Period (T) is the time it takes for one complete oscillation of an SHM system (measured in seconds). Frequency (f) is how many complete oscillations happen each second (measured in hertz, Hz). They’re reciprocals: T = 1/f. In SHM you often use angular frequency ω (radians per second), with T = 2π/ω and f = ω/(2π). For common systems: a mass on an ideal spring Ts = 2π√(m/k) and a small-angle simple pendulum Tp = 2π√(l/g). Note these periods (and thus f = 1/T) do not depend on amplitude for ideal SHM—they depend on system parameters (m, k, l, g). On the AP exam you may be asked to state or use these relations (CED 7.2.A.1 and subparts), so be ready to move between T, f, and ω in derivations or numeric problems. Want a quick refresher and practice? Check the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and more practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

For a mass on an ideal spring undergoing SHM, use the AP CED relation between angular frequency and period: T = 2π/ω and for a spring-mass oscillator ω = √(k/m). Combine them to get the standard formula: T = 2π √(m/k). Here m is the mass (kg) and k is the spring constant (N/m). The frequency is f = 1/T = (1/2π)√(k/m). Notice T depends only on m and k, not amplitude (that’s a key SHM result on the exam). Units: if m in kg and k in N/m, T is in seconds. Quick example: m = 0.50 kg, k = 200 N/m → T = 2π√(0.50/200) ≈ 0.31 s. This formula is in the CED (7.2.A.1.i). For more review and practice use the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and tons of practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Because simple harmonic motion (SHM) comes from a linear restoring force, the oscillation frequency depends only on the system parameters (m, k for a spring; l, g for a small-angle pendulum), not how far you pull it. For an ideal spring F = −kx leads to the equation ẍ + (k/m)x = 0 so ω = √(k/m) and T = 2π√(m/k) (CED 7.2.A.1.i). For a pendulum you linearize sinθ ≈ θ for small angles, get θ̈ + (g/l)θ = 0, so T = 2π√(l/g) (CED 7.2.A.1.ii). That linearization is the key: amplitude drops out of ω when the restoring force is proportional to displacement. If you go to large angles (sinθ ≠ θ), or include non-linear springs, damping, or driving forces, the period can depend on amplitude. For AP review see the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and more unit practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Short answer: no—they’re related but not the same. Regular frequency f is cycles per second (Hz). Angular frequency ω measures how quickly the motion’s phase angle changes (radians per second). They’re linked by the key SHM relation from the CED: T = 2π/ω = 1/f, so ω = 2πf. Why that matters: in SHM we usually write x(t) = A cos(ωt + φ). That ω tells you how fast the object moves through its cycle in radians—useful for equations of motion (a = −ω²x) and for deriving T for a spring (Ts = 2π√(m/k)) or a small-angle pendulum (Tp = 2π√(l/g)) as listed in LO 7.2.A.1. On the AP exam make sure you keep units straight: f in Hz, ω in rad/s, and use ω = 2πf when converting. For a quick review, see the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw). For extra practice, try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

