# Vibrating String

1. Nov 17, 2003

### bugler777

I am trying to understand the relationship between mass, tension and length in a vibrating string such as would be found on a violin or guitar. If the string has a uniform gauge throughout and is kept at a constant tension, then reducing the length by 1/2, for example, will cause the the pitch to double, or increase by an octave.

My question is, Is it incorrect to view this occurence as a reduction of vibrating mass by 1/2 rather than a halving of the length of the string? It seems to me that the real phenomena that is occuring is a reduction in the amount of mass that is vibrating and that (because the gauge of the string is uniform throughout) it is only coincidental that 1/2 length will also determine 1/2 mass.

I hope this makes sense.

2. Nov 17, 2003

### chroot

Staff Emeritus
Let's first talk about how a string vibrates. The two ends of the string are fixed to some kind of rigid supports, and thus cannot move. These points that cannot (or do not) move are called "nodes." The middle of the string is free to wag back and forth. The point(s) along the string with the largest motion are called antinodes.

In the simplest example, the string has two nodes, one at each end, and one antinode right in the middle. This wavelength is the so-called "fundamental" wavelength, or the "first mode."

Thus, the wavelength of the string's vibration is defined by the supports, and has nothing whatsoever to do with mass or elasticity.

Now, let's talk about the frequency of the vibration. The frequency and wavelength are related through the equation

$$\lambda \nu = v$$

where $$\lambda$$ is the wavelength, $$\nu$$ is the frequency, and $$v$$ is the velocity of wave propogation in the string.

The last quantity, the velocity of wave propagation, does depend upon the material properties of the string -- namely, it's linear mass density

$$\mu \left[ \frac{kg}{m} \right]$$

and it's elasticity, defined by a paramter

$$\tau \left[ \frac{ kg \cdot m } {s^2} \right]$$

Notice that $$\mu$$ has units of mass per unit length, and $$\tau$$ has units of force.

The velocity of a wave travelling through a medium is, in general, a function of both its mass (which "stores" kinetic energy in its motion) and its tension (which "stores" potential energy when the string is stretched). The equation looks like this:

$$\begin{equation*} \begin{split} v &= \sqrt{\frac{\textit{elastic property}}{\textit{inertial property}} }\\ &= \sqrt{\frac{\tau}{\mu} } \end{split} \end{equation*}$$

Combining these equations, you can see that the frequency of a wave as a function of the mass density of the string is given by

$$\nu = \frac{1}{\lambda} \sqrt{\frac{\tau}{\mu}}$$

In other words, if you double the mass density of the string, you are halving the quantity under the square root, and therefore making the new frequency only 70% of what it was originally.

As you can see it's not the mass density itself, but the inverse square root of the mass density, that matters. Doubling the mass does not halve the frequency; it reduces it by about 30%.

Does this make sense?

- Warren

Last edited: Nov 18, 2003
3. Nov 17, 2003

### bugler777

Does any of this relate to the fact that if a violin string is vibrating at a certain frequency and then is made half as long by moving the finger half way up the finger board, the frequency of vibration will double sounding a pitch an octave higher than the original. And wouldn't it be true that only half as much string mass will be vibrating at the new frequency?

4. Nov 18, 2003

### chroot

Staff Emeritus
No. While it is generally true that when a guitarist frets the string, only the side he plucks has any substantial vibration, this is not always the case.

A string can vibrate in many different ways -- each way is called a "mode." A string's first mode of vibration, as I said, has one node at each end, and one antinode exactly in the middle. The second mode (also called the second harmonic) has three nodes (one at each end, one in the middle) and two antinodes (at $$l / 4$$ and $$3 l / 4$$).

In a properly done experiment (where the node in the middle is supported properly), the entire string can vibrate. One half goes down while the other goes up. Obviously, the entire string is still vibrating, and therefore the mass is the same -- but now the frequency has doubled. This goes directly against what you are proposing.

