Oscillations and Gravitation — AP Physics C: Mechanics Phys C Mech Study Guide

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

Covers: Derivation of simple harmonic motion (SHM) from Hooke’s law, period and frequency calculation from SHM differential equations, energy conservation in SHM systems, gravitational potential energy and orbital mechanics, and full derivations of Kepler’s three laws of planetary motion.

You should already know: Strong calculus (concurrent OK), AP Physics 1 helpful.

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 papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Oscillations and Gravitation?

Oscillations describe periodic, restoring-force driven motion of systems around an equilibrium point, while gravitation is the universal attractive force between massive objects that governs orbital motion. Combined, these two topics make up ~20% of the AP Physics C: Mechanics exam weighting, appearing in both multiple-choice and free-response questions, often requiring calculus application and free-body diagram analysis.

2. Simple harmonic motion — derivation from $F=-kx$

Simple harmonic motion is defined as motion where the net restoring force acting on a system is directly proportional to displacement from equilibrium, and opposite in direction to the displacement vector. The canonical SHM system is a mass on a frictionless horizontal spring, which follows Hooke’s Law: $$F_{net}=-kx$$ where $k$ is the spring constant, and $x$ is displacement from the equilibrium position $x=0$.

Using Newton’s second law $F=ma$, where acceleration $a=\frac{d^{2}x}{d{t}^2}$, we substitute to get the second-order linear differential equation for SHM: $$m\frac{d^{2}x}{d{t}^2}+kx=0$$ We define the angular frequency of the system as $\omega =\sqrt{\frac{k}{m}}$, which simplifies the ODE to $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$. The general solution to this equation is: $$x(t)=A\cos (\omega t+\varphi )$$ where $A$ = amplitude (maximum displacement from equilibrium), and $\varphi$ = phase constant (determined by initial position and velocity at $t=0$).

Worked Example: A 0.5 kg mass is attached to a spring with $k=200$ N/m, displaced 0.02 m from equilibrium, and released from rest. Write the SHM displacement function for the system.

1. Calculate angular frequency: $\omega =\sqrt{\frac{200}{0.5}}=20$ rad/s
2. Apply initial conditions: At $t=0$, $x=0.02=A\cos (\varphi )$ and $v=0=-A\omega \sin (\varphi )$. This gives $\varphi =0$ and $A=0.02$ m.
3. Final function: $x(t)=0.02\cos (20t)$ meters. Exam tip: Examiners frequently ask you to derive this ODE for non-spring SHM systems (e.g., simple pendulum, physical pendulum), so always start with a free-body diagram to identify the restoring force.

3. Period and frequency from differential equation

Period $T$ is the time required for one full oscillation, measured in seconds. Frequency $f=\frac{1}{T}$, measured in Hz, describes the number of oscillations per second. These quantities are linked to angular frequency by: $$\omega =2\pi f=\frac{2\pi }{T}$$ The most powerful exam technique for calculating period for any SHM system is to extract $\omega$ directly from the rearranged ODE, rather than memorizing system-specific period formulas. For any ODE in the form $\frac{d^{2}y}{d{t}^2}+Cy=0$, $\omega =\sqrt{C}$, so $T=\frac{2\pi }{\sqrt{C}}$.

Worked Example: Derive the period of a simple pendulum of length $L$ for small angle displacements, where $\sin (\theta )\approx \theta$.

1. The restoring torque on the pendulum bob is $\tau =-mgL\sin (\theta )\approx -mgL\theta$
2. Using $\tau =I\alpha$, where moment of inertia $I=m{L}^2$ and angular acceleration $\alpha =\frac{d^2\theta }{d{t}^2}$, substitute to get: $$m{L}^2\frac{d^2\theta }{d{t}^2}=-mgL\theta ⟹\frac{d^2\theta }{d{t}^2}=-\frac{g}{L}\theta$$
3. Extract $\omega =\sqrt{\frac{g}{L}}$, so $T=2\pi \sqrt{\frac{L}{g}}$ Numerical check: For $L=1$ m, $g=9.8$ m/s², $T=2\pi \sqrt{\frac{1}{9.8}}\approx 2.01$ s, which matches experimental values.

4. Energy in SHM

SHM systems operate under conservative forces, so total mechanical energy is constant for the duration of oscillation. For a mass-spring system:

• Elastic potential energy: $U=\frac{1}{2}k{x}^2$ (maximum at maximum displacement, zero at equilibrium)
• Kinetic energy: $K=\frac{1}{2}m{v}^2$ (maximum at equilibrium, zero at maximum displacement)
• Total energy: $E=U+K=\frac{1}{2}k{A}^2$, since at $x=A$, $v=0$ and all energy is potential.

We can also derive the maximum velocity of the mass by setting $K_{max}=\frac{1}{2}m{v}_{max}^2=\frac{1}{2}k{A}^2$, giving $v_{max}=A\omega$, which matches the magnitude of the maximum value of the velocity function $v(t)=-A\omega \sin (\omega t+\varphi )$.

Worked Example: For the 0.5 kg mass-spring system from section 2 ($k=200$ N/m, $A=0.02$ m), calculate the velocity of the mass when displacement is 0.01 m.

