The synchronous speed of an AC motor is determined by the supply frequency and the number of poles in the machine. The formula for synchronous speed ($N_s$) is:
$N_s = \frac{120 \times f}{P}$
Where:
Given initial parameters:
Calculate the initial synchronous speed ($N_{s1}$):
$N_{s1} = \frac{120 \times 50}{4} = \frac{6000}{4} = 1500 \text{ rpm}$
The problem states that the frequency and pole number are halved. The load torque change does not affect the synchronous speed itself.
New parameters:
Calculate the new synchronous speed ($N_{s2}$):
$N_{s2} = \frac{120 \times f_2}{P_2} = \frac{120 \times 25}{2} = \frac{3000}{2} = 1500 \text{ rpm}$
After halving the frequency and the number of poles, the new synchronous speed of the motor remains 1500 rpm. For a synchronous motor, the operating speed is the synchronous speed.