f - f_0$$ we can obtain $$f_0 = \frac {EAα} {2μL^2f_0} {ΔT} + f.$$ With a little rearranging we obtain the quadratic form $$f_0^2 - f f_0 - \frac {EAα} {2μL^2} {ΔT} = 0$$ $$\left(ax^2 + bx + c = 0\right).$$ By setting our variables as
$$x = f_0 , a = 1 , b = -f , c = -\frac {EAα} {2μL^2} {ΔT}$$ we can obtain the equation $$f_0 = \frac f 2 + \sqrt{ \left( \frac f 2 \right)^2 + \frac {EAα} {2μL^2} {ΔT} }.$$ Now whenever $ΔT = 0, f_0 = f$.

…How did I do?

 

[NTL2009]

I see a few of us questioned being able to mathematically predict this based on formulas for the string tension/frequency versus temperature (almost 7 years ago!). Well, it's still interesting, and I still feel the same.

I agree that the double-locking tremolo probably can be ignored as a factor, but I don't think you can ignore the body/neck of the guitar - they will change length with temperature (but in the opposite direction, and ~ 1/3rd that of steel?), but that change in wood length might not (probably not?) be perfectly aligned with the strings. IOW, the body/neck might expand more/less on one side than the other, causing the neck to curve, and changing the tension on the higher/lower strings differently.

But another factor that I don't see discussed - the three lower pitched strings on an electric guitar are wound strings (typically the lower pitched 3 of the 6). Lower pitched strings are wound to add mass to achieve the lower pitch, w/o needing such a large diameter, stiff string. Only the core is really under tension, and (I'm not an ME) I'd think this makes the Young's Modulus different for each of these 3 wound strings (the ratio of wound mass to core mass is likely different for each?).

I'd think that much of this would need to be determined empirically, which l think just gets back to measuring the delta-F across a few typical expected delta-T and creating a table or formula from those measurements.

But if this is really just an interesting math exercise, carry on! :)

 

[NTL2009]

That reminded me - this may be interesting to anyone following this thread:

http://www.evertune.com/resources/introduction_to_evertune.php

Once the bridge is set up to the desired tuning by the user, it will stay in tune -- regardless of extreme playing techniques, huge bends, heavy picking, or drastic temperature or humidity changes.

It accomplishes this by applying a regulated tension to each string, rather than just stretching a string to a specific tension, and locking that tension.

 

[Chrono G. Xay]

IOW, the body/neck might expand more/less on one side than the other, causing the neck to curve, and changing the tension on the higher/lower strings differently.

On a good quality instrument I expect more of the twisting or warping will arise from particularly lopsided tension on the neck owing to the strings, the current tuning, and how the instrument has been setup. (On a related note, I’m curious to engineer a truss rod with a lateral axis of adjustment at the bottom of the neck to help more precisely balance torsional forces, be it for lower quality/strength woods or such as with multi-scale guitars and/or when strings have been selected for ‘progressive tension’.)

Anyway, there are plenty of guitar forums talking about neck twisting, like here:

Performing a 3D scan of the instrument and having the computer analyze it as if it’s a neck-thru construction would get most of the non-string ambiguity out of the way, no? From there we can just about plug in our ΔT string frequency equation.

But another factor that I don't see discussed - the three lower pitched strings on an electric guitar are wound strings (typically the lower pitched 3 of the 6). Lower pitched strings are wound to add mass to achieve the lower pitch, w/o needing such a large diameter, stiff string. Only the core is really under tension, and (I'm not an ME) I'd think this makes the Young's Modulus different for each of these 3 wound strings (the ratio of wound mass to core mass is likely different for each?).

I’m well aware of the wound strings. They’re actually part two of the string analysis I want to run; bare strings were always the starting point.

I remember reading that wound strings typically have a hexagonal core. I suspect that the wrapping wire(s) are under at least a low magnitude of tension when being wrapped around the core. I’m sure it results in the wound strings having their own effective values of A, E, and even α.

 

[NTL2009]

It's all interesting stuff. When I did a little deeper look into changing strings on my guitar, and the tension differences from tuning a semi or whole note lower, I learned that the tension on each of the 6 strings is about the same (there are tables and calculators on-line). The pitch differences come from the mass of the string. So that does help keep the forces equal from side to side.

Obvious I guess, I just had never thought about it before.

 

[Chrono G. Xay]

It's all interesting stuff. When I did a little deeper look into changing strings on my guitar, and the tension differences from tuning a semi or whole note lower, I learned that the tension on each of the 6 strings is about the same (there are tables and calculators on-line). The pitch differences come from the mass of the string. So that does help keep the forces equal from side to side.

Obvious I guess, I just had never thought about it before.

Look up what ‘progressive tension’ is and why it is (because it’s the motivation behind the construction of multi-scale guitars). By comparison the tension is very unbalanced.