Relationship of Induction Motor Frequencies
$$f = \frac {P\ N_S}{120}$$

\$f\$ = source frequency in Hz.  Your electrical stator frequency.
P = # of three-phase pole pairs (2, 4, 6, etc.).
\$N_S\$ = synchronous speed in rpm.

Synchronous speed at 60Hz will be:

3600 rpm for a 2-pole machine (2-pole/phase = 6 total poles).
1800 rpm for a 4-pole machine (12 total).


If we take a motor nameplate to illustrate the mechanical.
$$f_r = \frac {P\ N}{120}= \frac {4\ poles\times 1765\ rpm}{120} = 58.83Hz$$

\$f_r\$ = mechanical rotor frequency in Hz.
N = rotor speed in rpm.

\$f\$ and \$f_r\$ are related by slip.  For a motor to develop torque, \$f_r &lt; f\$ and \$N &lt; N_s\$.
$$ s = \frac {N_S - N}{N_S} = \frac {1800\ rpm - 1765\ rpm}{1800\ rpm} = 0.01944$$
Slip at rated motor load is 1.94%.  Slip and rotor speed actual depend on the load being driven with most motors having a slip between 0.5% and 5%.  Heavier the load, the larger the slip.
This is a 4-pole motor because:
$$P = \frac {120\ f}{N} = \frac {120 \times 60\ Hz}{1765\ rpm} = 4.079$$
Rounding down gives us 4-poles/phase.