Tension (and Frequency) with Change in Temperature

[Chrono G. Xay]

At one point I had been trying to construct an equation which would calculate the tension on a tuned string with a change in temperature (and therefore the fundamental frequency), but found my calculations were wrong. By extension, the purpose of the project was to be able to calculate what frequency the string should be tuned to beforehand so that with a change in temperature the change in tension will retune the string to the desired final frequency. For the sake of clarity and realism let's say that the string instrument is an electric guitar equipped with a double-locking tremolo where the tremolo is also 'blocked', which means that the tremolo does not move with a change in tension, and the effective length of string is merely the scale length of the instrument (i.e. the length of string from the locking but to the locking bridge). Let's also say that the body, neck, and headstock of the instrument remain immobilized even when all strings at their maximum tensile strength.

 

[Chestermiller]

Show us what you did?

Chet

 

[Chrono G. Xay]

Of course- one moment, please...

 

[Chrono G. Xay]

I started out with the equation for frequency:

fΔT = ( 1 / ( 2 * L ) ) * sqrt( G * F * ( m / L ) )

where

'fΔT' is the string's frequency,
'L' is the string's length (or rather, the distance from the edge of the string's slot in the nut to the slot on its corresponding saddle),
'G' is the force of gravity,
'F' is the force of tension on the string, and
'm' is the total mass of the vibrating portion of the string.

Then I considered how if 'F' is the force on the string after it has experienced a change in temperature, so 'F' became

F = F0 + FΔT
F0 => T0

where

T0 is the tension already on the string before the temperature change, and
FΔT is the extra force (or lessening thereof) with the change in temperature.

After that I used the equation for change in length resulting from a change in temperature, as well as the equation for the 'Tensile Modulus' (AKA "Young's Modulus")

ΔL = α * L * ΔT

where

'α' is the coefficient of thermal expansion, and
'ΔT' is the actual change in temperature ( T - T0, where T is the temperature the string will be exposed to)

E = ( F / A ) / ( ΔL / L )
F => F0 => T0
A => A0

=> ΔL = ( T0 * L ) / ( A0 * E )

where

'E' is the string material's modulus of elasticity, and
'A0' is the string's diameter before the change in temperature (but not before the string was initially pulled taught, so we still need to work our way back to those dimensions of the string that we can know existed before any tension was applied)

Anyway, I set both equations for ΔL as being equal to each other and solved for F (i.e. FΔT)

( F / L ) / ( A0 / E ) = ΔL = α * L * ΔT
F = α * ΔT * A0 * E

=> FΔT = α * ΔT * A0 * E

From there we start into the equation for area of a circle

A0 = ( π / 4 ) * d0^2

where

'd0' is the string's diameter before the change in temperature

Then I considered the fact that as tension is applied and the string's length changes, the string's diameter also changes

d0 = di + Δdi

where

'di' is the string's pre-taught (or 'initial') diameter, and
'Δdi' is the change in that pre-taught diameter

That being the case, the equation for change in diameter with change in length (i.e. Poisson's Ratio) would also come into play

Δdi = -di * ν * ( ΔL / L )

where

'ν' is the Poisson's Ratio of the string material

Thinking back, the expression '( ΔL / L )' sounded much like the earlier equation involving the modulus of elasticity, so I solved went back and solved that equation for '( ΔL / L )'

ΔL / L = T0 / ( E * Ai )

where

'Ai' is the pre-taught diameter of the string, and
Ai = ( π / 4 ) * di^2

Finally, the single equation I've forgotten until now: the equation for tension on a string

T0 = ( ( m / L ) ( 2 * L * f0 )^2 ) / G

where

'f0' is the frequency the string was tuned to before the change in temperature.

 

[Chrono G. Xay]

When I put all of this together for an equation that would, at the very least, tell me the frequency of the string after the change in temperature, I got this:

I was able to rewrite it so that I could do what I had outlined in my original post: what to tune a guitar string to so that the change in temperature brings the string 'in tune' (solve for f0), but I think it simply did not work...

 

[Chrono G. Xay]

Before anyone replies, please understand that I am an undergraduate Music Education major, studying percussion, but I have a passion for Physics and Mathematics (I took Calculus AB my senior year of high school, and am currently taking a Discrete Mathematics and Pre-Calculus class). I want to understand just why an instrument works and, more importantly, HOW it behaves (its quirks), for the purpose of being able to learn how to play an instrument MUCH faster and what balance of mechanics different pieces of music exercise and the more exact level of skill they require, as opposed to simply going through nothing but the 'traditional' methods. I would call myself a Musicologist (and inventor), weighted heavily towards physics).

