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

Assessments aligned to 2026 AP Physics 1 standards

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  |  8 points

▤ Scenario

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. The ball's circular path lies in a horizontal plane and its speed is constant. 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.” In a paragraph-length response, decide whether the period of this conical pendulum depends on the launch speed, justifying your decision from the quantities that appear in T = 2π√(L cos θ / g). Then state the property a good clock must have, and use it to compare the conical pendulum with the small-amplitude simple pendulum as both lose energy.

❖ 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.

Figure 1. The conical pendulum at θ = 53°, drawn in projection. The vertical depth L cos θ = 0.54 m is what sets the period; the readouts give L, θ, L cos θ and T.

Scoring (2 points):

▸ 1 point: Starts from T = 2π√(L cos θ / g) and identifies L cos θ — the vertical depth of the cone — as the effective pendulum length.

▸ 1 point: Substitutes L = 0.90 m, cos 53° = 0.60 and g = 10 m/s², and evaluates to T ≈ 1.46 s with units.

⚠︎ Common error (partial credit): Using sin 53° = 0.80 in place of cos 53°, giving T ≈ 1.69 s — the structural point stands, the substitution-and-evaluation point is lost; earns 1 of 2 points.

✗ Common error (no credit): Using the simple-pendulum period T = 2π√(L / g) and reporting 1.88 s — the cone angle plays no part, so neither criterion is met; earns 0 of 2 points.

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.

Figure 2. Period T against cone half-angle θ for L = 0.90 m and g = 10 m/s². The marked point is the part (a) case, θ = 53°, T = 1.46 s; T falls to zero as θ approaches 90°.

Scoring (2 points):

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

▸ 1 point: Justifies it from the functional dependence — T ∝ √(cos θ), and cos θ decreases as θ increases.

⚠︎ Common error (partial credit): States correctly that T decreases but justifies it only by “the string swings out further”, with no reference to cos θ — earns 1 of 2 points.

✗ Common error (no credit): Claiming T increases (“a wider circle is a longer trip”) — the circumference does grow, but the speed grows faster; 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 that the ball actually sweeps a horizontal circle. The conical pendulum's period is therefore 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.

Figure 3. The same string length L = 0.90 m used as a conical pendulum at θ = 30° and as a small-amplitude simple pendulum. Both strings are of equal length; they are drawn in projection and are not to a common scale.

Scoring (2 points):

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

▸ 1 point: States that θ = 0° is not achievable for a genuine conical pendulum, so T < T_s at every valid angle.

⚠︎ Common error (partial credit): Answering only that no such angle exists, without reducing to cos θ = 1 — earns 1 of 2 points for the physical conclusion; the derivation point is lost.

✗ Common error (no credit): Setting cos θ = 0 and reporting θ = 90° — inverts the condition, and that angle would give T = 0 rather than T_s; earns 0 of 2 points.

Part (d) — Model Answer

Relate the variable the student names to the variables the model actually contains.

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

The student is right, and the mechanism runs through the cone angle. The launch speed fixes θ: a larger v gives a larger θ, because v = √(g L sin θ tan θ) rises with θ. The period then follows from T = 2π√(L cos θ / g), so changing the launch speed changes the period.

A good clock must be isochronous — its period must not depend on amplitude or energy. A simple pendulum at small amplitude has that property: as it loses energy its amplitude decays but T = 2π√(L / g) still holds, because at small angular displacement the restoring torque is proportional to the displacement. (Beyond small amplitude even a simple pendulum's period drifts.) A conical pendulum has no such property: as it loses energy its speed drops, θ closes, cos θ rises, and T lengthens. That is why it is the poorer timekeeper.

Scoring (2 points):

▸ 1 point: Establishes the mechanism — a larger launch speed gives a larger cone angle θ, and T = 2π√(L cos θ / g) then gives a shorter period, so the period does depend on the launch speed.

▸ 1 point: Names isochronism — a period independent of amplitude or energy — and applies it both ways: the small-amplitude simple pendulum keeps its period as its amplitude decays, while the conical pendulum's period lengthens as θ closes.

⚠︎ Common error (partial credit): Establishes the speed–angle–period chain correctly but never names the isochronism requirement and never compares with the simple pendulum — earns 1 of 2 points.

⚠︎ Common error (partial credit): States the isochronism requirement and the comparison correctly but asserts the speed dependence without routing it through θ — earns 1 of 2 points.

✗ Common error (no credit): Claiming the period is independent of the launch speed because v does not appear in T = 2π√(L cos θ / g) — the launch speed enters through θ, which does appear; earns 0 of 2 points.