SOLUTION: Find a model for a simple harmonic motion that has initial displacement (t=0) of 0 inches, an amplitude of 5 inches, and a frequency 4/3 cycles per second.

Click here to see ALL problems on Trigonometry-basics

Question 1197654: Find a model for a simple harmonic motion that has initial displacement (t=0) of 0 inches, an amplitude of 5 inches, and a frequency 4/3 cycles per second.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

A trig template I'll be using is

where,
|A| = amplitude
B = used to determine the period T
C = phase shift
D = vertical shift and midline

The initial displacement of 0 inches tells us we don't have any vertical shifting going on. We go with D = 0.

The amplitude of 5 means A = 5.

We won't have any phase shifts here, so use C = 0
A useful property of the sine function is that its initial value, aka y intercept, is 0.

Now to find the value of B.

Frequency and the period are connected topics.
It turns out that
frequency = 1/period
and
period = 1/frequency

This means that we can apply the reciprocal to the frequency of "4/3 cycles per second" to get a period of "3/4 seconds per cycle".
Take note that the units swap positions.
The length of each cycle is 3/4 = 0.75 seconds.

Use the period T = 0.75 to find:
B = 2pi/T
B = 2pi/0.75
B = 8pi/3

Summary:
A = 5
B = 8pi/3
C = 0
D = 0

So we go from

to this

which simplifies to

which is a final answer. I say "a" instead of "the" because we could use cosine and have a nonzero phase shift involved, but it's easier to go with sine in my opinion.
There are infinitely many possible phase shifts with either sine or cosine, but this one involves a phase shift of 0 which is easiest.

I recommend to use graphing software such as Desmos or GeoGebra or whichever tool you like to confirm the answer.

One way to measure the period is to look at the horizontal distance from peak to neighboring peak.

Side note: a related concept to period is wave length.