Simple harmonic motion — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Definition of simple harmonic motion (SHM) via force and differential equation forms, general solution of the SHM ODE, energy conservation in SHM, angular frequency derivation for mass-spring and simple pendulum systems, phase and amplitude analysis.

You should already know: Newton's second law for translational motion. Solving second-order linear homogeneous differential equations. Conservation of mechanical energy.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Simple harmonic motion?

Simple harmonic motion (SHM) is periodic motion of an object where the restoring force is directly proportional to the displacement from a stable equilibrium position, and opposite in direction to displacement. It is the core topic of Unit 6 Oscillations, which accounts for 12-18% of the total AP Physics C: Mechanics exam score per the official Course and Exam Description (CED). SHM appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, and is often paired with energy, rotational motion, or forces for multi-point FRQ questions.

SHM is a special case of oscillatory motion: unlike damped or driven oscillation, undamped SHM has constant amplitude and period with no energy loss, making it the simplest oscillatory system to model mathematically. Common examples include horizontal/vertical mass-spring systems and small-angle simple pendulums, both of which approximate SHM under standard conditions. This guide follows AP exam notation conventions: $x(t)$ = displacement from equilibrium, $k$ = spring constant, $m$ = mass, $\omega$ = angular frequency, $A$ = amplitude, $\varphi$ = phase constant, $T$ = period, $f$ = frequency.

2. The Defining Differential Equation of SHM

SHM can be defined two equivalent ways: via force, or via a second-order differential equation. By the force definition: the net restoring force on the object is $F_{\text{net}}=-kx$, where $k$ is a positive constant of proportionality (the effective spring constant for the system) and $x$ is displacement from equilibrium. The negative sign confirms the force always points back toward equilibrium, pulling the object back when it is displaced.

Applying Newton's second law ($F_{\text{net}}=ma=m\frac{d^{2}x}{d{t}^2}$) rearranges this to the standard defining differential equation for SHM: $$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$$ where ${\omega }^2=\frac{k}{m}$ for a mass-spring system, and $\omega$ is the angular frequency of oscillation. Any motion that satisfies this differential equation is SHM, regardless of the physical system. The general solution to this ODE is $x(t)=A\cos (\omega t+\varphi )$ (equivalent to a sine + cosine combination via trigonometric identities), where $A$ is maximum displacement (amplitude) and $\varphi$ is the phase constant that adjusts for initial conditions. Angular frequency relates to period and frequency by $\omega =\frac{2\pi }{T}=2\pi f$.

Worked Example

A 2.0 kg block on a frictionless horizontal track is attached to a spring with $k=50$ N/m. Write the differential equation for the block's motion, then calculate $\omega$ and $T$.

1. Start with the net restoring force: $F_{\text{net}}=-kx=m\frac{d^{2}x}{d{t}^2}$.
2. Rearrange to standard SHM form: $\frac{d^{2}x}{d{t}^2}+\frac{k}{m}x=0$. Substitute values: $\frac{k}{m}=\frac{50}{2.0}=25$ s${}^{-2}$, so the differential equation is $\frac{d^{2}x}{d{t}^2}+25x=0$.
3. By definition, ${\omega }^2=\frac{k}{m}=25$, so $\omega =5$ rad/s.
4. Calculate period: $T=\frac{2\pi }{\omega }=\frac{2\pi }{5}\approx 1.26$ s.

Exam tip: If a problem gives you a non-standard system (e.g., a mass between two springs, a floating object oscillating), always derive the differential equation from Newton's second law to find $\omega$, do not guess it from memorized standard formulas.

3. Kinematics of SHM

Once you have the position function for SHM, velocity and acceleration are found via differentiation. For the standard position function $x(t)=A\cos (\omega t+\varphi )$, the first derivative (velocity) is: $$v(t)=\frac{dx}{dt}=-A\omega \sin (\omega t+\varphi )$$ and the second derivative (acceleration) is: $$a(t)=\frac{d^{2}x}{d{t}^2}=-A{\omega }^2\cos (\omega t+\varphi )=-{\omega }^{2}x(t)$$ Key relationships: maximum speed is $v_{\text{max}}=A\omega$, which occurs when the object passes through equilibrium ($x=0$), where all energy is kinetic. Maximum acceleration is $a_{\text{max}}=A{\omega }^2$, which occurs at maximum displacement ($x=\pm A$), where the restoring force is largest.

To find the amplitude and phase constant, you use initial conditions (position and velocity at $t=0$). The amplitude can be found from the identity $A^2=(A\cos \varphi {)}^2+{\left(\frac{A\sin \varphi \cdot \omega }{\omega }\right)}^2=x_0^2+{\left(\frac{v_0}{\omega }\right)}^2$, where $x_0=x(0)$ and $v_0=v(0)$. The phase constant is solved from $\tan \varphi =-\frac{v_0}{\omega x_0}$, and you must check the quadrant of $\varphi$ using the signs of $x_0$ and $v_0$.

