AP Physics 1 Unit 7: Simple Harmonic Motion (SHM) Foundations

Defining Simple Harmonic Motion

What simple harmonic motion is

Simple harmonic motion (SHM) is a special kind of back-and-forth (oscillatory) motion where the object’s acceleration is always directed toward an equilibrium position and has a magnitude proportional to how far the object is from equilibrium.

That sentence contains the whole idea, but it helps to unpack it carefully.

1. There is an equilibrium position (often called “zero” displacement). If the object is exactly at equilibrium and you let go, it stays there.
2. If the object is displaced (pulled or pushed away from equilibrium), the system produces a restoring effect that tries to bring it back.
3. In SHM specifically, the restoring effect is “perfectly linear”: doubling the displacement doubles the restoring influence.

The defining mathematical statement of SHM is that acceleration is proportional to displacement and opposite in direction:

$a=-\omega^2 x$

Here:

• $x$ is the displacement from equilibrium (positive on one side, negative on the other).
• $a$ is the acceleration.
• $\omega$ is a constant for that oscillator called the angular frequency (you’ll connect it to period and frequency in the next major section).
• The negative sign encodes “restoring”: if $x$ is positive, $a$ is negative (points back toward equilibrium), and vice versa.

You will also commonly see the SHM condition stated using force rather than acceleration. Since Newton’s second law is $F=ma$, SHM implies a restoring force proportional to displacement:

$F=-kx$

This is Hooke’s law, and it is the force model for an ideal spring.

Why SHM matters

SHM is a “core model” in physics because many real systems behave approximately like SHM when the motion stays small:

• A mass on a spring (very close to ideal SHM)
• A pendulum (approximately SHM for small angles)
• Vibrations of atoms in a solid (modeled as tiny springs)

In AP Physics 1, SHM shows up not just as a topic by itself, but as a place where multiple big ideas intersect:

• Newton’s laws (net force produces acceleration that changes with position)
• Energy (energy continuously trades between kinetic and potential)
• Graphs (how force, acceleration, velocity, and position relate)

A key reason teachers emphasize the “proportional and opposite” definition is that it lets you decide whether a situation truly is SHM rather than just “something that wiggles.” Not every oscillation is simple harmonic.

How SHM works (mechanism and relationships)

The equilibrium idea and sign conventions

In SHM, you choose equilibrium as $x=0$. Then:

• At $x=0$, the restoring force is $0$.
• At a positive displacement $x>0$, the restoring force is negative.
• At a negative displacement $x<0$, the restoring force is positive.

That is exactly what the minus sign in $F=-kx$ (or $a=-\omega^2 x$) is telling you.

A common confusion is to think “restoring” means “always the same direction in space.” It does not. It means “always points toward equilibrium.” The direction flips when the object crosses equilibrium.

Mass–spring system as the clearest SHM example

For a horizontal mass attached to an ideal spring (no friction, spring obeys Hooke’s law), the restoring force is

$F=-kx$

Apply Newton’s second law:

$\sum F=ma$

If the spring is the only horizontal force, then

$-kx=ma$

So

$a=-\frac{k}{m}x$

Comparing this with the SHM form $a=-\omega^2 x$, you identify

$\omega^2=\frac{k}{m}$

and therefore

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

Even if AP Physics 1 does not require calculus-based derivations, this algebraic comparison is important: it shows that SHM is not “magic sinusoid motion” first—it is Newton’s laws plus a linear restoring force.

Simple pendulum as approximate SHM

A simple pendulum is a mass (bob) on a light string of length $L$ swinging under gravity. The restoring effect comes from the component of gravity along the arc.

For small angles (in radians), the pendulum behaves approximately like SHM. The key approximation is that for small $\theta$,

$\sin\theta\approx\theta$

With that approximation, the tangential restoring acceleration becomes proportional to displacement along the arc, which leads to SHM behavior.

Important: pendulum motion is not perfectly SHM for large angles. On AP problems, if they say “small angle” or show a small oscillation, you are expected to use the SHM model.

What SHM looks like in multiple representations

You are often asked to connect the definition of SHM to graphs and qualitative reasoning.

Force–displacement graph

From Hooke’s law, $F=-kx$ is a straight line through the origin with slope $-k$. That straight-line relationship is the “signature” of SHM.

If the restoring force graph is not linear (for example, it curves), then the motion is not strictly SHM. It might still oscillate, but the period may depend on amplitude and the motion won’t match the SHM model perfectly.

Acceleration–displacement graph

From $a=-\omega^2 x$, the $a$ vs. $x$ graph is also a straight line through the origin with slope $-\omega^2$.

This is extremely useful because it lets you identify SHM without even mentioning springs:

• If acceleration is proportional to displacement and opposite direction, it is SHM.
• The slope magnitude tells you $\omega^2$.

