Exploring Frequency & Temperature: Theory & Experiment

[piyush_1903]

TL;DR Summary

Hey guys, I am a 16-year-old high school working on an experiment to investigate the relationship between the frequency of the guitar string and the temperature of the air around it. Please help me understand my findings and correct any mistakes.

So, my apparatus consists of a basic wooden plank with a nail on one end and a guitar tuner on the other. I plan on using a heating lamp to control the temperature of the apparatus, which is placed inside an insulation box, and use a non-contact infrared thermometer to measure the temperature of the air around the apparatus and the string itself. Lastly, I'm using audacity to graph a plot spectrum and measure the frequency produced by the guitar string. I am currently trying to work on a theoretical formula relating to frequency and temperature, as this would compliment my experimental findings and allow me to have a better understanding of my experimental values. Furthermore, I would like to see if I can explain any discrepancies between the predicted theoretical values and the experimental findings. Currently, my formula relating the two variables is:

• Assuming f0 to be the frequency at temperature t0 and tension Ti, and f0+Δf (new frequency) to be the frequency at t0+Δt (new temperature) and Ti+ΔT (new tension), the equation can be written as:
• Δf = (1/2L)*sqrt([Ti + σEAΔt]/μ) - 1/2L * sqrt[Ti/μ]
• ,σ is the coefficient of thermal expansion, E is Young’s modulus, and Δt is the change in temperature

 

[anorlunda]

I very much like your sensible approach to this project. Good for you.

Before answering, please tell us where you found that equation. Did you research the question? Where did you lool.

 

[piyush_1903]

I very much like your sensible approach to this project. Good for you.

Before answering, please tell us where you found that equation. Did you research the question? Where did you lool.

Hey!

Yes, I did, I have attached the file below. It shows me deriving my formula. Please do let me know if you think it can be simplified further or improved!

 

[anorlunda]

Please post that file again as a PDF file, or as a screen capture picture? You can paste a picture directly into the posts.

I worry about cybersecurity, so I will not open a DOC file sent by a stranger over the Internet.

 

[piyush_1903]

Please post that file again as a PDF file, or as a screen capture picture? You can paste a picture directly into the posts.

I worry about cybersecurity, so I will not open a DOC file sent by a stranger over the Internet.

 

[Chestermiller]

Shouldn't the tension decrease as the temperature is increased?

You seemed to switch from $\alpha$ representing the coefficient of linear expansion to $\sigma$, correct?

Have you had calculus yet?

 

[Z0dCHiY8]

I worry about cybersecurity, so I will not open a DOC file sent by a stranger over the Internet.

the're no problems to open unknown *.doc(x) safely == cloud viewer (like google docs) or offline in something like LibreOffice (it's very unlikely for crafted-to-MSOffice virus to be not ruined by unknown environment).

 

[Baluncore]

Is your string made of a piano wire, a mono-filament, or a spun fibre.

Your wooden plank will also be affected by temperature, humidity and time.

The tension in the string could rise or fall with temperature. It depends on if the wood or the string has the higher temperature coefficient, and on how long you allow it to equilibrate.

 

[piyush_1903]

Shouldn't the tension decrease as the temperature is increased?

You seemed to switch from $\alpha$ representing the coefficient of linear expansion to $\sigma$, correct?

Have you had calculus yet?

Hello,
Yes the tension of the string will decrease with a change in temperature, which I am trying to model mathematically by relating tension to youngs modulus, change in temperature and coefficient of thermal expansion.
And yes, my bad but I did switch signs there.
No I have not taken calculus, but I will be doing so in my upcoming Grade 12 year:)

 

[piyush_1903]

Is your string made of a piano wire, a mono-filament, or a spun fibre.

Your wooden plank will also be affected by temperature, humidity and time.

The tension in the string could rise or fall with temperature. It depends on if the wood or the string has the higher temperature coefficient, and on how long you allow it to equilibrate.

Yes, but I am trying to account for the change in tension with temperature by using the mathematical model to portray the final tension as a product of young's modulus(constant), linear density(constant) and change in temperature (measured w my infrared gun)

 

[Z0dCHiY8]

No I have not taken calculus, but I will be doing so in my upcoming Grade 12 year:)

You can use software to automate your calculations + global net has a lot of free tutorials (for instance, MIT lectures on YT).