Worked Example

For the 2.0 kg block from the previous example, initial conditions at $t=0$ are $x_0=0.10$ m and $v_0=-1.0$ m/s, and $\omega =5$ rad/s. Find $A$ and $\varphi$.

1. We know $x_0=A\cos \varphi =0.10$, and $v_0=-A\omega \sin \varphi =-1.0$, which rearranges to $A\sin \varphi =\frac{1.0}{5}=0.20$.
2. Calculate amplitude: $A^2=(A\cos \varphi {)}^2+(A\sin \varphi {)}^2=0.10^2+0.20^2=0.05$, so $A=\sqrt{0.05}\approx 0.22$ m.
3. Calculate phase constant: $\tan \varphi =\frac{A\sin \varphi }{A\cos \varphi }=\frac{0.20}{0.10}=2$. Both $A\cos \varphi$ and $A\sin \varphi$ are positive, so $\varphi$ is in the first quadrant: $\varphi =\arctan (2)\approx 1.11$ rad.
4. Final position function: $x(t)=0.22\cos (5t+1.11)$ m.

Exam tip: Always check the quadrant of the phase constant using the signs of both initial position and velocity. Calculators only return first/fourth quadrant values for arctangent, so you will get an incorrect result if $\varphi$ is in the second or third quadrant if you do not adjust.

4. Energy in Simple Harmonic Motion

For undamped SHM, total mechanical energy is conserved, with energy converting between kinetic energy of the mass and potential energy of the restoring force. For a mass-spring system, kinetic energy is $K=\frac{1}{2}m{v}^2$, and elastic potential energy is $U=\frac{1}{2}k{x}^2$. Substituting the SHM position and velocity functions, and using ${\omega }^2=\frac{k}{m}$, total energy simplifies to: $$E=K+U=\frac{1}{2}k{A}^2=\frac{1}{2}m{\omega }^2{A}^2$$ Total energy is constant, proportional to the square of the amplitude. At maximum displacement, $v=0$, so all energy is potential: $E=U_{\text{max}}=\frac{1}{2}k{A}^2$. At equilibrium, $x=0$, so all energy is kinetic: $E=K_{\text{max}}=\frac{1}{2}m{v}_{\text{max}}^2$, which matches $v_{\text{max}}=A\omega$.

A very useful result derived from energy conservation is the speed at any given displacement: $v=\omega \sqrt{A^2-x^2}$, which lets you calculate speed without working through kinematics or phase constants. This works for any undamped SHM system.

Worked Example

A 0.5 kg mass-spring system oscillates with SHM of amplitude 0.4 m and period 1.0 s. What is the total mechanical energy, and what is the speed when displacement is 0.2 m?

1. Calculate angular frequency: $\omega =\frac{2\pi }{T}=2\pi$ rad/s.
2. Total energy: $E=\frac{1}{2}m{\omega }^2{A}^2=0.5(0.5)(2\pi {)}^2(0.4{)}^2=0.16{\pi }^2\approx 1.58$ J.
3. Use the energy-derived speed formula: $v=\omega \sqrt{A^2-x^2}=2\pi \sqrt{0.4^2-0.2^2}=2\pi \sqrt{0.12}\approx 2.18$ m/s.
4. Sanity check: maximum speed is $A\omega =0.4(2\pi )\approx 2.51$ m/s, which is larger than our result, as expected for a point away from equilibrium.

Exam tip: If you are asked for speed at a given displacement, using energy conservation is almost always faster and less error-prone than differentiating the position function and calculating sine of the phase.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $x(t)=A\sin (\omega t)$ by default, even when initial position at $t=0$ is non-zero. Why: Students memorize the sine form for the specific case of starting at equilibrium, and forget to adjust for other starting positions. Correct move: Always start with the general form $x(t)=A\cos (\omega t+\varphi )$ and solve for $\varphi$ from initial conditions, regardless of the starting position.
• Wrong move: Using $\omega =\sqrt{g/L}$ for a pendulum displaced by 45 degrees from equilibrium. Why: Students memorize the simple pendulum result, but forget it only applies for small angles where $\sin \theta \approx \theta$ to produce the SHM differential equation. Correct move: For large-angle pendulum motion, explicitly derive the differential equation from Newton's second law and confirm it fits the SHM form before using SHM formulas.
• Wrong move: Forgetting the chain rule when differentiating $x(t)$, leading to $v(t)=-A\sin (\omega t+\varphi )$ (missing the $\omega$ factor). Why: Students focus on the derivative of cosine and forget to differentiate the inner linear term in the argument. Correct move: Always write the chain rule step explicitly: $\frac{d}{dt}f(\omega t+\varphi )=\omega f^{\prime }(\omega t+\varphi )$.
• Wrong move: For a vertical mass-spring system, changing $\omega$ to account for gravity. Why: Students think gravity changes the restoring force because it shifts the equilibrium position. Correct move: Recognize gravity only shifts the equilibrium position of a vertical mass-spring; the restoring force is still proportional to displacement from the new equilibrium, so $\omega =\sqrt{k/m}$ remains unchanged.
• Wrong move: Using the equivalent spring constant formula $1/k_{\text{eq}}=1/k_1+1/k_2$ for two springs in parallel. Why: Students confuse spring combinations with resistor combinations. Correct move: For parallel springs (displacement same for both springs), $k_{\text{eq}}=k_1+k_2$; for series springs (force same for both springs), $1/k_{\text{eq}}=1/k_1+1/k_2$.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A particle undergoes SHM described by $x(t)=3\cos (2t-\pi /4)$ (all units SI). What is the magnitude of the maximum acceleration of the particle? (A) 3 m/s² (B) 6 m/s² (C) 12 m/s² (D) 24 m/s²