 

[Chestermiller]

Before anyone replies, please understand that I am an undergraduate Music Education major, studying percussion, but I have a passion for Physics and Mathematics (I took Calculus AB my senior year of high school, and am currently taking a Discrete Mathematics and Pre-Calculus class). I want to understand just why an instrument works and, more importantly, HOW it behaves (its quirks), for the purpose of being able to learn how to play an instrument MUCH faster and what balance of mechanics different pieces of music exercise and the more exact level of skill they require, as opposed to simply going through nothing but the 'traditional' methods. I would call myself a Musicologist (and inventor), weighted heavily towards physics).

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:
$$f=\frac{2}{L}\sqrt{\frac{F}{μ}}$$
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
$$Δf=\frac{2}{L}\sqrt{\frac{F_0-(αEAΔT)}{μ}}-\frac{2}{L}\sqrt{\frac{F_0}{μ}}$$
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

 

[Chrono G. Xay]

Ahhh! I completely forgot that the only reason 'G' was in the string tension equation to begin with was to convert the units of force from Newtons into pounds (or kilograms), and is otherwise unnecessary!

I had just wanted to make sure I was being very accurate when I chose to take into account the change in cross-sectional area with respect to the change in tension.

Thank you very much!

 

[Chrono G. Xay]

If you don't mind me asking, would there not be an 'analog' to this equation that isn't that much more complicated but instead of strings deals with the change in frequency of a membrane with initial tension (and therefore initial frequency) experiencing a change in temperature ?

Note: When tuning a 'typical' drumhead (i.e. circular, with no waves or wrinkles intended) there is a term called 'clearing' where the tension all around the clamping ring (AKA 'rim') is 'calibrated' so that the frequencies of the same number of harmonic produced when struck anywhere around the edge of the membrane is as close to the same as possible for the most unified and therefore loudest sound (and I'm guessing this also maximizes the drumhead's sensitivity to being struck with varied magnitudes of force.

Anyway, I had looked at the equation for string tension, and after having gone back and reviewed just how the equation came to be, it seemed as though if one inserted the geometry of, say, a rectangle (which is the cross-section of a drumhead), they would get an equivalent equation for the tension of a--perfectly circular--membrane.

Let me build it up for you:

1) Start with the equation for frequency

f = sqrt( T / μ ) / ( 2 * L )

2) Solve for T

T = m * L * ( 2 * f )^2

3) Define mass

m = V * ρ

4) Define Volume (this is more like where we really begin)

V (string) = As * L
V (membrane) = Am * h

where

'As' is the cross-section of a string,
'L' is the vibrating length of string,
'Am' is the cross-section of a membrane, and
'h' is the thickness of the membrane.

(They still look similar, in that one is a long, thin cylinder, and the other is a thin, large cylinder.)

A = ( π / 4 ) * D^2

The perhaps most noticeable difference between the two comes when for the membrane

L => D

which makes their respective equations

Ts = π * ρ ( L * D * f )^2
Tm = π * ρ * h * f^2 * D^3

(The user then has the option to divide by gravity to convert the units of force from Newton into pounds or kilograms.)

Something bugs me, though... If this equation is true, would we not actually take the tension and divide it by the number of pairs of tuning bolts (AKA 'lugs') or even the circumference of the vibrating area of the membrane?

 

[Nidum]

Not exactly what you want but interesting to read :

https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membranehttps://courses.physics.illinois.edu/phys193/lecture_notes/p193_lect4_ch4_part2.pdfThe vibration characteristics of an actual flat plates or instrument body can be analysed quite well using Finite Element methods .

 

[Chestermiller]

Nidum's Wiki article gives the frequency for the ideal case where the membrane tension is circumferentially symmetric, and is thus analogous to your guitar string situation. But the tuning bolts have to be done just right to achieve the symmetry.