 

[piyush_1903]

Yes, but I am trying to account for the change in tension with temperature by using the mathematical model to portray the final tension as a product of young's modulus(constant), linear density(constant) and change in temperature (measured w my infrared gun)

Oh my bad, my string is a typical steel E guitar string. I also have access to "inextensible" fishing line string which is nylon-based with some plastic.

 

[piyush_1903]

Is your string made of a piano wire, a mono-filament, or a spun fibre.

Your wooden plank will also be affected by temperature, humidity and time.

The tension in the string could rise or fall with temperature. It depends on if the wood or the string has the higher temperature coefficient, and on how long you allow it to equilibrate.

its a classic steel based guitar string. I also have a nylon based inesxtensible fishing line string

 

[Chestermiller]

So you have $$\Delta f=\frac{1}{2L\sqrt{\mu}}(\sqrt{T_i-\alpha EA\Delta t}-\sqrt{T_i})$$What do you get if multiply numerator and denominator of the right hand side by $$\sqrt{T_i+\alpha EA\Delta t}+\sqrt{T_i}$$and neglect the quadratic term in $\Delta t$?

 

[piyush_1903]

So you have $$\Delta f=\frac{1}{2L\sqrt{\mu}}(\sqrt{T_i-\alpha EA\Delta t}-\sqrt{T_i})$$What do you get if multiply numerator and denominator of the right hand side by $$\sqrt{T_i+\alpha EA\Delta t}+\sqrt{T_i}$$and neglect the quadratic term in $\Delta t$?

I'm sorry, I would end up with a quadratic right? But why do I need to do so, is ther esomething wrong with my current formula?

 

[Baluncore]

Be prepared for a surprise. I believe the CTE of wood is typically 30 ppm / °C along the grain. Piano wire is only about 10 ppm. The wood will get longer faster than the wire, so you can expect tension in the wire will increase with temperature.

 

[piyush_1903]

Be prepared for a surprise. I believe the CTE of wood is typically 30 ppm / °C along the grain. Piano wire is only about 10 ppm. The wood will get longer faster than the wire, so you can expect tension in the wire will increase with temperature.

Interesting, I did not consider that! I am conducting the experiment tommorow, will share the results!

 

[Chestermiller]

I'm sorry, I would end up with a quadratic right? But why do I need to do so, is ther esomething wrong with my current formula?

Right. There would be no quadratic term. So what do you get?

 

[Chestermiller]

Here's another interesting approach to consider. Since the initial frequency and tension are related by:
$$f_0=\frac{1}{2L}\sqrt{\frac{T_i}{\mu}}$$and the fréquency and tension after any temperature change are related by: $$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$$if we divide the 2nd equation by the 1st equation, we obtain:$$\frac{f}{f_0}=\sqrt{\frac{T}{T_0}}=\sqrt{\frac{T_i-\alpha EA\Delta t}{T_i}}=\sqrt{1-\frac{\alpha EA\Delta t}{T_i}}$$Squaring both sides then gives:$$\left(\frac{f}{f_0}\right)^2=1-\frac{\alpha EA\Delta t}{T_i}$$Solving the first equation for $T_i$ in terms of $f_0$ and substituting into this equation then yields: $$\left(\frac{f}{f_0}\right)^2=1-\frac{\alpha EA}{4\mu L^2 f_0^2}\Delta t$$or $$f^2=f_0^2-\frac{\alpha EA}{4\mu L^2 }\Delta t$$Note that this equation no longer contains the initial tension or the present tension as a parameter, meaning that the tension does not need to be measured. The equation indicates that a plot of $f^2$ vs $\Delta t$ should be a straight line with an intercept of $f_0^2$ and a slope of $-\frac{\alpha EA}{4\mu L^2 }$ (which only involves the length and material parameters of the wire).

Assuming that @Baluncore's assessment is correct, the same kind of relationship will apply (i.e., s straight line relationship between $f^2$ and $\Delta t$), except that the slope can be positive (or less negative), and will depend on the properties of both the wood and the wire.

