Arsthbt

What advantage does ultrasound have over sound between 20-20000Hz that it is used in medical imaging over sound in that frequency range?

I think the simple answer here is resolution.

Generally when imaging with waves (including light) the limit to resolution is a length that is similar to the wavelength, $lambda$.

If $f$ is the frequency and $c$ is the speed of the wave then the wavelength is given by

$$lambda = c over f $$

so the higher we make $f$ the smaller $lambda$ becomes and the better the resolution and the more detail we can see in scans....

The speed of sound in water is ~1500 m/s and with say 1.5 MHz = 1 500 000 Hz frequency we calculate

$$lambda = 0.001 rm m = 1 rm mm$$

At 20000 Hz $lambda = 75$ mm

Higher frequency provides higher resolution.

Using waves for imaging, you want to distinguish points in the specimen. Consider the simplest case: you'd like to distinguish 2 points.

The waves are coming from behind the specimen passing through the points or the waves are reflected on the points. Considering the phase relationship of the waves at the detector (i.e. investigating interference) because of the diffraction on the 1 point, even with focusing, one cannot get a single point on the detector. Take a look on the single slit diffraction and on the 2D version, on the Airy disk.

On the focused image you will get concentric circles. For $alpha$, the angular distance between the two most visible, inner ones one gets
$$lambda /left( 2d right) = sin left( alpha right) approx alpha ,$$
where $lambda$ corresponds to the wavelength and $d$ is the size of point.

Two points are called resolvable if their picture on the detector are distinguishable, for example, if their Airy-disk are non-overlapping. That means, it will limit how close the distinguishable points can be. With a given pixel and size properties of the detector you can only modify $lambda$, and as it can be seen, lowering $lambda$ makes it possible to resolve even closer objects.

And as it is known,
$$lambda = c/f,$$
where $c$ is the speed of the wave and $f$ is frequency, the higher the frequency, the smaller the wavelength is, increasing the achievable resolution.