A question of physics..

The pitch of a string is determined by the mass, the tension, and the string length. By mass, I mean mass per inch of string. Thicker strings are generally lower in pitch, because they have more mass. Given six strings of equal length, as on a guitar, you can lower the pitch by either increasing the mass per inch of a particular string, or lowering the tension. Since strings with different tensions feel very different, it's done with mass. To increase the mass you can either make the string of a more massive material, or you can increase the thickness of the string. Steel strings are done by increasing their diameter, because it's a bit difficult to change tha mass of steel. Nylon strings use changes in string material a bit more, so they can get away with diameters closer to that of the other strings.

To answer these questions directly:
1) If we make a set of strings in which two strings of equal mass have different diameters, which of the two will have the lower frequency? Why?

2) If we make a set of strings where two strings of the same diameter have different masses which will have the lower frequency? Why?
1) They will be exactly the same. The thicker string will probably have a different tone than the thinner one, but it will have the same pitch if both strings are at the same tension.

2) The more massive string will have the lower frequency.

The reason a more massive string vibrates at a lower pitch is because of its inertia. Any string with tension on it has a force pulling it straight, which is proportional to the tension on the string. When you pluck it, the force pulls it back. A string with greater mass accelerates on the way back more slowly because is has more inertia resisting that acceleration. Since it changes direction more slowly, it takes longer to return to center, pass across, continue past center, and come back to center again. More time to complete one oscillation. The time taken is called the period (T) of the oscillation. A longer period means a lower frequency (the frequency is the inverse of the period. f=1/T).

Greybeard-

Energy goes in; energy comes out. Given purely mechanical considerations, it is not obvious by a string of higher mass would continue to vibrate longer. If the endpoints are perfect (no energy loss), the places were energy leaves the system are primarily two: friction in the flexing string ends (heat loss) and energy transfer to the air, in both frictional losses and acoustic coupling. Given all this, the string provided with the most vibrational energy will probably vibrate the longest. Put the same energy into two strings of different mass/length, the one with lower mass-per-length will have to undergo larger transverse (side-to-side) amplitude to store the same energy as the other string. Now it could be argued that since frictional losses are highly non-linear and related to transverse velocity as well as string flex angle, that the string undergoing the largest vibrational amplitude might tend to decay (lose energy) faster. But this ends up being offset by two other factors: Thinner strings (less mass=per-length) will have lower frictional losses in bending and experience lower fictional loss and lower acoustic coupling to the air. So again, it's not obvious that more mass-per-length wins an equal energy sustain contest.

Now if we're talking same offset (i.e., plucking displacement), the the denser string sustains longer since it received more energy to begin.

Laz's explanation is the important one here. Most of the decay occurs due to coupling of the string's energy into the guitar -- through the nut/fret and bridge (bigtime in acoustic guitars) and the magnetic drag of the pups.

-Greg

another addition to sustain is how much feedback(vibrations) the body projects in the bridge, the strings are vibrating the bridge and the rest of the guitar, but the bridge is vibrating the strings after you have strummed or plucked them, this presents achallenge to guitar makers, because less reinforcement on the bridge will give the guitar more sustain, but too little could collapse the guitar

Now if we're talking same offset (i.e., plucking displacement), the the denser string sustains longer since it received more energy to begin.

Greg,
That's all I said "All other things being equal a string of greater mass will vibrate longer than one of less mass"
Laz's explanation is the important one here. Most of the decay occurs due to coupling of the string's energy into the guitar -- through the nut/fret and bridge (bigtime in acoustic guitars) and the magnetic drag of the pups.

Which is why I didn't mention it - it has nothing to do with the original question.

Now if we're talking same offset (i.e., plucking displacement), the the denser string sustains longer since it received more energy to begin.

Greg,
That's all I said "All other things being equal a string of greater mass will vibrate longer than one of less mass"

G-

You've chosen the "equal" to be string displacement. I choose to it to be the energy input. This is where we differ.

-G

As musicians, we intuitively give more energy to thicker strings... it's the only way to maintain the same volume when you change strings

:)

I'm kind of amazed at how far this thread went. Who would have thought?

Greg, I do have a window to out, but a man has to save something to do for the afternoons. :lol: