Simple Harmonic Motion Examples to Know for AP Physics 1 (2025)

What You Need to Know

Simple Harmonic Motion (SHM) is the motion you get when a system has a restoring influence that is proportional to displacement from equilibrium and points back toward equilibrium:

$F_{\text{net}} = -k_{\text{eff}} x \quad\Rightarrow\quad a = -\omega^2 x$

Why it matters in AP Physics 1: SHM shows up everywhere—springs, pendulums, energy graphs, period questions, and “is this SHM?” conceptual traps.

The big idea (how you recognize SHM)

A system undergoes SHM when (for small displacements):

• There is a stable equilibrium position.
• The net force (or net torque) behaves like $-\text{constant}\times (\text{displacement from equilibrium})$.

Then the motion is sinusoidal with:

$\omega = \sqrt{\frac{k_{\text{eff}}}{m}}, \quad T = \frac{2\pi}{\omega}$

Critical reminder: Displacement $x$ in SHM is measured from equilibrium, not necessarily from where the spring is unstretched.

The “examples you must know” list

These are the high-yield SHM systems for AP Physics 1:

• Horizontal mass–spring (classic)
• Vertical mass–spring (equilibrium shift!)
• Simple pendulum (small-angle) (uses $\sin\theta\approx\theta$)
• Springs in series/parallel leading to an effective spring constant $k_{\text{eff}}$
• Small oscillations about equilibrium (e.g., block in a shallow bowl / track approximated as parabolic, or any situation where you can linearize the restoring force)

Step-by-Step Breakdown

Use this every time you see “oscillates,” “vibrates,” “small displacement,” or “period.”

A. Decide if it’s SHM (force test)

1. Find equilibrium: where $F_{\text{net}}=0$ (or $\tau_{\text{net}}=0$).
2. Define a displacement variable from equilibrium:
   • Springs: $x$ (meters)
   • Pendulum: either angle $\theta$ (radians) or arc length $s=L\theta$
3. Write the net restoring influence in terms of that displacement.
4. Check if it’s (or can be approximated as):

$F_{\text{net}}\approx -k_{\text{eff}}x \quad \text{or} \quad \tau_{\text{net}}\approx -\kappa\,\theta$

If yes, it’s SHM with:

$\omega = \sqrt{\frac{k_{\text{eff}}}{m}} \quad \text{(translational)} \qquad \text{or} \qquad \omega = \sqrt{\frac{\kappa}{I}} \quad \text{(rotational)}$

Decision point: If you see $\sin\theta$, SHM only holds for **small angles** where $\sin\theta\approx\theta$ (with $\theta$ in radians).

B. Find the period quickly (pattern recognition)

1. If it’s a mass–spring system:

$T = 2\pi\sqrt{\frac{m}{k_{\text{eff}}}}$

2. If it’s a simple pendulum (small-angle):

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

3. If multiple springs are involved:
   • Compute $k_{\text{eff}}$ (series/parallel) first.

C. Use energy to get speed at a position (fast on exams)

For ideal SHM (no damping): total mechanical energy is constant.

Mass–spring:

$E = \frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

So:

$v = \omega\sqrt{A^2-x^2}$

Pendulum (small-angle): you can use energy too, but be careful whether you’re using angle or arc length.

Mini worked method (annotated)

A block of mass $m$ attached to a spring oscillates. You’re told “amplitude $A$” and asked for $v$ when displacement is $x$.

1. Recognize mass–spring SHM: $\omega=\sqrt{k/m}$.
2. Use SHM speed relation:

$v = \omega\sqrt{A^2-x^2}$

3. If they ask “maximum speed,” set $x=0$:

$v_{\max}=\omega A$

Key Formulas, Rules & Facts

SHM core relationships (know cold)

Spring systems you must recognize

Simple pendulum (small-angle) essentials

“Small oscillations about equilibrium” (linearization idea)

If for small displacements the restoring force can be approximated as:

$F(x)\approx -\left(\frac{dF}{dx}\bigg|_{x=0}\right)x = -k_{\text{eff}}x$

then it behaves like SHM with:

$\omega=\sqrt{\frac{k_{\text{eff}}}{m}}$

This shows up when they give you a weird force function but tell you “small oscillations.”

Examples & Applications

Example 1: Classic horizontal mass–spring (most common)

A $0.50\,\text{kg}$ block on a frictionless surface is attached to a spring with $k=200\,\text{N/m}$. It’s pulled to amplitude $A=0.10\,\text{m}$ and released.

Setup & key insight: Identify $\omega$ and then use speed/energy relationships.

• $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.50}}=\sqrt{400}=20\,\text{rad/s}$
• Period:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{20}=0.314\,\text{s}$

• Max speed (at equilibrium):

$v_{\max}=\omega A=(20)(0.10)=2.0\,\text{m/s}$

• Total energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(200)(0.10)^2=1.0\,\text{J}$

Exam variation: They may give you a graph of $x(t)$; you read $A$ from peak and $T$ from peak-to-peak, then compute $\omega$.

Example 2: Vertical mass–spring (equilibrium shift trap)

A mass $m$ hangs from a vertical spring $k$ and oscillates with small amplitude.

Setup & key insight: Gravity changes the equilibrium position but does not change the period.