Position, velocity, and acceleration relationships (conceptual)

Even without memorizing sinusoidal equations, SHM has consistent “phase” relationships:

• At maximum displacement (turning points), speed is zero but acceleration magnitude is maximum.
• At equilibrium, acceleration is zero but speed magnitude is maximum.

That fits the restoring idea: when you are far from equilibrium, the restoring force is strongest, so the acceleration is largest.

Example 1: Deciding whether a motion is SHM from a force model

A cart is attached to a device that exerts a force given by

$F=-12x$

with $F$ in newtons and $x$ in meters.

Is the motion SHM? If so, what is the spring constant equivalent?

Reasoning: The defining SHM requirement in force form is $F=-kx$. The given force matches that form exactly with a constant of proportionality 12.

So the motion is SHM, and

$k=12\ \text{N/m}$

What to notice: You did not need any information about amplitude, velocity, or energy. The linear restoring force is enough.

Example 2: Using an acceleration–displacement graph idea

Suppose an oscillator has an acceleration related to displacement by

$a=-9x$

with $a$ in $\text{m/s}^2$ and $x$ in $\text{m}$.

Because this matches $a=-\omega^2 x$, you have

$\omega^2=9$

so

$\omega=3\ \text{rad/s}$

You haven’t found period yet (that comes next), but you have already captured the system’s “stiffness per mass” behavior through $\omega$.

Misconceptions to watch for (built into the definition)

• “Any back-and-forth motion is SHM.” Not true. SHM requires a restoring acceleration proportional to displacement. Many oscillations (like a bouncing ball) are periodic but not SHM.
• “The minus sign means acceleration is negative.” Not always. It means acceleration points opposite displacement. If $x$ is negative, then $a$ is positive.
• “Pendulums are always SHM.” Only approximately, and only for small angles.

Exam Focus

• Typical question patterns
  • You’re given a force law or a graph (such as $F$ vs. $x$ or $a$ vs. $x$) and asked whether the motion is SHM and what parameter (like $k$ or $\omega^2$) the slope represents.
  • You’re asked conceptual questions about where speed/acceleration are largest or zero (turning points vs. equilibrium).
  • You’re asked to justify why a pendulum is approximately SHM only for small angles.
• Common mistakes
  • Treating a nonlinear restoring force as SHM just because the motion repeats.
  • Mixing up displacement direction and acceleration direction (forgetting that acceleration points toward equilibrium).
  • Assuming the pendulum SHM model applies at large amplitude without a “small-angle” cue.

Frequency and Period of SHM

What period and frequency mean

For any repeating motion, two closely related quantities describe “how fast it repeats”:

• Period $T$: the time for one complete cycle (one full back-and-forth), measured in seconds.
• Frequency $f$: the number of cycles per second, measured in hertz, where $1\ \text{Hz}=1\ \text{s}^{-1}$.

Their relationship is:

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

This is not specific to SHM—it’s true for any periodic motion. But SHM is especially nice because its period and frequency are often constant (in the ideal model) and can be written in terms of physical parameters like mass, spring constant, length, and gravity.

Why period and frequency matter in SHM

In SHM problems, you often want to predict how changing the system changes the motion:

• If you increase the mass on a spring, does it oscillate more slowly or more quickly?
• If you use a stiffer spring, what happens to the period?
• Does amplitude affect the period?

SHM provides clean, testable answers for ideal systems. Those answers become powerful tools on AP-style questions, especially when the problem is more conceptual than computational.

Angular frequency and notation connections

In oscillations, it is common to use angular frequency $\omega$ (units rad/s). It connects to $f$ and $T$ by:

$\omega=2\pi f$

and

$\omega=\frac{2\pi}{T}$

This can feel like “extra notation,” but it matters because the SHM defining equation uses $\omega$:

$a=-\omega^2 x$

So $\omega$ is the bridge between “how the force depends on position” and “how fast the motion repeats.”

Key SHM period formulas (AP Physics 1)

AP Physics 1 focuses on two canonical SHM systems.

1) Mass–spring oscillator

For an ideal mass–spring system,

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

and therefore

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Variables:

• $m$ is the mass (kg)
• $k$ is the spring constant (N/m)

How to interpret the dependencies (conceptually):

• Larger $m$ means more inertia, so it accelerates less for the same force, making oscillations slower and $T$ larger.
• Larger $k$ means a stronger restoring force for the same displacement, making oscillations faster and $T$ smaller.

A major SHM takeaway is that, in the ideal model, the mass–spring period does not depend on amplitude. Pull it a little or a lot (without stretching the spring beyond the linear Hooke’s law region), and the period is the same.

2) Simple pendulum (small-angle approximation)

For a simple pendulum oscillating with small amplitude,

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

and

$f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}$

Variables:

• $L$ is the pendulum length (m)
• $g$ is gravitational field strength (approximately $9.8\ \text{m/s}^2$ near Earth’s surface)

How to interpret the dependencies:

• Larger $L$ gives a slower swing (larger $T$).
• Larger $g$ gives a faster swing (smaller $T$).
• In the small-angle SHM model, the period does not depend on the bob’s mass.

