Physics C: Mechanics · Unit 6: Oscillations and Gravitation · 16 min read · Updated 2026-05-11
Oscillations and Gravitation — AP Physics C: Mechanics
AP Physics C: Mechanics · Unit 6: Oscillations and Gravitation · 16 min read
1. Simple Harmonic Motion Derivation from Hooke's Law★★★☆☆⏱ 4 min
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 $a = \frac{d^2x}{dt^2}$, we substitute to get the second-order linear differential equation for SHM:
m \frac{d^2x}{dt^2} + kx = 0
We define the angular frequency of the system as $\omega = \sqrt{\frac{k}{m}}$, which simplifies the ODE to $\frac{d^2x}{dt^2} = -\omega^2 x$. The general solution to this equation is:
x(t) = A\cos(\omega t + \phi)
where $A$ = amplitude (maximum displacement from equilibrium), and $\phi$ = phase constant (determined by initial position and velocity at $t=0$).
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.
2. Period and Frequency from SHM Differential Equations★★☆☆☆⏱ 3 min
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 $\frac{d^2y}{dt^2} + Cy = 0$, $\omega = \sqrt{C}$, so $T = \frac{2\pi}{\sqrt{C}}$.
3. Energy Conservation in SHM★★★☆☆⏱ 3 min
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: $U = \frac{1}{2}kx^2$ (maximum at maximum displacement, zero at equilibrium)
Kinetic energy: $K = \frac{1}{2}mv^2$ (maximum at equilibrium, zero at maximum displacement)
Total energy: $E = U + K = \frac{1}{2}kA^2$, since at $x=A$, $v=0$ and all energy is potential
We can also derive the maximum velocity of the mass by setting $K_{max} = \frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$, giving $v_{max} = A\omega$, which matches the magnitude of the maximum value of the velocity function $v(t) = -A\omega\sin(\omega t + \phi)$.
4. Gravitational Potential Energy and Orbital Mechanics★★★★☆⏱ 4 min
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, with zero potential energy defined at $r = \infty$, is:
U(r) = -\frac{GMm}{r}
For circular orbits, gravitational force provides the centripetal force required for uniform circular motion:
Total mechanical energy for a circular orbit is $E = K + U = \frac{1}{2}mv_{orb}^2 - \frac{GMm}{r} = -\frac{GMm}{2r}$, 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.
5. Kepler's Laws of Planetary Motion★★★★☆⏱ 3 min
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:
**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.
**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: $L = mr^2 \frac{d\theta}{dt} = \text{constant}$, so $\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$.
**Law of Periods**: The square of the orbital period $T$ is proportional to the cube of the semi-major axis $a$ of the orbit: $T^2 = \frac{4\pi^2}{GM}a^3$.
6. Exam-Style Practice Problems★★★★☆⏱ 4 min
Common Pitfalls
Why: Students memorize the surface-level formula without noting its limited range of applicability.
Why: Most textbook examples release masses from rest at maximum displacement, which gives $\phi=0$, but this is not general.
Why: Students mix up altitude (height above the surface) with distance from the planet's center of mass.
Why: The mass-spring period formula includes mass, so students incorrectly generalize this to all SHM systems.
Why: The two quantities share the same symbol and units (rad/s), leading to mix-ups.