Question

An object undergoing simple harmonic motion takes 0.23 s to travel from one point of zero velocity to the next such point. The distance between those points is 33 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.

Answer 1

The object will be stationary at two points in each cycle. The "positive" displacement and the "negative" displacement.
so the time of 0.23 seconds is only half the time of the whole cycle.
The whole cycle must be 2 * 0.23 s= .46 sec

The frequency is how many cycles per second. If it takes 0.5 s ( a half a second for ONE cycle) how many of these can you get in a second?
f = 1/T = 1/ 0.46= 2.174 Hz

And if the amplitude is defined as the furthest distance from the mean or midpoint of the oscillation then how does this compare to the distance between the two extremes?

Clearly it must be half the distance between the extremes i.e 33 / 2 c=16.5 cm