FRQ: Conical Pendulum Period — Designing a Conical-Pendulum Timer

Assessments aligned to 2026 AP Physics 1 standards — 11 points

Question Type: Qualitative/Quantitative Translation (QQT)

▤ Scenario (Live simulation follows text)

An engineer is evaluating whether a conical pendulum can serve as the timing element of a slow clock. A ball hangs from a fixed pivot by a light string of length L = 0.90 m and is driven in a horizontal circle at constant speed. The cone half-angle θ — the angle between the string and the vertical — can be set to any value between 0° and 90°. The period of the conical pendulum is T = 2π√(L cos θ / g). For comparison, a simple pendulum of the same length L, swinging with small amplitude, has period T_s = 2π√(L / g). Use g = 10 m/s².

Use the ground frame; treat every string as massless and the ball as a point particle in uniform circular motion. Angles are measured from the vertical; reason about magnitudes only.

✎ Free Response Questions

(a) For a cone half-angle θ = 53°, calculate the period T of the conical pendulum.

(b) Without recalculating from scratch, predict how the period T changes as the engineer increases θ beyond 53° toward larger angles. Justify your prediction using the functional dependence of T on θ.

(c) The engineer asks whether an angle θ can be chosen so that the conical pendulum has exactly the same period as the simple pendulum of the same length. Determine whether such an angle exists, state its value, and state whether it is physically achievable for a genuine conical pendulum.

(d) A student argues: “A conical pendulum makes a poor clock because its period depends on how fast you launch the ball.” Evaluate whether the dependence the student names is the correct one, then construct a complete argument for the real reason a conical pendulum is a poor timekeeper, comparing it with the small-angle simple pendulum.

❖ Answer Key & Scoring Guide

▸ earns credit  ⚠ common error, partial credit  ✗ common error, no credit

Part (a) — Model Answer

Use cos 53° = 0.60.

T = 2π√(L cos θ / g) = 2π√( (0.90)(0.60) / 10 )

T = 2π√(0.054) = 2π(0.2324) = 1.46 s

The period is T ≈ 1.46 s.

The vertical depth of the cone, L cos θ, acts as the effective pendulum length—dictating the timing of the orbit regardless of how wide the horizontal radius gets.

Scoring (3 points):

▸ 1 point: Correct substitution of L = 0.90 m, cos 53° = 0.60, g = 10 into T = 2π√(L cos θ/g).

▸ 1 point: Correct quantity inside the root, L cos θ/g = 0.054 s², giving √ = 0.232 s.

▸ 1 point: Correct period T ≈ 1.46 s, with units.

⚠︎ Common error (partial credit): Using sin 53° = 0.80 instead of cos 53° (giving T ≈ 1.69 s) — earns 1 of 3 points, for correctly applying the formula's structure; the substitution and final-value points are lost.

Part (b) — Model Answer

Examine how T = 2π√(L cos θ/g) responds to θ with L and g fixed.

T ∝ √(cos θ) ⇒ as θ increases, cos θ decreases, so T decreases

As θ increases, cos θ decreases, so L cos θ decreases and the period T = 2π√(L cos θ/g) gets shorter. The period decreases as the cone opens wider.

A wider angle demands drastically more tension to sustain the required centripetal acceleration, forcing the mass to clear a larger circumference in significantly less time.

Scoring (2 points):

▸ 1 point: States that T decreases as θ increases.

▸ 1 point: Justifies via functional dependence — cos θ decreases with θ, so L cos θ and hence T (∝ √cos θ) decrease.

✗ Common error (no credit): Claiming T increases (“bigger circle, longer trip”) — ignores that the speed rises too; earns 0 of 2 points.

Part (c) — Model Answer

Set the two periods equal and solve for θ.

2π√(L cos θ / g) = 2π√(L / g)

cos θ = 1 ⇒ θ = 0°

The only solution is θ = 0°, which is not physically achievable: a genuine conical pendulum needs θ > 0° so the ball actually sweeps a horizontal circle. Therefore the conical pendulum's period is always less than that of a simple pendulum of equal length, approaching it only in the limit θ → 0°.

A standard simple pendulum represents the absolute upper limit for the period; the moment the mass sweeps a horizontal plane, the cycle accelerates.



Scoring (2 points):

▸ 1 point: Sets T = T_s and reduces to cos θ = 1, i.e. θ = 0°.

▸ 1 point: Concludes θ = 0° is not achievable for a real conical pendulum (no circular motion), so T < T_s for all valid angles.

⚠︎ Common error (partial credit): Answering only “no” without the θ = 0° limiting-case derivation — earns 1 of 2 points.

Part (d) — Model Answer

Relate the named variable (launch speed) to the model's actual variables (L, θ).

T = 2π√(L cos θ / g) (no m, no v)    with v = √(g L sin θ tan θ)

The student is practically correct because the launch speed determines the cone angle θ — a higher launch speed v results in a larger cone angle θ. Since T = 2π√(L cos θ / g), changing the launch speed changes θ, which changes the period. A good clock must be isochronous: its period must be independent of amplitude or energy. As a simple pendulum loses energy, its amplitude decreases but T remains constant. If a conical pendulum loses energy, its speed drops, its angle decreases, and its period increases — making it a poor timekeeper.

Scoring (4 points):

▸ 1 point: Connects launch speed to the resulting cone angle θ (higher v ⇒ larger θ).

▸ 1 point: Connects the change in angle θ to the resulting change in period T via T = 2π√(L cos θ/g).

▸ 1 point: States that a useful clock must have a period independent of its amplitude or energy (isochronism).

▸ 1 point: Contrasts the conical pendulum's shifting period with the simple pendulum's constant period as energy is lost.