Not sure if I can give you an audio example easily, but numerically I will give a go.
If we process the pixels (or samples, but I will refer to pixels) from left to right with a low pass filter as I described with F=0.9
pixel number 1 (n1) = n1 * 0.9 + n0 * 0.1
n2 = n2 * 0.9 + n1 * 0.1
n3 = n3 * 0.9 + n2 * 0.1
etc…
but the effects of previous calculations accumulate so if effectively becomes:
n1 = n1 * 0.9
n2 = n2 * 0.9 + (n1 * 0.9) * 0.1
n3 = n3 * 0.9 + (n2 * 0.9 + (n1 * 0.9) * 0.1) * 0.1
n4 = n4 * 0.9 + ((n2 * 0.9 + (n1 * 0.9) * 0.1)* 0.1) * 0.1
or (if I have done my maths right)
n1 = n1 * 0.9
n2 = n2 * 0.9 + n1 * 0.09
n3 = n3 * 0.9 + n2 * 0.09 + n1 * 0.009
n4 = n4 * 0.9 + n3 * 0.09 + n2 * 0.009 + n1 * 0.0009
So each pixel is the average of all the preceding pixels weighted by distance from the current pixel.
By adjusting the proportion of current and previous samples in the calculation (F) you can determine how much effect distant pixels have on the current pixel. Similar to a Gaussian blur radius.
High frequencies in images is related to what photographers call sharpness. That is, difference in adjacent pixels. Mid frequencies are related to ‘local contrast’ that is a difference in pixel values over several pixels.
A small blur radius reduces high frequencies/sharpness, and larger blur radius affects mid-frequencies/local-contrast. So cut-off frequency is comparable to blur radius.