
Samuel B. answered 03/25/24
Precalculus, Geometry, and Algebra 2 teacher for 5 years
There are three aspects of this question that might be a bit tricky. I'll try to explain the answer in pieces so you can jump into your own discovery once you can see the question clearly.
First, there's a bit of a perspective shift because the whole problem is vertical instead of horizontal like we're used to with sine waves. We could try to visualize our sine wave vertically, but it's probably easiest to think of the input as time and the output as a horizontal swing (measured in degrees left or right).
Secondly, the degrees mentioned in the problem have nothing to do with the input of the sine function, which is what we're used to. It might be easy to get confused here, but don't confuse the degree measurements with inputs for our function.
Finally, we have to modify the sine equation to model the situation.
f(t) = Asin[B(x-C)] + D
where A is the amplitude, B is related to the period, C is the phase shift, and D is the vertical shift.
If the pendulum starts at rest at the center of it's swing and then begins by swinging to the right, it isn't being shifted along the input or output, so we can simplify the equation to: f(t) = Asin(Bx)
What is the amplitude in this question?
What is the period?
Amplitude: the pendulum swings from a maximum of 12 degrees to a minimum of -12 degrees, so its amplitude is 12.
Period: the pendulum has a period of 2 seconds. The period = 2π/B so B = 2π/period. (If you want to use degrees, it would be 360/period)
Our final equation will look like this: f(t) = 12sin(πt)
Plug in some possible values (making sure to use the radian mode on your calculator to see how this would align with the information that you're given).