# Homework Help: Vibrating string amplitude

1. Dec 31, 2009

### kevin36

1. The problem statement, all variables and given/known data

guitar scale= 25.5inch
pitch A=440Hz
string gauges 0.056inch, 0.053inch

Hello folks, can anyone help me with formula that relates string tension, scale length, pitch and string gauge.. I need to be able to calculate the amplitude of different string gauges.

kevin

2. Jan 1, 2010

### Stonebridge

I'm not sure what you mean by calculating the amplitude from these values. The amplitude of vibration of a guitar string will depend on how hard you strum or pick the string. This information is not given.
On the other hand, how hard you need to strum in order to get the same loundness does depend on the gauge, material and tension of the string.
Could you be more specific?

3. Jan 1, 2010

### kevin36

Thanks for the reply. I was hoping to test the theory that higher gauge strings allow lower action (and less buzz), due to the higher tension (and smaller amplitude).

Let's pick a random number for strumming force (same for both string gauges). How would I calculate amplitude of different string gauges (weights)?

thank you,
kevin

4. Jan 1, 2010

### Stonebridge

There are a couple of things going on here that need to be separated. The 1st is clear enough.

Using a higher (thicker) gauge string increases u, the mass per unit length of the string. (Assuming the same material). For a given length (l) of guitar string, the frequency (f) of the note will depend on the tension, T, and u. The standard formula is f = (1/2l) * sqrt(T/u)
So for a given f, increasing the gauge (u) requires an increase in tension.
This is a standard observation for guitarists. Thicker strings require a higher tension.

The 2nd factor requires a little more analysis. There is no reference to amplitude here.
You can play without buzz so long as you don't strum too hard. Strumming hard gives more amplitude. So the question becomes, why do you need to strum harder with thinner strings? Well, you don't. Except that the guitar is quieter for the same "amount of strum". So to get a similar volume of sound you need to strum harder and give more amplitude to the strings. Too much and you get buzzing. On any guitar, if the action is too low you will get buzzing on the frets.
So the big question is; why to thinner strings with lower tension produce a quieter sound than thicker strings with higher tension - other things being equal?
Now over to you. What physical parameters actually determine the perceived loudness of the guitar? Do you know any formula including, or any possible relationship with, the quantities we mention (T, u, l)? A clue. The energy of a wave here will depend on amplitude. The loudness of a wave will depend on its energy.

5. Jan 1, 2010

### kevin36

Yes, that makes sense.

Except for the "frequency (f) of the note will depend on the tension" part. We are tuning both string gauges to the same note, standard A440 tuning.

The thicker gauge is tuned to say E, is under specific tension, and vibrates X mm under the S "amount of strum".

Now let's change the string for a lighter gauge, tune it to the same E note and subject it to the same "amount of strum".

How do you calculate the amplitude of vibration?

Plus, let's change the amount of strum to equal the sound volume. How do you now calculate the amplitude?

(I know I'm lost, this has to be really easy..)

6. Jan 1, 2010

### Stonebridge

The problem is not one of calculating amplitude as such, but of determining why you have to produce a greater amplitude (via more "strum") when the string is of lower tension and thinner gauge. You only get buzzing if you strum too hard, so the question is about why you have to do this. The answer is that the thinner lighter string produces less sound.
The mathematics you need to show this, is about the energy in the vibrating string.
More accurately, it's about the energy that is turned into sound.
How does the (mechanical) energy of the vibrating string depend on its tension, thickness and amplitude. You need to show either a) that a thicker string under more tension has more energy than a thinner string under less tension at the same amplitude; or b) what amplitudes would be needed to produce notes of the same "loudness" from two different strings.
There is also the problem of the transmission of the sound to the body of the guitar. The sound you hear is produced mostly there - not in the string itself. Does this transmission, via the bridge, depend on tension or thickness?
Experimentally and theoretically, "amount of strum" could be difficult to quantify.
Do you know a formula for the k.e. and p.e. of a mass that is oscillating with SHM?
This is where you need to start.

7. Jan 1, 2010

### kevin36

No, unfortunately I don't have that formula.

The theory says that you can get lower action with heavier strings, because they provide more tension and you can get them lower without buzzing since they "do not move as much".

And it does work. You can set the heavy gauge very low and hit it hard- there's very little or no buzzing.

You can't seem to do that with lighter gauge. You get buzz even with medium strum.

How can I plug this into an equation?