For a simple (small-angle) pendulum, T = 2π√(l/g) (CED 7.2.A.1.ii). If you double the length l → 2l, the period becomes T' = 2π√(2l/g) = √2 · 2π√(l/g) = √2 · T. So the period increases by a factor of √2 ≈ 1.414—about a 41% longer period. The frequency f = 1/T therefore decreases by 1/√2 (about 0.707× the original). Remember this result assumes the small-angle approximation (simple pendulum model). For a quick refresher tied to the AP CED and practice problems, see the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and try related practice at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Short answer: the math comes from the different restoring forces and how mass appears in the equation of motion. Spring: Hooke’s law gives F = −kx. Newton’s 2nd law: m d²x/dt² = −kx → d²x/dt² + (k/m) x = 0 so ω = sqrt(k/m) and T = 2π sqrt(m/k). Mass stays in the numerator because the spring’s stiffness k is a property of the spring while the mass resists acceleration (inertia). Pendulum: for small angles the restoring torque comes from gravity: τ ≈ −mgLθ. Rotational form of Newton’s 2nd law: Iα = τ with I = mL² and α = d²θ/dt², so mL² d²θ/dt² = −mgL θ → d²θ/dt² + (g/L) θ = 0 so ω = sqrt(g/L) and T = 2π sqrt(L/g). The mass cancels because gravity provides the restoring torque (∝ m) and the pendulum’s rotational inertia (∝ m) both scale with m. Note the small-angle approximation is required for the pendulum formula (CED 7.2.A.1.i–ii). For a quick refresher, see the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw). For practice derivations like these (useful on the AP exam), check Fiveable’s unit review (https://library.fiveable.me/ap-physics-c-mechanics/unit-7) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Think of it as “which restoring force matters?” For a mass on a spring the restoring force is Hooke’s law, F = −kx, so the dynamics give ω = √(k/m) and T_s = 2π√(m/k). For a simple pendulum (small-angle approx) the restoring torque comes from gravity: τ ≈ −mgℓθ → angular frequency ω = √(g/ℓ) and T_p = 2π√(ℓ/g). Quick memory tricks: - Spring → k and m appear → T ∝ √(m/k). - Pendulum → g and length ℓ appear → T ∝ √(ℓ/g). Also remember: pendulum formula needs small-angle approximation (CED 7.2.A.1.ii). On the AP exam they expect you to pick the correct model (mass-spring vs simple pendulum) and use T = 2π/ω or T = 1/f if given ω (CED 7.2.A.1). If you want a short study refresher, check the Topic 7.2 study guide on Fiveable (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw). For more practice problems across Unit 7, see the unit page (https://library.fiveable.me/ap-physics-c-mechanics/unit-7) and practice bank (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Period T, frequency f, and angular frequency ω for SHM are just different ways to measure how fast the oscillation goes: - T = period = time for one full cycle (seconds). - f = frequency = cycles per second = 1/T (Hz). - ω = angular frequency = 2πf = 2π/T (rad/s). So the compact relations are: T = 1/f and T = 2π/ω (equivalently f = ω/2π). Apply these to common AP systems (from the CED): for a mass–spring, Ts = 2π√(m/k); for a small-angle simple pendulum, Tp = 2π√(l/g). Those come from ω = √(k/m) for springs and ω = √(g/l) for small-angle pendula, then using ω = 2π/T. Topic 7.2 (7.2.A.1) expects you to know these forms and switch between T, f, and ω on the exam—practice problems: (https://library.fiveable.me/practice/ap-physics-c-mechanics). For a quick study summary see the Topic 7.2 guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw).

Short answer: gravity changes the equilibrium angle of a spring–mass (if vertical) but doesn’t change the spring’s linear restoring force, while for a pendulum gravity provides the restoring torque that causes the oscillation. Why: for a mass on an ideal spring the restoring force is Hooke’s law, F = −kx, so small oscillations about the equilibrium satisfy m d²x/dt² = −kx → ω = √(k/m) and T = 2π√(m/k). Gravity just shifts the equilibrium position (mg = kΔx) but cancels out of the linearized motion. For a simple pendulum the restoring torque comes from gravity: τ ≈ −mgL sinθ ≈ −mgL θ for small θ, giving mL² d²θ/dt² = −mgL θ → ω = √(g/L) and T = 2π√(L/g). So g appears in the pendulum period because gravity is the source of the restoring “force”; for springs the spring does that. These are exactly the Topic 7.2 formulas in the CED (Tspring = 2π√(m/k), Tpendulum = 2π√(L/g)). For a quick review, check the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw). For extra practice, see the Unit 7 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-7) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Because the restoring torque for a simple pendulum comes from gravity, the mass cancels out in the equation of motion. For small angles (small-angle approximation: sinθ ≈ θ), the tangential restoring force is F ≈ −mgθ, and the equation for angular acceleration becomes m·(l)·θ̈ = −mgθ → θ̈ + (g/l)θ = 0. The mass m appears on both sides and drops out, so the angular frequency ω = √(g/l) and the period T = 2π/ω = 2π√(l/g) (CED 7.2.A.1.ii). That’s why, for a light string and a point-mass bob displaced by a small angle, the period depends only on length l and g, not on mass. Limitations: this holds only for small angles and an ideal pendulum (massless string, point mass). For large angles, extended bobs, or a heavy rod (physical pendulum), mass distribution or nonlinear terms change the period. For more practice and the AP-aligned derivation, see the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and AP Physics C practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Start with Hooke’s law and Newton’s 2nd law. For a mass m on an ideal spring, the restoring force is F = −kx, so m d²x/dt² = −kx. Rearrange to get the standard SHM differential equation: d²x/dt² + (k/m) x = 0. Compare to d²x/dt² + ω² x = 0, so ω² = k/m and ω = sqrt(k/m). The period T is related to angular frequency by T = 2π/ω, so T = 2π sqrt(m/k). That’s the AP C essential result (CED 7.2.A.1.i). Note: T depends only on m and k, not on amplitude or phase. For a quick refresher or practice problems, see the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and the Unit 7 overview (https://library.fiveable.me/ap-physics-c-mechanics/unit-7). For extra practice, try problems at (https://library.fiveable.me/practice/ap-physics-c-mechanics).