Chet

 

[Chrono G. Xay]

I've seen many of those diagrams before, save (I think) the bells and the rectangular sign charts of plates, but it's really nice seeing all of them together in one article. :)

 

[Chrono G. Xay]

I understand that they have to be done just right, and that's why there are music devices designed to make sure the desired result is very close:

http://tune-bot.com

Oh- @Nidum, I think it's a funny coincidence that the first image in the link you posted was a 3-D graph of a deformed circular membrane, because another project I'm working on deals with attempting to calculate the 'feel' of a circular membrane. I mean, a taught membrane can be thought of as a spring, right? capable of storing kinetic energy (say, from an implement angularly-accelerated starting a relatively short distance away, if you get my drift) as potential energy by experiencing a transverse displacement originating at a point a distance from its center. It is this measure of displacement that I wish to predict, and the model mentioned above, which could be likened to revolving the graph of sqrt(x) about the y-axis, is exactly what I had been thinking a drumhead does, given that the force lines of a taught membrane don't just intersect the point of contact, but 'go all over'.

(I do understand, though, that the tip of a drumhead is not a pinpoint, but many different shapes and orientations, depending on how they make contact.)

However, I am guessing that fewer lines of force intersect the contact point, which is why that point experiences a displacement which could be thought of as being exponentially greater, if you know what I mean.

Sufficed to say, the many projects I have set for myself involve primarily membranes and strings, from behavior in different weather to cavity resonance to displacement in response to a point force to 'sustain' after having been set in motion (my next 'big' project, considering how it deals with calculating the damping coefficient of a string or membrane, and I feel like that's math I simply do NOT understand yet). I am just hoping that the members of Physics Forums will be patient with me as I stumble through them! Haha

 

[Chrono G. Xay]

Just to make sure I'm understanding you, Chet, are you saying that the equation for predicting the frequency of an ideal string experiencing a change in temperature is given by

$$f_{ΔT} = f_0 + Δf$$

where 'Δf' is the non-linearized form you showed me earlier?

Furthermore, if this is true, all that would need to be done in order to predict the needed initial frequency for the desired final frequency is re-evaluate for ' f₀ '?

Something else I want to make sure of in particular is that there wasn't an accident when you wrote what I'm assuming was supposed to be the equation for predicting the frequency of an ideal string, which I have always understood to be

$$f = \frac{1}{2L}\sqrt{\frac{F}{μ}}$$

 

[Chestermiller]

Just to make sure I'm understanding you, Chet, are you saying that the equation for predicting the frequency of an ideal string experiencing a change in temperature is given by

$$f_{ΔT} = f_0 + Δf$$
No. This is just the definition of the change in frequency.

where 'Δf' is the non-linearized form you showed me earlier?

Yes.

Furthermore, if this is true, all that would need to be done in order to predict the needed initial frequency for the desired final frequency is re-evaluate for ' f₀ '?

Yes, if what you mean is that you want to specify the final frequency and the temperature change. But I think you will find that the behavior will be typically linear enough so that the linearized equation is adequate. Also, if you are using the linearized equation, it won't matter much if you use f or f₀ in the equation for the change in frequency.

Something else I want to make sure of in particular is that there wasn't an accident when you wrote what I'm assuming was supposed to be the equation for predicting the frequency of an ideal string, which I have always understood to be

$$f = \frac{1}{2L}\sqrt{\frac{F}{μ}}$$

I don't quite understand what you are driving at. Your original equation assumed that the force was expressed in pounds force, so you needed a factor of g_c to make the units came out correctly. If one is using the metric system, there is no need for such a factor. This equation assumes that all the units are metric.

I'm really impressed by your ability and your determination. You seem to have what it takes to really be a great physicist or engineer. Have you ever considered studying Mechanical Engineering. There are many people in ME that specialize in mechanical vibrations of systems. I know you love music, but the talent out there in music is so terrific that it is hard to make money because of the competition. Any chance of combining ME with your music pursuit?

Chet

 

[Chrono G. Xay]

I openly concede the 'g', which was a brain fart on my part when I first referenced the equation for tension.

All I'm saying is that for whatever reason I'm having a bad mental 'disconnect' seeing \frac{2}{L} instead of \frac{1}{2L}, and maybe it's just a formatting error between what you wrote and what my browser is choosing to display, but the best I can guess, otherwise, with what little experience I have with physics, let alone math, is it seems as though sometime, somewhere, you multiplied the equation for frequency (and Δf) by four, and if that was supposed to occur in the process of solving for f_{ΔT} (which I would be happy to concede by our relative difference in experience), I want to understand where that happened so I know better moving forward.

