AP Physics 1 : Introduction: Simple Harmonic Motion Study Notes

Kinematics of Simple Harmonic Motion

As you can see from the uniform circular motion graph above, the $y$-position of an object in simple harmonic motion as a function of time is the function sine or cosine of the angle around the circle, depending on where you declare the starting position to be. From calculus, the general equations of periodic or oscillating motion are therefore:
position: $\quad x=A \cos (\omega t+\phi)$
velocity: $\quad v=-A \omega \sin (\omega t+\phi)=\frac{d x}{d t}$
acceleration: $a=-A \omega^2 \cos (\omega t+\phi)=\frac{d v}{d t}=\frac{d^2 x}{d t^2}$
Where:
$\begin{aligned} x & =\text { displacement from the equilibrium point } x=0 \\ A & =\text { amplitude (maximum displacement) } \\ \omega & =\text { angular frequency } \\ t & =\text { time } \\ \phi & =\text { phase angle (offset) }\end{aligned}$

The phase angle $\phi$ or offset is the position where the cycle starts, relative to the equilibrium (zero) point.
Because many simple harmonic motion problems (including $A P^{\oplus}$ problems) are given in terms of the frequency of oscillation (number of oscillations per second), we can multiply the angular frequency by $2 \pi$ to use $f$ instead of $\omega$, i.e., $\omega=2 \pi f$.
On the $A P^{\oplus}$ formula sheet, $\phi$ is assumed to be zero, resulting in the following version of the position equation:
$
x=A \cos (2 \pi f t)
$
You are expected to be able to understand and use the position equation above, but simple harmonic problems that involve velocity and acceleration equations are beyond the scope of this course and have not been seen on the $A^{\oplus}$ exam.
* The derivatives $\frac{d x}{d t}, \frac{d v}{d t}$, and $\frac{d^2 x}{d t^2}$ are from calculus. The velocity and acceleration equations are beyond the scope of the $A P^{\oplus}$ Physics course.