Indeed, you can play harmonics on a guitar by striking an open string (to excite several modes, but mostly the fundamental mode) and then gently touching the string right in the middle, over the 12th fret. By gently touching the string, you are killing the fundamental-mode vibration, and keeping the second-mode vibration. When you remove your finger -- viola -- you have the entire string vibrating at twice its fundamental frequency. Once again, this goes directly against what you are proposing.

Let's look again at that last equation,

$$\nu = \frac{1}{\lambda} \sqrt{\frac{\tau}{\mu}}$$

What you seem to be missing is this piece of wisdom: the frequency of the vibrating string does not depend upon the mass of the entire string. Instead, it depends upon the linear mass density, the mass per unit length. The mass of the entire string never enters the equation above.

It is true that halving the string's length doubles its frequency (if you halve $$\lambda$$, you double $$\nu$$), but it has nothing to do with half the string having half the mass. Instead, you've just decreased the wavelength by two, and therefore increased the frequency by two. The total mass just has nothing to do with it.

As I explained in my previous post, doubling the mass of the string (while keep its wavelength constant) only reduces its frequency by about 30%.

- Warren

Last edited: Nov 18, 2003
5. Nov 18, 2003

### bugler777

Warren,

You have taken the time to address my question at length and I thank you very much. Very impressive. I will continue to try to fully understand what you're saying scientifically/mathematically, but it seems I will have to readdress the approach I was taking with another issue, a much more complicated one.

I had hoped that the physics of the vibrating string might add some direction to understanding, more specifically, the issues involved with vibrating lip tissue in a brass instrument. Perhaps, a varying amount of lip tissue mass will be relevant, but it appears that if that proves to be true, the relevance won't be due to any similarities with a vibrating string.

Charles

6. Nov 18, 2003

### chroot

Staff Emeritus
Well, there are some similarities. The lips are supported all around by the mouthpiece itself, meaning that the vibration has to be (nearly) zero there. The mouthpiece creates a two-dimensional (circular) node around the lips.

The lips vibrate mostly up and down, I suppose -- essentially closing and opening. The wavelength of the vibration is fixed by the mouthpiece, but the frequency is not. The musician can vary both the mass and the tension he exerts.

Remember that

$$\nu = \frac{1}{\lambda} \sqrt{\frac{\tau}{\mu}}$$

If you tighten your lips, you're going to increase $$\tau$$, and therefore increase the frequency. If you have fat lips, you're going to increase $$\mu$$, and therefore decrease the frequency.

(If you do such experiments, make sure that you don't connect an instrument to the mouthpiece! The instrument is designed to allow only particular patterns of vibration, called standing waves. If you begin blowing a middle C, and tighten your lips up, eventually most of the vibration will excite the next note up in the overtone series. As a bugler, I assume you are dearly familiar with this!)

As a result, I'd say that a person with fat lips probably has a harder time hitting those high notes -- but if they work on their embouchure, they may be able to make up for it with higher lip tension.

- Warren

7. Nov 20, 2003

### turin

Yes, it is incorrect. I am in no way disagreeing with chroot, but maybe a different angle will help you better understand.

The entire mass does not move all at once. If you have a decent first semester physics book, it probably derives the freqency of a vibrating spring. Look carefully at the diagrams (if any). The model is that of a tiny little segment of the string. Then, there are two forces on the tiny segment that oscillate (the tension on either end). Since the model isolates the tiny segment from the rest of the string, then it is the mass of the tiny segment that is used in the frequency calculation, and the rest of the string is just the tensions on the ends of the segment. That is why the linear mass density is used, because the mass of the tiny segment is the linear mass density times the length of the tiny segment.

The model makes the length of every tiny segment the same, regardless of the length of the string. So, changing the length of the string does not change the mass that shows up in the derivation.

