AP Physics 1: Simple Harmonic Motion

Simple harmonic motion (SHM) is the most important model for oscillations in AP Physics 1 because it connects a clear physical idea, a restoring force that pulls a system back toward equilibrium, to a precise mathematical description of periodic motion. Springs, pendulums (at small angles), and many vibrating systems behave approximately like SHM. Once you understand the defining force law and how energy moves between kinetic and potential forms, you can analyze period, frequency, amplitude, and phase in a consistent way.

What counts as simple harmonic motion?

An object is in simple harmonic motion when the net force is proportional to its displacement from equilibrium and points back toward equilibrium:

$$F=-kx$$

Here, $x$ is displacement from equilibrium, $k$ is a positive constant that measures the “stiffness” of the system, and the negative sign indicates the restoring direction. Combining Newton’s second law with this force law gives:

$$m\frac{d^{2}x}{d{t}^2}=-kx\Rightarrow \frac{d^{2}x}{d{t}^2}=-\frac{k}{m}x$$

The acceleration is also proportional to displacement and opposite in direction:

$$a=-{\omega }^{2}x$$

where the angular frequency $\omega$ is:

$$\omega =\sqrt{\frac{k}{m}}$$

This is the core of SHM: the motion repeats because the system continually accelerates back toward equilibrium, overshoots due to inertia, and repeats the cycle.

Describing periodic motion: amplitude, period, and frequency

Oscillations repeat in time. The period $T$ is the time for one full cycle. The frequency $f$ is the number of cycles per second:

$$f=\frac{1}{T}$$

For a mass-spring system that behaves ideally, the period is:

$$T=2\pi \sqrt{\frac{m}{k}}$$

Several important conclusions follow:

• Increasing mass $m$ increases the period. The oscillator is “slower.”
• Increasing spring constant $k$ decreases the period. A stiffer spring oscillates faster.
• In ideal SHM, period does not depend on amplitude. A bigger oscillation has more energy, but it does not take longer just because it is larger.

The amplitude $A$ is the maximum displacement from equilibrium. It is set by how the system is released (how far it is pulled or pushed and with what initial speed), not by $m$ or $k$ alone.

A common position model for SHM is:

$$x(t)=A\cos (\omega t+\varphi )$$

where $\varphi$ is a phase constant that depends on initial conditions. You do not need calculus to use this form conceptually in AP Physics 1, but it helps you reason about where the object is in its cycle at a given time.

Connecting SHM to uniform circular motion

A powerful way to visualize SHM is to relate it to uniform circular motion. Imagine a point moving around a circle of radius $A$ at constant angular speed $\omega$. The projection of that point onto a diameter (say, the x-axis) moves back and forth in SHM.

This connection explains why cosine and sine show up naturally and why $\omega$ is called angular frequency. It also clarifies phase: phase is literally the angle (in radians) describing where the rotating point is on its circular path. The “shadow” on the axis is the oscillator’s position.

This is not just a mathematical trick. It provides intuition for key SHM facts:

• Maximum speed occurs at equilibrium ($x=0$), where the “shadow” crosses the center fastest.
• Acceleration is largest in magnitude at the endpoints ($x=\pm A$), where the projection has the greatest displacement and must be pulled back most strongly.

Springs: force, acceleration, and motion

For a horizontal mass on a spring (no friction), the restoring force is simply $F_s=-kx$. At different points in the motion:

• At $x=\pm A$: velocity is zero, acceleration has maximum magnitude $a=\mp {\omega }^{2}A$ toward the center.
• At $x=0$: force and acceleration are zero, speed is maximum.

If the oscillator is vertical, gravity shifts the equilibrium position but does not change the period. The spring stretches until the spring force balances weight, $k\Delta x=mg$. Oscillations occur about this new equilibrium. The dynamics around equilibrium still follow $F=-kx$ with $x$ measured from equilibrium, leading to the same $T=2\pi \sqrt{m/k}$.

Pendulums: when they approximate SHM

A simple pendulum (mass on a light string of length $L$) is not exactly SHM for all angles. The restoring force component along the arc is:

$$F_t=-mg\sin \theta$$

For small angles (in radians), $\sin \theta \approx \theta$, which makes the restoring “force” proportional to displacement along the arc. Under the small-angle approximation, the motion is approximately SHM with period:

$$T=2\pi \sqrt{\frac{L}{g}}$$

Key implications:

• A longer pendulum has a longer period.
• A stronger gravitational field (larger $g$) makes the period shorter.
• For small angles, the period is approximately independent of amplitude. At larger angles, the approximation breaks down and the period increases slightly.

Energy transformations in SHM

One of the cleanest ways to analyze SHM in AP Physics 1 is with energy. In an ideal oscillator, the total mechanical energy is constant:

$$E=K+U=\text{constant}$$

Mass-spring energy

For a spring:

• Spring potential energy: $$U_s=\frac{1}{2}k{x}^2$$
• Kinetic energy: $$K=\frac{1}{2}m{v}^2$$
• Total energy: $$E=\frac{1}{2}k{A}^2$$ (since at maximum displacement, $x=A$ and $v=0$)

As the mass moves:

• At the endpoints ($x=\pm A$): $U_s$ is maximum, $K=0$.
• At equilibrium ($x=0$): $U_s=0$, $K$ is maximum.

This energy picture lets you solve for speed at a given position without using kinematics:

$$\frac{1}{2}k{A}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{\frac{k}{m}(A^2-x^2)}$$

So speed depends on where you are in the oscillation: largest at the center, zero at the turning points.

Pendulum energy (small-angle SHM)

For a pendulum, gravitational potential energy converts to kinetic energy. Taking the lowest point as $U_g=0$, the total energy depends on the release height. In SHM problems, you often track energy between the endpoints (maximum height, zero speed) and the bottom (maximum speed).

Even when you use the SHM period formula for a pendulum, energy methods remain valid without needing the small-angle approximation, since conservation of energy is more general. The approximation mainly affects whether the motion matches the sinusoidal SHM model and whether $T=2\pi \sqrt{L/g}$ holds.

Practical problem-solving habits for AP Physics 1

Identify the equilibrium and define displacement correctly

In SHM, $x$ must be measured from equilibrium, not from an arbitrary reference point. For a vertical spring, equilibrium is below the unstretched length due to gravity. For a pendulum, equilibrium is at the lowest point.

Distinguish period from amplitude and energy

Amplitude affects energy: $E=\frac{1}{2}k{A}^2$ for a spring. But in ideal SHM, it does not affect period. On exams, a common trap is assuming a larger amplitude means a longer period.

Use energy when forces are messy

If you want speed at a position, conservation of energy is often faster and less error-prone than trying to use acceleration and kinematics.

Check limiting cases

If $m$ increases, the period of a spring oscillator should increase. If $k$ increases, period should decrease. If a pendulum gets longer, period should increase. These checks catch algebra mistakes quickly.

Why SHM matters beyond the unit

SHM is the foundation for understanding resonance, waves, vibrations in structures, and many real systems that are only approximately harmonic. In AP Physics 1, it also serves as a unifying model: the same ideas about restoring forces, periodic motion, and energy transformations apply across springs and pendulums, even though the physical setups look different. Mastering SHM means you can move smoothly between force-based reasoning, energy reasoning, and the mathematics of periodic motion.