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Consider a string under tension, for example, a string on a guitar. When a guitar string is plucked, it vibrates at a certain frequency. When the tension on the string is increased by twisting the tuning peg at the end of the neck, the frequency of the string increases. According to this video, increasing the tension on a string also increases the speed of the wave moving through it. According to this article, the speed of a wave is higher in a more dense medium.

Does this mean that increasing the tension also increases the density?

This does not seem very logical to me, since density is the "compactness" of a substance, or how close together the individual molecules are. By twisting the tuning peg on a guitar, you rotate the cylinder that the string is fixed to, wrapping it around more tightly. This pulls more of the string around that cylinder, similarly to when you reel in a fishing rod. Since more of the string's molecules are now wrapped around the peg, there should be fewer molecules over the same distance between the bridge and the nut (the two ends of the vibrating portion of the string), therefore bringing the density down.

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  • $\begingroup$ Increasing the density of a continuous medium would correspond to increasing the thickness of a vibrating string. $\endgroup$
    – DWin
    Mar 12, 2015 at 1:20

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Your intuition is right: the density of the string goes down a little bit when you increase the tension.

HOWEVER: the wave in a string is a transverse wave which depends on the tension and the mass per unit length. If you double the tension the mass per unit length goes down by a small amount (the string gets a bit "thinner" because it gets longer) . Both these things increase the velocity of the transverse wave which is given by

$$v = \sqrt{\frac{T}{\rho}}$$

Where T is the tension and $\rho$ the mass per unit length. Finally, the fundamental frequency is determined as the reciprocal of the round trip time of the wave along the string:

$$T = \frac{2 \ell}{v}$$

so that

$$f = \frac{1}{T} = \frac{v}{2\ell} = \frac{\sqrt{\frac{T}{\rho}}}{2\ell}$$

So to raise the frequency by an octave you need four times the tension (don't try this - you might break the neck) or a string that's half the diameter (one quarter of the area - so one quarter of the mass per unit length). This explains why different strings on the guitar have different gage - you would need too much tension to get the high range from a long thick string.

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The article you refer to is talking about the speed of sound (or speed of longitudinal wave vibrations) within a material and how that relates to volumetric mass density. If you were transmitting sound from one end of a guitar string to another, this would be relevant.

But the sound produced from a string is not related to the speed that it travels within the material. Instead, it only depends on the speed of vibration, which is mainly dependent on the linear mass density and tension.

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The characteristic speed, $S$ of transverse waves on a taut string or wire with tension $T$ and density per unit length, $\rho$ is

$$S=\sqrt \frac T\rho$$

So indeed as you remove wire from play by wrapping it around the capstan, you decrease the wires density and increase tension. This increases velocity.

and since

$$\omega=\frac v\lambda$$

frequency (pitch) is increased

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    $\begingroup$ Why does removing wire change its density? $\endgroup$
    – BowlOfRed
    Mar 12, 2015 at 2:24
  • $\begingroup$ @BowlOfRed The wire is stretched, i.e. less mass in the same length. density per longitudinal centimeter diminishes. $\endgroup$
    – anna v
    Mar 12, 2015 at 4:18
  • $\begingroup$ Ah, I think you are referring to the linear density, while I think the OP (via the linked article) was referring to the (volumetric) density. I agree the first will decrease, but I think the second may not. $\endgroup$
    – BowlOfRed
    Mar 12, 2015 at 5:21
  • $\begingroup$ Suggest that you are consistent with use of either $S$ or $v$ and that you use the "frequency" in the musical sense (complete cycles) $f=\frac{\omega}{2\pi}$ $\endgroup$
    – Floris
    Mar 12, 2015 at 19:30
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For linear deformations, when a material is compressed/stretched in one direction, it usually tends to expand/contract in the other two directions perpendicular to the direction of compression/stretching. This is called the Poisson effect, and the Poisson ratio is a measure of this effect. For some materials the Poisson ratio can be close to 0.5 (rubber) which means that the material tends to behave as incompressible under this type of deformation. So for a rubber string there is no change of the mass density (per volume) as it is being stretched. But for the stainless steel the Poisson's ratio is about 0.3 which means that the mass density decreases as the string is being stretched. For most string instruments it is the transverse modes that produce the main sound, and it is the mass density per length rather than per volume what matters. However, in some musical instruments the string longitudinal modes can be audible too. Unlike the transverse modes that depend on the string tension, for the longitudinal modes the propagation speed is given by $\sqrt{Y/\rho}$, where $Y$ is the Young modulus and $\rho$ is the mass density per volume, so the change of $\rho$ can be in principle a detectable effect.

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