1. Total energy: $E=\frac{1}{2}(200)(0.02{)}^2=0.04$ J
2. Potential energy at $x=0.01$ m: $U=\frac{1}{2}(200)(0.01{)}^2=0.01$ J
3. Kinetic energy: $K=E-U=0.03$ J
4. Solve for velocity: $v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.03)}{0.5}}\approx 0.346$ m/s Exam note: FRQs often ask you to plot $U$, $K$, and $E$ as functions of $x$ or $t$. $E$ is a horizontal line, $U$ is an upward-opening parabola, and $K$ is a downward-opening parabola, intersecting $U$ at $x=\pm \frac{A}{\sqrt{2}}$.

5. Gravitational potential energy and orbits

Newton’s Law of Universal Gravitation states that the attractive force between two point masses $M$ and $m$ separated by distance $r$ is: $$F_g=\frac{GMm}{r^2}$$ where $G=6.67\times 10^{-11}$ N·m²/kg² is the universal gravitational constant. The gravitational potential energy of the two-mass system is: $$U(r)=-\frac{GMm}{r}$$ with the zero point defined at $r=\infty$. The surface-level potential energy formula $U=mgh$ is only valid for heights $h<<R_{planet}$, where $g$ is approximately constant.

For circular orbits, gravitational force provides the centripetal force required for uniform circular motion: $$\frac{GMm}{r^2}=\frac{m{v}^2}{r}⟹v_{orb}=\sqrt{\frac{GM}{r}}$$ Total mechanical energy for a circular orbit is $E=K+U=\frac{1}{2}m{v}_{orb}^2-\frac{GMm}{r}=-\frac{GMm}{2r}$, which is negative for bound orbits (objects trapped in the gravitational field). The escape velocity, the minimum speed required to escape the gravitational field (so $E=0$), is $v_{esc}=\sqrt{\frac{2GM}{R}}$, where $R$ is the radius of the central body.

Worked Example: A 1000 kg satellite orbits Earth ($M=5.97\times 10^{24}$ kg, $R=6.37\times 10^6$ m) at an altitude of 400 km. Calculate its orbital speed and total energy.

1. Orbital radius: $r=6.37\times 10^6+4\times 10^5=6.77\times 10^6$ m
2. Orbital speed: $v=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{6.77\times 10^6}}\approx 7660$ m/s
3. Total energy: $E=-\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 1000}{2\times 6.77\times 10^6}\approx -2.94\times 10^{10}$ J

6. Kepler's laws derivation

Kepler’s three laws of planetary motion were originally empirical, but can be derived directly from Newton’s law of universal gravitation and conservation laws:

1. Law of Ellipses: Planets orbit the Sun in elliptical paths, with the Sun at one focus of the ellipse. The inverse-square nature of gravitational force results in conic-section orbit solutions; bound orbits (negative total energy) are ellipses, while unbound orbits are parabolas or hyperbolas.
2. Law of Equal Areas: A line connecting a planet to the Sun sweeps out equal areas in equal time intervals. This is a direct consequence of conservation of angular momentum: $L=m{r}^2\frac{d\theta }{dt}=constant$. The area swept in time $dt$ is $dA=\frac{1}{2}{r}^{2}d\theta$, so $\frac{dA}{dt}=\frac{L}{2m}=constant$.
3. Law of Periods: The square of the orbital period $T$ is proportional to the cube of the semi-major axis $a$ of the orbit: $T^2=\frac{4{\pi }^2}{GM}{a}^3$. For circular orbits, $a=r$; substituting $T=\frac{2\pi r}{v_{orb}}$ and $v_{orb}=\sqrt{\frac{GM}{r}}$ gives the formula directly, which also holds for elliptical orbits.

Worked Example: A comet has a perihelion (closest to Sun) distance of $r_p=1\times 10^{11}$ m and speed $v_p=5\times 10^4$ m/s. What is its speed at aphelion (farthest from Sun) $r_a=5\times 10^{11}$ m? Angular momentum is conserved: $m{r}_p{v}_p=m{r}_a{v}_a⟹v_a=\frac{r_p}{r_a}{v}_p=\frac{1\times 10^{11}}{5\times 10^{11}}\times 5\times 10^4=1\times 10^4$ m/s. Exam tip: For solar system orbit calculations, use units of AU (astronomical units, $1AU=1.5\times 10^{11}$ m) and Earth years: for Earth, $a=1$ AU and $T=1$ year, so $T^2=a^3$ directly with no need to calculate $GM$.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using $U=mgh$ for orbital or large-height gravitational energy calculations. Why students do it: They memorize the surface-level formula without noting its limits. Correct move: Use $U(r)=-\frac{GMm}{r}$ for all cases where height is >1% of the planet’s radius, or for orbits around non-Earth bodies.
• Wrong move: Assuming the phase constant $\varphi =0$ for all SHM problems. Why students do it: Most textbook examples release masses from rest at maximum displacement, which gives $\varphi =0$. Correct move: Always solve for $\varphi$ using both initial position and initial velocity at $t=0$, never assume it is zero.
• Wrong move: Using surface radius instead of orbital radius (surface radius + altitude) in orbit calculations. Why students do it: They mix up altitude (height above surface) with distance from the planet’s center. Correct move: Mark orbital radius clearly on your free-body diagram, and explicitly add the planet’s radius to given altitude before plugging into formulas.
• Wrong move: Adding mass to the simple pendulum period formula by mistake. Why students do it: The mass-spring period includes mass, so students incorrectly generalize. Correct move: Always derive $\omega$ from the second-order ODE for the system, so you don’t have to memorize which period formulas include mass.
• Wrong move: Confusing SHM angular frequency $\omega$ with rotational angular velocity. Why students do it: They share the same symbol and units (rad/s). Correct move: Label the two quantities separately in your work; $\omega$ for SHM is a constant system parameter, while rotational angular velocity can change with time.