Worked Solution: For any SHM, acceleration satisfies $a(t)=-{\omega }^{2}x(t)$, so maximum acceleration magnitude is $a_{\text{max}}={\omega }^{2}A$, where $A$ is amplitude and $\omega$ is angular frequency. From the given position function, we identify $A=3$ m and $\omega =2$ rad/s. Substituting gives $a_{\text{max}}=(2{)}^2(3)=12$ m/s². The most common mistake is forgetting to square $\omega$, which gives the incorrect result 6 m/s². The correct answer is C.

Question 2 (Free Response)

A 1.5 kg block is attached to two identical springs each with spring constant $k=30$ N/m. The block sits between two fixed walls, with one spring attached to each end of the block, and both springs are at natural length when the block is at $x=0$. (a) Show that the block undergoes SHM when displaced from $x=0$, and derive the angular frequency of the motion. (b) At $t=0$, the block is pulled to $x=0.2$ m and released from rest. Write the complete position function $x(t)$ for the block. (c) Calculate the total energy of the system, and the kinetic energy of the block when $x=0.1$ m.

Worked Solution: (a) When the block is displaced $x$ to the right, the left spring stretches by $x$ and exerts a force $-kx$ left. The right spring compresses by $x$ and also exerts a force $-kx$ left. Net force: $F_{\text{net}}=-2kx=m{d}^{2}x/d{t}^2$, which rearranges to $d^{2}x/d{t}^2+(2k/m)x=0$, the standard SHM differential equation. ${\omega }^2=2k/m=(2\cdot 30)/1.5=40$, so $\omega =\sqrt{40}\approx 6.32$ rad/s. (b) Initial conditions: $x(0)=0.2$ m, $v(0)=0$. General form: $x(t)=A\cos (\omega t+\varphi )$. At $t=0$, $x(0)=A\cos \varphi =0.2$, $v(0)=-A\omega \sin \varphi =0$, so $\sin \varphi =0$, $\varphi =0$, $A=0.2$ m. Final position function: $x(t)=0.2\cos \left(\sqrt{40}t\right)$ m. (c) Total energy: $E=\frac{1}{2}{k}_{\text{eq}}{A}^2=\frac{1}{2}(60)(0.2{)}^2=1.2$ J. By conservation of energy: $E=K+\frac{1}{2}{k}_{\text{eq}}{x}^2$, so $K=1.2-\frac{1}{2}(60)(0.1{)}^2=1.2-0.3=0.9$ J.

Question 3 (Application / Real-World Style)

A car's suspension can be modeled as a single mass-spring system undergoing SHM. After hitting a bump, the empty 1200 kg car oscillates with an angular frequency of 2.0 rad/s. When carrying a 500 kg load of passengers and luggage, what is the new angular frequency of the oscillation (to 2 significant figures), assuming the effective spring constant of the suspension remains constant? Explain what the change means for the car's ride.

Worked Solution: For a mass-spring SHM system, $\omega =\sqrt{k/m}$. First solve for the effective spring constant from the empty car: $k={\omega }^{2}m=(2.0{)}^2(1200)=4800$ N/m. The new total mass with the load is $m_{\text{new}}=1200+500=1700$ kg. Substitute to find new angular frequency: ${\omega }_{\text{new}}=\sqrt{k/m_{\text{new}}}=\sqrt{4800/1700}\approx 1.7$ rad/s. This lower angular frequency corresponds to a longer period of oscillation, meaning the loaded car oscillates more slowly after hitting a bump, resulting in a softer, slower ride than the empty car.

7. Quick Reference Cheatsheet

8. What's Next

This chapter gives you the foundational model for all oscillatory motion in AP Physics C: Mechanics, and the differential equation approach used here is applied to many other dynamic systems across the course. Next in Unit 6 Oscillations, you will study damped and driven harmonic motion, which extend the SHM model to include energy loss and external driving forces. Without mastering the differential equation, kinematics, and energy relations of SHM, you will not be able to solve the more complex ODEs for damped motion, or understand resonance, a commonly tested AP topic. SHM also connects directly to uniform circular motion (SHM is the projection of uniform circular motion onto a line) and energy conservation, core concepts for the entire exam.

• Damped and driven harmonic motion
• Uniform circular motion
• Conservation of mechanical energy