A simple pendulum is a point mass (bob) hanging from a massless string of length L that, for small angular displacements, undergoes simple harmonic motion with period T = 2π√(L/g). The period depends only on length and gravitational field strength, not on the bob's mass or (for small angles) amplitude.
A simple pendulum is the idealized version of anything that swings. The model strips it down to a point mass on a massless, unstretchable string of length L, pivoting from a fixed point. Pull the bob to the side and gravity provides a restoring force component of -mg sin θ along the arc, always pointing back toward the lowest point.
Here's the move that makes it AP Physics C material. The restoring force depends on sin θ, which is NOT proportional to displacement, so a pendulum is not truly simple harmonic motion. But for small angles (roughly under 15°), sin θ ≈ θ, and the equation of motion becomes the SHM equation: angular acceleration proportional to negative angular displacement. That small-angle approximation gives you ω = √(g/L) and T = 2π√(L/g). Notice what's missing from those formulas. Mass cancels out (gravity is both the driving cause and proportional to m), and amplitude doesn't appear at all. A heavier bob or a slightly bigger swing takes the same time per cycle.
The simple pendulum lives in the oscillations unit of AP Physics C: Mechanics, alongside the spring-mass system as one of the two canonical SHM examples. The CED expects you to derive the period from first principles, not just memorize T = 2π√(L/g). That derivation is a classic calculus-based exercise: apply Newton's second law (or torque) to the bob, invoke the small-angle approximation, recognize the resulting differential equation as SHM, and read off ω. It also rolls up earlier units into one problem. You use force decomposition from dynamics, energy conservation to find the speed at the bottom of the swing, and the SHM framework to describe the motion over time. If you can fully analyze a pendulum, you've shown you can connect half the course.
Keep studying AP Physics C: Mechanics Unit Ro4fvf7uNKWRj5Zi
Restoring Force (Unit 7)
SHM requires a restoring force proportional to displacement. For the pendulum, gravity supplies -mg sin θ, which only behaves linearly when θ is small. The small-angle approximation is literally the step that turns a pendulum into a simple harmonic oscillator.
Spring Constant (Unit 7)
The spring-mass system is the pendulum's twin. Compare T = 2π√(m/k) with T = 2π√(L/g) and the pattern jumps out: period is always 2π√(inertia term / stiffness term). For the pendulum, mg/L plays the role of an effective spring constant, which is why mass cancels.
Angular Frequency (Unit 7)
Solving the pendulum's equation of motion gives θ(t) = θ_max cos(ωt + φ) with ω = √(g/L). Once you have ω, every SHM tool transfers directly, including maximum speed, maximum acceleration, and period T = 2π/ω.
Torque and Rotational Dynamics (Unit 5)
You can also derive pendulum motion using τ = Iα about the pivot, with τ = -mgL sin θ and I = mL². This is the gateway to the physical pendulum, where the swinging object has extended mass and you must use its actual rotational inertia.
On multiple choice, pendulum questions test whether you know what the period does and doesn't depend on. Expect stems like "the length is quadrupled, what happens to the period?" (it doubles, since T ∝ √L) or "the same pendulum is taken to the Moon" (g drops, so T increases). On free response, the pendulum shows up as a derivation or experimental design task. You might be asked to derive T = 2π√(L/g) starting from Newton's second law, clearly stating the small-angle approximation, or to design a lab measuring g by timing oscillations and linearizing data (plot T² versus L, and the slope is 4π²/g). Energy conservation between the highest and lowest points of the swing is another frequent FRQ piece. The single most common trap is applying the SHM formulas to a large-amplitude swing without acknowledging the small-angle condition.
A simple pendulum concentrates all its mass at a point a distance L from the pivot, so I = mL² and T = 2π√(L/g). A physical pendulum is any real extended object swinging about a pivot (a rod, a meter stick, a swinging sign), so you must use its actual rotational inertia and the distance d from pivot to center of mass: T = 2π√(I/mgd). If an AP problem says "uniform rod pivoted at one end," it's a physical pendulum, and plugging into the simple pendulum formula is the classic wrong answer.
A simple pendulum is a point mass on a massless string, and it only undergoes simple harmonic motion when the angular displacement is small enough that sin θ ≈ θ.
The period is T = 2π√(L/g), which depends on string length and gravitational field strength but not on the mass of the bob or the amplitude of small swings.
Quadrupling the length doubles the period, and moving the pendulum somewhere with weaker gravity (like the Moon) makes the period longer.
On FRQs, you should be able to derive the period by applying Newton's second law (or τ = Iα), using the small-angle approximation, and matching the result to the standard SHM differential equation.
A pendulum lab measuring g is a common experimental design question, and plotting T² versus L gives a straight line with slope 4π²/g.
If the swinging object has spread-out mass, it's a physical pendulum and you need T = 2π√(I/mgd) instead of the simple pendulum formula.
It's an idealized point mass (bob) hanging from a massless string of length L. For small angles it undergoes simple harmonic motion with period T = 2π√(L/g), making it one of the two standard SHM systems in the course alongside the spring-mass oscillator.
No. Mass appears in both the restoring force (mg sin θ) and the inertia (ma), so it cancels in the equation of motion. T = 2π√(L/g) contains no mass at all, which is exactly why this is a favorite MCQ trap.
Only approximately. The true restoring force goes as sin θ, not θ, so a pendulum is not exact SHM. For small angles (roughly under 15°), sin θ ≈ θ and the SHM equations work well. AP graders expect you to state this small-angle approximation explicitly in derivations.
A simple pendulum has all its mass at a single point, giving T = 2π√(L/g). A physical pendulum is an extended object (like a rod pivoted at one end) and requires its actual rotational inertia, giving T = 2π√(I/mgd). Using the simple formula on an extended object is a common FRQ error.
Apply Newton's second law along the arc (or τ = Iα about the pivot) to get angular acceleration equal to -(g/L) sin θ. Use the small-angle approximation sin θ ≈ θ, recognize the SHM form α = -ω²θ, identify ω = √(g/L), and then T = 2π/ω = 2π√(L/g).