Is there a direct relationship between frequency and density of a liquid in the singing wine glass experiment? I know that as density increases,...

The singing wine glass is a special case of the more general phenomenon of a resonance tube. It's easier to model if we assume that the resonance tube is a perfect cylinder, so it isn't varying in width.

Let's suppose we have a resonance tube that is a perfect cylinder, partially filled with water, and we set it to vibrating; what pitch (i.e. frequency) will be the resulting sound?

The density of the liquid is actually relatively unimportant, so long as it is much higher than the density of air. The surface of the liquid constrains the vibration to have a node, which means that the effective length of the resonance tube we care about is the height of the air column between the surface of the liquid and the top of the glass.

If we call that distance `L ` and the allowable wavelengths `lambda` , in order to have a node at the end of the air column, the length of the column must therefore be an odd harmonic of `lambda/4` , i.e.:

`L = 1/4 (2n+1) lambda`

for some nonnegative integer `n` . Given the speed of sound in air, this is the primary determinant of the frequency of the resulting sound.

However, I don't want to leave you hanging completely on the density part.

The velocity of sound `v` as it passes through a fluid---and in this case therefore the frequency, since wavelength is constrained here, is inversely proportional to the square root of the density `rho` , with an extra proportionality term `C` that is determined by some subtler elastic properties of the fluid:

`v = sqrt{C/rho}`

You can also think of this as making the kinetic energy per unit volume constant:
`K/V = 1/2 rho v^2 = 1/2 rho sqrt{C/rho}^2 = 1/2 sqrt{C}`