Also: in real pendulums, larger amplitudes make the period slightly larger. On AP Physics 1, that effect is usually ignored unless they explicitly discuss “non-small angles” or deviations from SHM.

Notation reference (to keep symbols straight)

How you use these relationships in problems

A productive way to approach SHM timing questions is:

1. Identify the oscillator type (spring vs. pendulum).
2. Decide what physical parameter is changing (mass, spring constant, length, gravity).
3. Use proportional reasoning when possible (often faster than plugging numbers).
4. If needed, compute $T$, then get $f$ from $f=1/T$ (or vice versa).

Proportional reasoning examples (very AP-style)

From $T=2\pi\sqrt{m/k}$:

• If $m$ increases by a factor of 4, then $T$ increases by a factor of $\sqrt{4}=2$.
• If $k$ increases by a factor of 9, then $T$ decreases by a factor of $\sqrt{9}=3$.

From $T=2\pi\sqrt{L/g}$:

• If $L$ increases by a factor of 4, then $T$ increases by a factor of 2.
• If $g$ decreases by a factor of 4, then $T$ increases by a factor of 2.

A common mistake is to think period scales directly with the parameter (like “double the mass, double the period”). The square root is the key feature.

Example 1: Finding a spring constant from a measured period

A cart of mass $m=0.50\ \text{kg}$ oscillates on a horizontal spring with period $T=1.40\ \text{s}$. Find the spring constant $k$.

Step 1: Choose the correct model. This is a mass–spring oscillator, so

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

Step 2: Solve for $k$.

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

Square both sides:

$\left(\frac{T}{2\pi}\right)^2=\frac{m}{k}$

Rearrange:

$k=\frac{m}{\left(\frac{T}{2\pi}\right)^2}$

Step 3: Substitute values.

$k=\frac{0.50}{\left(\frac{1.40}{2\pi}\right)^2}$

Compute the intermediate ratio:

$\frac{1.40}{2\pi}\approx0.223$

Square it:

$0.223^2\approx0.0497$

Now:

$k\approx\frac{0.50}{0.0497}\approx10\ \text{N/m}$

Interpretation: A relatively small $k$ means a “soft” spring, which matches a longer period.

Example 2: Comparing pendulum periods without heavy computation

Pendulum A has length $L$. Pendulum B has length $4L$. Both swing with small amplitude at the same location. How do their periods compare?

Use

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

Let $T_A=2\pi\sqrt{\frac{L}{g}}$.

Then

$T_B=2\pi\sqrt{\frac{4L}{g}}=2\pi\sqrt{4}\sqrt{\frac{L}{g}}$

Since $\sqrt{4}=2$,

$T_B=2T_A$

Meaning: Quadrupling length doubles the period.

Connecting back to the definition of SHM

It’s worth explicitly linking this timing information to the SHM defining statement $a=-\omega^2 x$.

From $\omega=2\pi/T$, a larger $\omega$ means a smaller $T$ (faster oscillations). So if a system has a larger restoring “strength per inertia,” it should have a larger $\omega$.

That’s exactly what you see:

• For a spring, $\omega=\sqrt{k/m}$ (bigger $k$ or smaller $m$ increases $\omega$).
• For a pendulum (small-angle), $\omega=\sqrt{g/L}$ (bigger $g$ or smaller $L$ increases $\omega$).

This is why the SHM condition is so powerful: once you know the proportionality constant between acceleration and displacement, you can infer timing behavior.

Common misconceptions specific to period and frequency

• Mixing up period and frequency units. Period is seconds per cycle; frequency is cycles per second. If your answer for $T$ comes out in $\text{s}^{-1}$, something is inverted.
• Forgetting the square root. The dependence on $m$, $k$, $L$, and $g$ is through a square root.
• Assuming amplitude changes the period in ideal SHM. For an ideal spring and a small-angle pendulum model, amplitude does not affect $T$. If a question says the amplitude changes and asks what happens to $T$, the intended answer is often “no change,” provided the SHM conditions still hold.

Exam Focus

• Typical question patterns
  • Compute $T$ or $f$ for a mass–spring or small-angle pendulum using the standard formulas, sometimes after finding $k$ from a graph or from energy information given elsewhere.
  • Ratio/proportional reasoning questions: “If $m$ doubles, what happens to $T$?” or “If $L$ increases by a factor of 9, what happens to $f$?”
  • Convert between $T$, $f$, and $\omega$, especially when $\omega$ is obtained from an $a$ vs. $x$ relationship.
• Common mistakes
  • Using the pendulum formula without the small-angle assumption (or applying it to a system that is not a simple pendulum).
  • Treating $\omega$ as the same as $f$ (they differ by a factor of $2\pi$).
  • Plugging into $f=1/T$ backward (reporting frequency when asked for period, or vice versa), often detectable by unreasonable magnitude or units.