8. Dec 5, 2003

### bugler777

A couple of comments that may or may not be of interest.

chroot - The lips vibrate mostly up and down, I suppose -- essentially closing and opening.

bugler - It is controversial whether the lips actually close during one cycle of vibration. I happen to believe they do but there are others who claim otherwise and insist they can prove their point with new photo images.

chroot - The wavelength of the vibration is fixed by the mouthpiece,

chroot - The wavelength of the vibration is fixed by the mouthpiece,

bugler - The actual length of the aperture of the lips inside the mouthpiece changes. IOW, the aperture does not extend the length of the diameter of the inside of the mouthpiece.

chroot - The musician can vary both the mass and the tension he exerts. Remember that

***(equation for frequency)***

If you tighten your lips, you're going to increase "t", and therefore increase the frequency. If you have fat lips, you're going to increase "mass" , and therefore decrease the frequency.

Bugler – It appears that there is very little lip tissue that is involved in the actual vibration. Large lips may not bring any significant amount of additional lip tissue into the mix than smaller lips. Again, controversial.

chroot - If you do such experiments, make sure that you don't connect an instrument to the mouthpiece! The instrument is designed to allow only particular patterns of vibration, called standing waves. If you begin blowing a middle C, and tighten your lips up, eventually most of the vibration will excite the next note up in the overtone series. As a bugler, I assume you are dearly familiar with this!

Bugler – Yes, the harmonics associated with the fundamental pitch. The standing waves, however (as I understand it), end at the concert pitch F over high C on a Bb trumpet.

Pitches can also be “bent” (forced) between notes of the harmonic series.

And the trumpet brings a resistence (support) to the vibrating lips that doesn't exist without the trumpet being present.

chroot - As a result, I'd say that a person with fat lips probably has a harder time hitting those high notes –

Bugler – And yet in the actual world, large lips don’t seem to be a disadvantage to high note playing. There are many players with large lips who play easily in the extreme high register and there are many, many players with smaller to thin lips who struggle constantly with the upper range. And vice versa.

A lot of time, research and money has been spent (supposedly more than any other musical instrument) trying to determine what is going on with the lips during sound production on a trumpet, and yet the answers continue to be elusive. There are even some who defend the position that the lips are not even vibrating!

I was hoping there would be merit in viewing varying mass of lip tissue as a factor. So, back to the drawing boards. Thank you for your help.

Charles

9. Dec 7, 2003

### zoobyshoe

When you tighten or loosen a string are you changing the elasticity or the linear mass density or both?

10. Dec 7, 2003

### chroot

Staff Emeritus
Technically, a little bit of both -- but I assume that the tension is the overriding change.

- Warren

11. Dec 7, 2003

### turin

I agree with chroot. For metals, Young's modulus is on the order of GN/m2, so making any kind of appreciable increase in length (reduction in &mu;) would require a tremendous tension.

12. Nov 17, 2007

### ben_gal

Chroot!
You say:
"It is true that halving the string's length doubles its frequency "

why the distance betwen the guitare frets are not the same?

Another question: how can I calculate the distance between holes and the dimensions of the mouthpiece in a flute ?

How can I predict the quality of a guitar using chladney figures for the upper plate and what is the influence of the ribs and hole?

Last edited: Nov 17, 2007
13. Jul 23, 2011

### mln326

Ok, if a string is between two nodes.... What is the Node?

14. Jul 23, 2011

### sophiecentaur

The musical scale is based on ratios of lengths (in the case of strings). Very roughly, the frequency ratio of a semitone is the twelfth root of two. i.e. after moving up by twelve semitones, you get to double the frequency. That leads to the frets having uneven spacing; they are actually the same proportion of the string, at any length.
Edit - not 'very roughly' - it's exact for the well tempered scale.

As to the quality of a guitar - I think you have to play it and use your ears. Those chaldney diagrams show only a part of the sound 'quality'. I guess they would reveall a really poor guitar - but that's probably all.

Last edited: Jul 23, 2011