(btw, I really appreciate your time!)

kevin

8. Jan 2, 2010

### Stonebridge

The equation in post 4, f= (1/2L) * Sqrt(T/u)
tells us half the story. If we consider 2 strings of the same length L vibrating at the same frequency f, then increasing u means you must increase T.
The relationship between u and the gauge of the string is simple. u is mass per unit length. The mass, m, of a length l of guitar string (assuming it to be a cylinder) is density (p) times volume (v), which is p * (Pi*r^2) * l (Volume is cross section x length, and cross section is Pi r squared, where r is radius of wire)
As wire gauge is a measure of its diameter (d), the formula becomes
u = m/l = p * Pi * (d/2)^2 (because r=d/2)
The equation at the top tells us that sqrt (T/u) is a constant (for the same f and length of string) therefore T/u is constant. (The value of the constant is (2fL)^2)
So T=u times this constant
u=p * Pi * (d/2)^2
so
T =
If you do this you get a formula that calculates the tension (in newton) in the string for a given gauge (d), frequency (f), length (L) and density (p)
Can you get an expression for T?
If the tension in a string is greater, then you need more force to displace it to one side in order to set it in motion. More "strum". So heavier gauge strings do not "move" so much for the same amount of applied strumming force. Or, looking at it another way, you can strum them harder and get the same displacement to one side.
This displacement of the string to one side is closely related to its amplitude once it starts vibrating. So you can strum harder without buzzing, caused by too great amplitude.
The mathematics that deals with how this string displacement then relates to the amplitude of the stable vibration is rather complex. However, in simple terms it can be said to be more or less proportional.
It is that amplitude that then determines the loudness of the sound.
Do you follow the reasoning so far?
The next stage is to look at what determines the energy of this vibration. If the problem is simplified somewhat, it can be considered to depend on the product of u, frequency squared and amplitude squared.
This then gives the relationship between the amplitude (a) of the string and the gauge, for strings of the same material, length and frequency vibrating with the same energy (loudness).
In other words, all other things being constant
amplitude squared is proportional to 1/ u
but u is proportional to gauge (d) squared, so
the amplitude is inversely proportional to the gauge.
What this means is that, all other things being equal (strings have same length, material, frequency and energy(loudness) then you can say that increasing the gauge decreases the (required) amplitude. In fact, if you double the one you half the other.
From this you can work out what changing gauge does to the amplitude.

I would stress that this is a very simplified analysis and hasn't taken account of a large number of other factors. If anyone else wants to come in on this and comment on my reasoning I would be very grateful.

9. Jan 2, 2010

### kevin36

Ok, it's finally beginning to sink in. I got the first part.

I also found some info on T.

UW = unit weight .056" = .00063477 lb/in
UW = unit weight .053" = .00056962 lb/in

L = 25.5in (scale length)
F = 82.4Hz (frequency or pitch)

T(tension) = (UW x (2 x L x F)^2)/386.4

to convert the result into Newtons multiply by 4.45

UW(unit weight) = (T x 386.4)/(2 x L x F)^2

I'm still working on the second equation (amplitude).

Thank you so much for your help. I really appreciate it.

Kevin

10. Jan 2, 2010

### Stonebridge

Kevin, be careful with the units!
For tension to be in newton, you must have all the other units in the same system.
L must be metre
density kg/m3
gauge metre
mass per unit length kg/m
I recommend converting.

I'm still not sure if I've nailed this problem. There seems to be too many variables and it requires quite a bit of simplification to come up with a meaningful and useful answer for you.
It's easy enough to show that the tension is (needs to be) greater for thicker strings.
It's also easy to show that tighter strings do not displace so much when you pluck them.
It's also true that the thinner strings, plucked to the same amplitude, will produce less energy of vibration than thicker ones. And a quieter sound.
Plucking the thinner strings harder (more amplitude) to compensate will cause buzzing if the guitars "action" is too low.
This seems to me to be the intuitive physics here.

Imagine you have an "action" such that your string is 5mm clear of the fretboard at the middle (12th) fret. Let's say this gives you a safe amplitude of 4mm without buzzing.
Let's assume you have a light E string of gauge 0.046".
If safe amplitude is inversely proportional to gauge then the constant (mixing units!) is
4 x 0.046 = 0.178
Change this to a heavy string of 0.054 gauge.
amplitude = constant / 0.054
amplitude = 3.3mm
This would imply that, all other things being equal, to produce the same loudness you only need 3.3mm amplitude with the heavier string.
You could lower the action accordingly.

Good luck and pleasant tunes.

11. Jan 3, 2010

### kevin36

I think you've nailed it. Thank you so much for your time.

kevin

12. Jun 23, 2011

### caveatemptor

Hey, really interesting stuff.
I have another question, I won't be able to understand the equations since I haven't read physics in 10 years but I hope someone will be able to answer.

If we tune down the thickest string to (in my case) C# (69.3hz)
and use 0.054 string i should get the same tension as if I was using 0.046 in standard tuning (E).

(46/E) T = [0.00038216 x (2 x 24.75 x 82.4)2 / 386.4 = 16.45 lbs
(54/C#) T = [0.00053838 x (2 x 24.75 x 69.3)2 / 386.4 = 16.40 lbs
(56/C#) T = [0.00057598 x (2 x24.75 x 69.3)2 / 386.4 = 17.54 lbs

How does this affect action?

My first thought was that since the heavier string is thicker it will need more space to vibrate even with the same tension. But this might not be true since you measure action from the bottom of the string to the fret and not from the middle.

Or does more physics come to play?