IB Physics - 40 multiple choice question paper with working - Version 2 - Part 4

This is the SECOND mock exam work-through of an IB level multiple choice exam. Part 4 of 4.
NOTE: THESE QUESTIONS ARE NOT ALL FROM IB SO YOU MIGHT NEED A CALCULATOR. Remember, IB paper 1 does not allow a calculator normally.
The original un-marked question paper is here - drive.google.com/open?id=11AT...
Full playlist with all parts here: • IB 40 mark multiple ch...

[@user-jh3kz7dp2z]

To my understanding, the period measures the time it would take for one complete oscillation. However, this does not mean that the angular velocity throughout the mass' travel is constant. This only allows the average angular velocity of the mass to be calculated. Due to the gravitational field strength that acts vertically downwards, it would be impossible for the net acceleration of the mass to be directly pointing towards the center of the circle at all times, or perpendicularly to the direction of the mass' motion.
For example, take the point P. The mass at P has an instantaneous velocity directly upwards and will have a force from the tension of the string pointing directly to the center of the circle. However, there is also an unbalanced force vertically downward, and the superposition principle would indicate that the net force acting on the mass at point P will be diagonally downwards. Therefore, since the condition for uniform circular motion is not actually satisfied by the system, it is wrong to assume that the angular velocity (or speed) will be constant throughout the mass' trajectory. Thus, the kinetic energy cannot possibly be the same at Q and P.
Hopefully, I haven't made any mistakes in my logic here, I'm still an IB student at the time of this comment so please correct me if there are flaws in my argument.