Would it be disrespectful of me to ask if I may see some of the work you went through?

 

[Chestermiller]

I openly concede the 'g', which was a brain fart on my part when I first referenced the equation for tension.

All I'm saying is that for whatever reason I'm having a bad mental 'disconnect' seeing \frac{2}{L} instead of \frac{1}{2L}, and maybe it's just a formatting error between what you wrote and what my browser is choosing to display, but the best I can guess, otherwise, with what little experience I have with physics, let alone math, is it seems as though sometime, somewhere, you multiplied the equation for frequency (and Δf) by four, and if that was supposed to occur in the process of solving for f_{ΔT} (which I would be happy to concede by our relative difference in experience), I want to understand where that happened so I know better moving forward.

Would it be disrespectful of me to ask if I may see some of the work you went through?

Oh. It's just my writing the 2 in the wrong place. Sorry. I didn't even notice my error. Thanks for spotting it. If you'd like, I'll go back and correct my earlier post.

Chet

 

[Jean Jack]

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
$$Δf=\frac{2}{L}\sqrt{\frac{F_0-(αEAΔT)}{μ}}-\frac{2}{L}\sqrt{\frac{F_0}{μ}}$$
If we linearize this with respect to the temperature change ΔT, we obtain:
$$Δf=-\frac{EAα}{L\sqrt{F_0μ}}ΔT$$

Hi, Chet, I am 17 years old and I have tried hard but I can't figure out how you linearised this equation, could you be so kind as to show me the steps, or to start me off?
Also I have been looking everywhere for the thermal expansion equation ΔF=−αEAΔT , and I have been unable to find it, could you show me how you have derived this?
Many thanks.
Jean.

 

[Jean Jack]

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:
$$f=\frac{2}{L}\sqrt{\frac{F}{μ}}$$
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
$$Δf=\frac{2}{L}\sqrt{\frac{F_0-(αEAΔT)}{μ}}-\frac{2}{L}\sqrt{\frac{F_0}{μ}}$$
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

Hi, sorry I'm new to this forum, not certain how to use everything, could I direct you to my question i posted earlier today on this thread please? many thanks in advance, it is extremely important for me!

 

[Chestermiller]

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
$$Δf=\frac{2}{L}\sqrt{\frac{F_0-(αEAΔT)}{μ}}-\frac{2}{L}\sqrt{\frac{F_0}{μ}}$$
If we linearize this with respect to the temperature change ΔT, we obtain:
$$Δf=-\frac{EAα}{L\sqrt{F_0μ}}ΔT$$

Hi, Chet, I am 17 years old and I have tried hard but I can't figure out how you linearised this equation, could you be so kind as to show me the steps, or to start me off?

If we factor out $\sqrt{\frac{F_0}{\mu}}$ from $\sqrt{\frac{F_0-(αEAΔT)}{μ}}$, we obtain:
$$\sqrt{\frac{F_0-(αEAΔT)}{μ}}=\sqrt{\frac{F_0}{\mu}}\sqrt{1-\frac{αEAΔT}{F_0}}$$
From the binomial expansion, we know that, if x is small, to linear terms in x, $$\sqrt{1-x}\approx 1-\frac{x}{2}$$

Also I have been looking everywhere for the thermal expansion equation ΔF=−αEAΔT , and I have been unable to find it, could you show me how you have derived this?

Think of it as a two step process.

Step 1. The rod is heated by $\Delta T$, and is allowed to freely expand without any constraint

Step 2. The heated bar is compressed by applying forces at its end large enough to return it to its original length.

What is the change in length in step 1? What is the compressive force in step 2?

 

[Jean Jack]

If we factor out $\sqrt{\frac{F_0}{\mu}}$ from $\sqrt{\frac{F_0-(αEAΔT)}{μ}}$, we obtain:
$$\sqrt{\frac{F_0-(αEAΔT)}{μ}}=\sqrt{\frac{F_0}{\mu}}\sqrt{1-\frac{αEAΔT}{F_0}}$$
From the binomial expansion, we know that, if x is small, to linear terms in x, $$\sqrt{1-x}\approx 1-\frac{x}{2}$$

Think of it as a two step process.

Step 1. The rod is heated by $\Delta T$, and is allowed to freely expand without any constraint

Step 2. The heated bar is compressed by applying forces at its end large enough to return it to its original length.

