# School investigation: How does the size of a resonance box affect the sustain of a guitar string?

TL;DR for a lab report, how does the volume of a resonance chamber underneath the guitar string affect its sustain i.e. time it stays above a volume threshold x. Is there an equation to model this relationship?

For a physics Lab investigation for school, I decided to investigate the following question: How does the volume of a resonance chamber underneath a guitar string affect the time it stays above a volume threshold of x decibels when struck using a constant force?
In terms of the control variables: the thickness of the string, force the string is struck with, the pitch of the string... and a few others.
The independent variable would be the volume of the chambers, which I will adjust by having two boxes, where, one is in the other with their openings are facing each other. So, I cab adjust the volume by pulling them apart or closing in on them.
Finally, the dependent variable is the time the string will be above x decibels. I will measure this by keeping a mic next to the string at a constant distance and measuring its sound. I will start a timer once I strike the string, and stop it when the sound measured goes underneath x decibels.

I am unable to find any resources to help me with this. Could anyone explain the physics behind this, if there is any, and help me derive an equation to represent the relationship between my IV and DV?

• We do not normally solve homework problems or give recommendations. But I would direct you to anything written by Fletcher and Rossing. They are world experts in acoustics and the physics of musical instruments. Some of their books are very mathematical but there are a few on the physics of instruments that are ideal for musicians and sound engineers to understand the basics. The problem with your question is that is does not address the other properties of the acoustic guitar, wood, bracing, glue, etc. These things will change with size too and you probably can't isolate one variable.
– user196418
Jun 6, 2021 at 20:49

First, the size of the air cavity affects the low frequency of the guitar (see this answer), but the most important part for mid and high frequencies is the top plate, and its interaction with the resonant air.

The air enclosed in your air cavity works as a Helmholtz resonator, which frequency is given by

$$f = \frac{v}{2\pi}\sqrt{\frac{A}{V L_{eq}}}\, ,$$

where $$v$$ is the speed of sound (close to 340 m/s), $$A$$ is the cross-section area of the hole, $$V$$ is the volume of air enclosed, and $$L_{eq}$$ is the equivalent length of air vibrating. In this case I think that $$L_{eq} = 0.6D$$, where $$D$$ is the diameter of the hole.

Regarding your project, I think that you can analyze the change in timbre with the change in volume of the resonance box. You can do this by recording the signals and analyzing them later on using, for example, Audacity.

You have chosen a very complicated system to analyze. In general: increasing the box volume decreases its resonant frequency, and matching the string frequency to the box frequency will decrease the sustain time. Writing an equation expressing this is a term project task for an upper division engineering student.

• Just wondering, is this a sort of linear relationship, where as the resonant frequency gets closer to the string frequency, the sustain time decreases? Jun 4, 2021 at 20:44