 

[piyush_1903]

Here's another interesting approach to consider. Since the initial frequency and tension are related by:

f0=12L√Tiμf0=12LTiμ​

and the fréquency and tension after any temperature change are related by:

f=12L√Tμf=12LTμ​

if we divide the 2nd equation by the 1st equation, we obtain:

ff0=√TT0=√Ti−αEAΔtTi=√1−αEAΔtTiff0=TT0=Ti−αEAΔtTi=1−αEAΔtTi​

Squaring both sides then gives:

(ff0)2=1−αEAΔtTi(ff0)2=1−αEAΔtTi​

Solving the first equation for TiTi in terms of f0f0 and substituting into this equation then yields:

(ff0)2=1−αEA4μL2f20Δt(ff0)2=1−αEA4μL2f02Δt​

or

f2=f20−αEA4μL2Δtf2=f02−αEA4μL2Δt​

Note that this equation no longer contains the initial tension or the present tension as a parameter, meaning that the tension does not need to be measured. The equation indicates that a plot of f2f2 vs ΔtΔt should be a straight line with an intercept of f20f02 and a slope of −αEA4μL2−αEA4μL2 (which only involves the length and material parameters of the wire).

Assuming that @Baluncore's assessment is correct, the same kind of relationship will apply (i.e., s straight line relationship between f2f2 and ΔtΔt), except that the slope can be positive (or less negative), and will depend on the properties of both the wood and the wire.

Wow, this took me a while to understand but yes this is a lot easier! It's just really complex algebra but it will remove one of the experimental parameters, therefore reducing uncertainty. And tension had been a really hard thing to measure for me because I was measuring the initial frequency to find the initial tension and then substituting for final tension for initial tension plus thermal stress. I will give this approach a shot too!

Plus, you are right it a linear function is a lot easier to analyze!
Oh also, I noticed I had a stray variable A in my initial calculations that does not signify anything. The formula for thermal stress is simply young's modulus, coeffecient of expansion and change in temperature

 

[Dullard]

if your 'infrared gun' is a handheld unit with a laser dot, be aware:

The gun probably has a much wider field of view than you think. People often assume that the reported temperature is the temperature of the 'dot' area - it's actually the average temp of a much larger area (you can figure out the FOV of yours with a little experimentation). It will work just fine as long as you make sure that everything (including the string) in the FOV is at the same temperature.

 

[piyush_1903]

if your 'infrared gun' is a handheld unit with a laser dot, be aware:

The gun probably has a much wider field of view than you think. People often assume that the reported temperature is the temperature of the 'dot' area - it's actually the average temp of a much larger area (you can figure out the FOV of yours with a little experimentation). It will work just fine as long as you make sure that everything (including the string) in the FOV is at the same temperature.

Making the Purchase rn, thank you for your advice I wouldn't have known this. Making a really simple "isolation box" rn. Its like a greenhouse, but it will preserve heat and all the apparatus inside to reach equilibrium with the surroundings. Its just a shoe box with aluminium foil on the inside and cling wrap over the top. when i measure temperature ill take the cling wrap out and immediately take the temperature

 

[piyush_1903]

Be prepared for a surprise. I believe the CTE of wood is typically 30 ppm / °C along the grain. Piano wire is only about 10 ppm. The wood will get longer faster than the wire, so you can expect tension in the wire will increase with temperature.

Hello!

Surprised I was because the tension did in fact drop! I think I figured out why but please correct me if I am wrong. So my understanding is that the coefficient of thermal expansion measures the expansion per degree kelvin raised. Therefore, per degree kelvin raised, wood expands 3 times more than steel. However, steel has a far lower specific heat capacity than wood, therefore it raises it's temperature a lot faster given the same amount of energy as the wood (energy is constantly leaving the heat lamp the same to both objects). This may explain why the string still drops in frequency, albeit only after significant heating.

 

[Baluncore]

It is not all to do with heat capacity.
Wood is a good insulator compared to steel. The wood is also thicker, so it takes much longer for heat to enter the full depth of the wood than the steel wire.
I think the wire tension will depend on the time scale of the experiment. If you are quick you will see only the steel CTE, but for longer times the wooden CTE will come to dominate the tuning of your one string instrument.

 

[piyush_1903]

It is not all to do with heat capacity.
Wood is a good insulator compared to steel. The wood is also thicker, so it takes much longer for heat to enter the full depth of the wood than the steel wire.
I think the wire tension will depend on the time scale of the experiment. If you are quick you will see only the steel CTE, but for longer times the wooden CTE will come to dominate the tuning of your one string instrument.