What is the change in length in step 1? What is the compressive force in step 2?

Thank you sooo much I really really appreciate it, I had a look back at the equation linking Tension and Temperature, and I understand it completely now!
I also managed to linearize the equation thanks to your tip!

However it is to my understanding that you accidentally used:
2/L* sqrt(F0/u)
when it should be:
1/2L * sqrt(f0/u)

therefore in the final equation, we should get:

where there is a 4L on the denominator, although this doesn't change the relationship but just the coefficient, I would like to point that out :)
again thank you for the help I much appreciate it!

 

[Tobias T]

If we factor out $\sqrt{\frac{F_0}{\mu}}$ from $\sqrt{\frac{F_0-(αEAΔT)}{μ}}$, we obtain:
$$\sqrt{\frac{F_0-(αEAΔT)}{μ}}=\sqrt{\frac{F_0}{\mu}}\sqrt{1-\frac{αEAΔT}{F_0}}$$
From the binomial expansion, we know that, if x is small, to linear terms in x, $$\sqrt{1-x}\approx 1-\frac{x}{2}$$

Think of it as a two step process.

Step 1. The rod is heated by $\Delta T$, and is allowed to freely expand without any constraint

Step 2. The heated bar is compressed by applying forces at its end large enough to return it to its original length.

What is the change in length in step 1? What is the compressive force in step 2?

Sorry I'm still not sure about what components are in what step of the change in tension formula. Is the coefficient of thermal expansion and temperature change the account of step 1, while the Young's modulus and cross sectional area step 2?

 

[Chestermiller]

Sorry I'm still not sure about what components are in what step of the change in tension formula. Is the coefficient of thermal expansion and temperature change the account of step 1, while the Young's modulus and cross sectional area step 2?

No. I explained precisely what I meant in post #20. If a bar has a length L and we heat it unconstrained by $\Delta T$, its length increases as $$\Delta L=L\alpha \Delta T$$. This involves a strain of $\alpha \Delta T$. That's step 1. If we now compress the bar back to its original length, the compressive stress would be $\sigma=E\epsilon=E\alpha \Delta T$. This is step 2.

 

[NTL2009]

I am very impressed with the OP's determination and knowledge, and @Chestermiller 's informed replies. However, since the goal appears to be a practical application of tuning an instrument at one temperature in anticipation of performance at a different temperature, I think you could augment the theoretical with the practical.

I'd suggest measuring the delta F under expected delta T. There are other effects, the expansion /contraction of other elements, such as the guitar neck, that your formulas do not account for. Measurements would take this into account, and a predictive formula could be derived from them.

 

[azutou]

The equation for the change in tension is: Δ F = − α E A Δ T

What would the A mean here? I'm sorry if I'm missing something obvious.

Also, would the relationship between the change in temperature and change in frequency be negatively linear, and between change in frequency and change in tension a sqrt one?

 

[azutou]

ΔF=−αEAΔT

Why is this negative? I have found the axial force equation which is F = E α dt A, but don't understand why in this case it is negative (even though it makes sense). Thank you!

 

[Chestermiller]

Why is this negative? I have found the axial force equation which is F = E α dt A, but don't understand why in this case it is negative (even though it makes sense). Thank you!

If you heat it at constant length, it lowers the tension.

 

[Xilus]

Thermal Expansion equations?
steel string or nylon?
for most materials they say delta L is directly proportional to delta Temperature.
we just need an equation to convert Delta Temp to Tension

so a certain tension creates a certain note.
thats what you guys were working on right?

 

[Xilus]

Nylon 50-90 M/(MxK)
what kind of unit is M/(MxK)?
the website says:
"The calculation is: (given factor) x 10-6 x length x change in temperature °C"
but I don't know how to interpret it.

 

[Chestermiller]

Nylon 50-90 M/(MxK)
what kind of unit is M/(MxK)?
the website says:
"The calculation is: (given factor) x 10-6 x length x change in temperature °C"
but I don't know how to interpret it.
View attachment 213586

Those units mean "meter per meter per degree K." The first M refers to the change in length of the sample, the 2nd M refers to the original length of the sample, and the K refers to the change in temperature of the sample. Obviously the M's can be canceled (since we are really talking about the strain in the sample, which is dimensionless), and we are left with units of 1/K.

 

[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’.