The voltage divider, with resistors of equal value, reduces the 5v source to a 2.5v “base” at the junction of the two 47K ohm resistors – let’s call them R1 (nearer positive) and R2 (nearer ground). However, as soon as the 47K resistor straddles the piezo – let’s call that resistor Rin, it forms a parallel circuit with R2, reducing both resistances from 47K to 23.5K.
R2||Rin = (R2 * Rin) / (R2 + Rin) = (47K * 47K) / (47K + 47K) = 23.5K

So now the anticipated 2.5v ends up as 1.66v.
Vb = (Vs * R2||Rin) / (R1 + R2||Rin) = (5v * 23.5K) / (47K + 23.5K) = 1.66v

That is still enough to keep the piezo’s +.010v to -.010v excursions from giving any pin a negative value. Arduino pins can either source (put out conventional current) or sink (take it in), but not both (refer to OUTPUT and INPUT). The analog pins measure fluctuating current (okay), not alternating current (danger!).

So to be martinet and get back the 2.5v junction, here’s how to do it (keeping in mind that R2 = Rin).
First, let’s find R2, given R1:
R1 = (R2 * Rin) / (R2 + Rin)
47K = (R2 * R2) / (R2 + R2)
47K = (R2 squared) / (2R2)
47K * 2R2 = R2 squared
94KR2 = R2 squared
94K = R2, which also equals Rin

Then, find the value for R2 and Rin in parallel:
R2||Rin = (R2 * Rin) / (R2 + Rin) = (94K * 94K) / (94K + 94K) = 47K

So now the anticipated 2.5v ends up as 2.5v.
Vb = (Vs * R2||Rin) / (R1 + R2||Rin) = (5v * 47K) / (47K + 47K) = 2.5v