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

Question 1 (MCQ Style)

A 2 kg mass attached to a spring of $k=800$ N/m undergoes SHM with amplitude 0.05 m. What is the magnitude of the mass’s acceleration when its kinetic energy equals its potential energy? A) 5 m/s² B) 10 m/s² C) 14 m/s² D) 20 m/s² Worked Solution:

1. When $K=U$, $U=\frac{1}{2}E=\frac{1}{2}(\frac{1}{2}k{A}^2)=\frac{1}{4}k{A}^2$
2. Set $U=\frac{1}{2}k{x}^2=\frac{1}{4}k{A}^2⟹x=\frac{A}{\sqrt{2}}=\frac{0.05}{\sqrt{2}}\approx 0.03535$ m
3. Acceleration magnitude: $|a|=\frac{|F|}{m}=\frac{kx}{m}=\frac{800\times 0.03535}{2}\approx 14.1$ m/s² Answer: C

Question 2 (FRQ Part Style)

A simple pendulum of length 1.5 m is displaced 5 degrees from equilibrium and released. (a) Derive the period of the pendulum for small angles. (b) Calculate the maximum speed of the bob. Worked Solution: (a) 1. Tangential restoring force: $F_t=-mg\sin (\theta )\approx -mg\theta$ for small $\theta$. 2. Displacement along the arc: $s=L\theta$, so $F_t=-\frac{mg}{L}s$. 3. Apply $F=ma$: $-\frac{mg}{L}s=m\frac{d^{2}s}{d{t}^2}⟹\frac{d^{2}s}{d{t}^2}=-\frac{g}{L}s$. 4. Extract $\omega =\sqrt{\frac{g}{L}}$, so $T=2\pi \sqrt{\frac{L}{g}}$. (b) 1. Max potential energy at maximum displacement: $U=mgL(1-\cos (\theta ))$. $\theta =5^{\circ }=0.0873$ rad, $\cos (\theta )\approx 0.9962$. 2. Set $U=K_{max}=\frac{1}{2}m{v}^2$: $v=\sqrt{2gL(1-\cos (\theta ))}=\sqrt{2\times 9.8\times 1.5\times 0.0038}\approx 0.334$ m/s.

Question 3 (FRQ Style)

A 500 kg satellite is placed in circular orbit around Mars ($M=6.42\times 10^{23}$ kg, $R=3.39\times 10^6$ m) at an altitude of 200 km. (a) Calculate orbital speed. (b) Calculate orbital period in hours. (c) What minimum delta-v (change in speed) is required for the satellite to escape Mars’ gravity? Worked Solution: (a) Orbital radius $r=3.39\times 10^6+2\times 10^5=3.59\times 10^6$ m. $v_{orb}=\sqrt{\frac{GM}{r}}=\sqrt{\frac{6.67\times 10^{-11}\times 6.42\times 10^{23}}{3.59\times 10^6}}\approx 3450$ m/s. (b) $T=\frac{2\pi r}{v_{orb}}=\frac{2\pi \times 3.59\times 10^6}{3450}\approx 6540$ s $\approx 1.82$ hours. (c) Escape speed $v_{esc}=\sqrt{\frac{2GM}{r}}=\sqrt{2}\times 3450\approx 4880$ m/s. Delta-v = $4880-3450=1430$ m/s.

9. Quick Reference Cheatsheet

10. What's Next

Oscillations and gravitation are among the most heavily tested units on the AP Physics C: Mechanics exam, and they form a foundation for related topics in AP Physics C: Electricity and Magnetism, where Coulomb’s inverse square electrostatic force mirrors Newton’s gravitation, and oscillating LC circuits follow the exact same differential equations as mass-spring SHM systems. On the Mechanics exam, you will frequently encounter FRQs that combine SHM with kinematics, work-energy principles, or torque, so make sure you can apply the derivations you learned here to unfamiliar systems like vertical mass-spring systems or physical pendulums.

If you struggle with any of the calculus derivations, energy calculations, or orbital mechanics problems covered in this guide, you can ask Ollie for step-by-step walkthroughs, additional practice problems, or targeted review of prerequisite concepts like differential equations or conservation of angular momentum. Head to the homepage, to test your knowledge with full-length AP Physics C: Mechanics practice exams aligned to the latest College Board syllabus, or to access more study guides for every unit on the exam.