I see! Okay I will try a range of temperatures at equilibrium. It will interesting the explain the change in frequency that way in my experiment.

Also just to clarify, since I am justifying it as a matter of specific heats, doesn't the specific heat determine whether or not material qualifies as an insulator or conductor? Or is there some other quantity that determines it? Like, i meant the same thing. Wood requires a lot more energy to raise its temperature by one kelvin compared to steel (because its a better insulator/higher heat capacity). But do tell me if something else justifies that classification of an object as an insulator?

 

[Chestermiller]

I see! Okay I will try a range of temperatures at equilibrium. It will interesting the explain the change in frequency that way in my experiment.

Also just to clarify, since I am justifying it as a matter of specific heats, doesn't the specific heat determine whether or not material qualifies as an insulator or conductor? Or is there some other quantity that determines it? Like, i meant the same thing. Wood requires a lot more energy to raise its temperature by one kelvin compared to steel (because its a better insulator/higher heat capacity). But do tell me if something else justifies that classification of an object as an insulator?

The heat transfer is determined by the thermal diffusivity which is equal to the thermal conductivity divided by the product of density and heat capacity.

 

[piyush_1903]

The heat transfer is determined by the thermal diffusivity which is equal to the thermal conductivity divided by the product of density and heat capacity.

Nice, these are new concepts. I shall see if I need to explore them in this study.

 

[Baluncore]

Heat capacity and insulation are quite different things.

Specific heat capacity is the thermal energy needed to heat a unit mass by one unit of temperature. It is only dependent on materials.
https://en.wikipedia.org/wiki/Specific_heat_capacity

Insulation is the ratio of temperature difference to heat energy flow between two surfaces. It is the reciprocal of thermal conductivity. It is highly dependent on the structure between the surfaces, but it also dependent on the material. Wool, foams and double glazing are examples of how structure affects thermal insulation.
https://en.wikipedia.org/wiki/Thermal_conductivity
https://en.wikipedia.org/wiki/R-value_(insulation)

 

[Sephen Smith]

Summary: Hey guys, I am a 16-year-old high school working on an experiment to investigate the relationship between the frequency of the guitar string and the temperature of the air around it. Please help me understand my findings and correct any mistakes.

So, my apparatus consists of a basic wooden plank with a nail on one end and a guitar tuner on the other. I plan on using a heating lamp to control the temperature of the apparatus, which is placed inside an insulation box, and use a non-contact infrared thermometer to measure the temperature of the air around the apparatus and the string itself. Lastly, I'm using audacity to graph a plot spectrum and measure the frequency produced by the guitar string. I am currently trying to work on a theoretical formula relating to frequency and temperature, as this would compliment my experimental findings and allow me to have a better understanding of my experimental values. Furthermore, I would like to see if I can explain any discrepancies between the predicted theoretical values and the experimental findings. Currently, my formula relating the two variables is:

• Assuming f0 to be the frequency at temperature t0 and tension Ti, and f0+Δf (new frequency) to be the frequency at t0+Δt (new temperature) and Ti+ΔT (new tension), the equation can be written as:
• Δf = (1/2L)*sqrt([Ti + σEAΔt]/μ) - 1/2L * sqrt[Ti/μ]
• ,σ is the coefficient of thermal expansion, E is Young’s modulus, and Δt is the change in temperature

I love your project. I was struck by the same idea this morning age 66 not a youth!

I have these ideas to offer: I’m not sure which gauges of guitar string shows the best responsiveness to temperature changes for which temperature ranges. Maybe heavier gauge strings are more sensitive than light gauge at say -10 to + 20C.

It might be nice to have some software that could convert oscilloscope readings into a temp readout. Also if you had say six strings tuned to an ‘open tuning’ such as ‘Open G’ you might have a thermometer that both sounds sweet and has higher overall accuracy.

I can imagine this as a wall mounted instrument that looks beautiful and would sell well on Etsy. I wish you every success. This is not just good physics it could be good commerce.

 

[JT Smith]

I've often wondered how good of a thermometer my guitar is. The tuning typically changes diurnally, going up in pitch in the morning when it's cold and then back down in pitch when it's warm later in the day. A guitar is perhaps more complicated than a wooden plank with a string attached but I would imagine it would behave similarly. Maybe I'm wrong though.

Good luck!