1. Find equilibrium extension $x_0$ from the spring’s natural length:

$kx_0=mg \Rightarrow x_0=\frac{mg}{k}$

2. Define displacement from equilibrium: $y=x-x_0$.
3. Net force (taking downward positive) near equilibrium:

$F_{\text{net}}=mg-kx=mg-k(y+x_0)=mg-ky-kx_0=-ky$

So it’s SHM in $y$ with:

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

Exam variation: They’ll give “stretched an additional $\Delta x$ then released.” Your amplitude is $A=\Delta x$ about equilibrium, not from the unstretched spring.

Example 3: Simple pendulum (small-angle) + speed at bottom

A simple pendulum of length $L=1.2\,\text{m}$ is released from rest at a small angle $\theta_0=0.20\,\text{rad}$.

Setup & key insight: Period depends only on $L$ and $g$ (small-angle), and speed at the bottom comes from converting gravitational potential to kinetic.

• Period:

$T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{1.2}{9.8}}\approx2.20\text{ s}$

• Approximate maximum speed at bottom (use energy): height drop

$\Delta h=L(1-\cos\theta_0)$

For small $\theta_0$, you can use $\cos\theta\approx 1-\frac{\theta^2}{2}$, so:

$\Delta h\approx L\left(1-\left(1-\frac{\theta_0^2}{2}\right)\right)=\frac{L\theta_0^2}{2}$

Energy: $mg\Delta h=\frac{1}{2}mv_{\max}^2$:

$v_{\max}\approx \sqrt{2g\left(\frac{L\theta_0^2}{2}\right)}=\theta_0\sqrt{gL}$

Plug in:

$v_{\max}\approx (0.20)\sqrt{(9.8)(1.2)}\approx 0.69\,\text{m/s}$

Exam variation: They may ask “is it SHM?” You must say: only approximately, for small angles where $\sin\theta\approx\theta$.

Example 4: Two springs (effective spring constant)

A mass $m$ is attached to two identical springs (each $k$) in parallel, oscillating horizontally.

Setup & key insight: In parallel, both springs stretch the same $x$, so forces add.

• Restoring force: $F=-kx-kx=-2kx$ so $k_{\text{eff}}=2k$.
• Period:

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

Series version (common twist): If the two identical springs are in series, then $k_{\text{eff}}=\frac{k}{2}$ and:

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

Common Mistakes & Traps

1. Using displacement from the wrong reference point (vertical spring)

   • What goes wrong: you plug $x$ measured from the unstretched length into $F=-kx$ and treat it as SHM.
   • Why wrong: SHM equation needs displacement from equilibrium, where net force is zero.
   • Fix: find $x_0=\frac{mg}{k}$, then use $y=x-x_0$.

2. Forgetting the small-angle condition for pendulums

   • What goes wrong: you claim a pendulum is always SHM.
   • Why wrong: exact restoring torque involves $\sin\theta$, which is not linear for large angles.
   • Fix: explicitly state “small-angle” and use $\sin\theta\approx\theta$ (radians).

3. Dropping the $2\pi$ in period formulas

   • What goes wrong: you use $T=\sqrt{m/k}$ or $T=\sqrt{L/g}$.
   • Why wrong: that’s missing the sinusoidal timing factor.
   • Fix: memorize $T=2\pi\sqrt{\cdot}$ for both main systems.

4. Mixing up frequency and angular frequency

   • What goes wrong: you treat $\omega$ (rad/s) like $f$ (Hz).
   • Why wrong: $\omega=2\pi f$.
   • Fix: if you find $\omega$, compute $f$ by $f=\frac{\omega}{2\pi}$.

5. Wrong effective spring constant for series/parallel

   • What goes wrong: you add series constants or invert parallel constants.
   • Why wrong: series springs share force; parallel springs share displacement.
   • Fix: remember: parallel adds $k$, series adds reciprocals.

6. Saying max speed occurs at amplitude

   • What goes wrong: you say it’s fastest at the endpoint because “it traveled far.”
   • Why wrong: at amplitude, it turns around so $v=0$.
   • Fix: max speed at equilibrium; max acceleration at endpoints.

7. Using degrees inside trig approximations

   • What goes wrong: you use $\sin\theta\approx\theta$ with $\theta$ in degrees.
   • Why wrong: approximation only works when $\theta$ is in radians.
   • Fix: convert to radians or keep it symbolic.

8. Confusing amplitude with peak-to-peak distance on graphs

   • What goes wrong: you read peak-to-peak as $A$.
   • Why wrong: peak-to-peak is $2A$.
   • Fix: amplitude is from equilibrium line to a peak.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can state the SHM condition: $F\propto -x$ (or for pendulum: $\sin\theta\approx\theta$).
• You always define displacement from equilibrium (especially for vertical springs).
• You know the two must-know periods:
  • $T=2\pi\sqrt{\frac{m}{k_{\text{eff}}}}$
  • $T=2\pi\sqrt{\frac{L}{g}}$ (small-angle)
• You can compute $k_{\text{eff}}$ for series and parallel springs.
• You know where the motion is fastest/slowest:
  • $v_{\max}$ at $x=0$, $v=0$ at $x=\pm A$
  • $|a|$ max at $x=\pm A$
• You can use energy to get speeds without solving for $t$:
  • $\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$
• You won’t mix $\omega$ and $f$: $\omega=2\pi f$.

You’ve got this—if you can spot $F=-k_{\text{eff}}x$ and measure from equilibrium, SHM questions become plug-and-play.