It means you replace sinθ with θ (θ in radians) because for small angles sinθ ≈ θ. For a pendulum the tangential acceleration is −g sinθ; using the small-angle approximation gives −gθ, which turns the nonlinear equation into a linear SHM equation with ω = √(g/l) and T = 2π√(l/g) (this is why the CED’s period formula for a simple pendulum assumes a “small angle”). How small is “small”? Rough guide: θ ≲ 0.1 rad (≈6°) gives very small error; θ up to ~0.2 rad (≈11°) is often acceptable in problems. For larger amplitudes the approximation breaks down: the motion is not exactly simple harmonic and the period depends on amplitude. If you want a quick recap tied to the CED, check the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw). For practice, use Fiveable’s practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

Short answer: the period gets shorter. Why: For a mass-on-a-spring the CED gives T_s = 2π√(m/k). Increasing k (a stiffer spring) makes the ratio m/k smaller, so the square root and therefore T decrease. Equivalently, the angular frequency ω = √(k/m) increases with k, and T = 2π/ω, so bigger k → bigger ω → smaller T. Physically, a stiffer spring produces a larger restoring force for the same displacement, so the mass oscillates faster. Tip for the AP exam: be ready to use and manipulate T = 2π√(m/k) or ω = √(k/m) to predict how T or f change when m or k change (CED 7.2.A.1.i). For a quick review, check the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics).

You see 2π in every SHM period formula because T counts full cycles while ω (angular frequency) measures how fast the motion goes in radians per second. One full cycle = 2π radians, so T = (angle for one cycle)/(angular speed) = 2π/ω. In SHM the motion is mathematically the same as projection of uniform circular motion, so displacement often looks like x(t)=A cos(ωt+φ)—ωt is in radians, and one full oscillation happens when ωt increases by 2π. That’s why the mass-spring and small-angle pendulum periods include 2π after you substitute ω: - ω = √(k/m) → Ts = 2π√(m/k) - ω = √(g/l) → Tp = 2π√(l/g) This is exactly what the CED lists in 7.2.A.1 and the two essential formulas (7.2.A.1.i and 7.2.A.1.ii). For a quick refresher, see the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-mechanics) to reinforce converting between ω, f, and T.

Use the formulas from the CED and inspect the functional dependence. - Mass–spring: T = 2π√(m/k) so f = 1/(2π)√(k/m). If you increase m, f decreases like 1/√m. Example: doubling m → f goes to f/√2 (≈0.707·f). Halving m → f increases by √2 (≈1.414·f). - Simple pendulum (small-angle): T = 2π√(l/g) so f = 1/(2π)√(g/l). If you increase l, f decreases like 1/√l. Example: quadrupling l → f goes to f/2. Changing mass for an ideal simple pendulum doesn’t change f (mass cancels out). Notes: amplitude (small-angle) doesn’t change these formulas; for a real large-angle pendulum the small-angle approximation breaks down. These relations are exactly what AP Topic 7.2 tests—practice predicting factor changes by algebraically substituting new m or l. For a quick review use the Topic 7.2 study guide (https://library.fiveable.me/ap-physics-c-mechanics/unit-7/2-frequency-and-period-of-shm/study-guide/ZP4XBQORYFR7bHpw) and try more problems at Fiveable practice (https://library.fiveable.me/practice/ap-physics-c-mechanics).