Bring out a vertical spring with a 200 g mass hanging on it and set it oscillating with an amplitude of about 5 cm. Ask students to watch ONE specific moment: the instant the mass passes through equilibrium. Ask: 'Right now, at this equilibrium moment — is the mass speeding up, slowing down, or moving at constant speed for that instant? And what is the net force on it right now?' Take a quick hand vote on three options: (A) moving fastest, zero net force; (B) momentarily stopped, maximum net force; (C) moving fastest, maximum net force. Do not resolve — tell students that by the end of class they will be able to defend the correct answer using three time graphs. Then flip the question: 'What about at the very top of the bounce? Speed? Force?' This surfaces the misconception that velocity and force peak together. Keep the spring oscillating in view for the rest of the lesson.

1. 7m

   Position as a sinusoid — reading x(t)

   Content

   For an ideal spring–mass oscillator released from rest at x = +A, the position is x(t) = A cos(ωt), where A is the amplitude in meters and ω = 2π/T is the angular frequency in rad/s. The graph is a cosine curve that peaks at +A at t = 0, crosses zero at t = T/4, reaches −A at t = T/2, crosses zero again at t = 3T/4, and returns to +A at t = T. Worked example: a glider on a spring with T = 2.0 s and A = 0.10 m gives ω = π rad/s and x(t) = 0.10 cos(πt) m. At t = 0.50 s, x = 0.10 cos(π·0.5) = 0 m — the equilibrium crossing. At t = 1.0 s, x = −0.10 m — the far turning point. The amplitude is set by how far you pulled the mass; the angular frequency is set by k and m through ω = √(k/m) and does NOT depend on A.

   Delivery

   Anchor students on the two features they will read off every graph: amplitude (vertical extent) and period (horizontal repeat). Have them call out T and A from the projected curve before you name them. Emphasize that ω uses radians, not degrees — a frequent calculator error. Ask, 'At t = T/4, where is the mass physically?' to link the graph to the spring you left oscillating. Pre-empt the confusion between frequency f and angular frequency ω = 2πf.

2. 7m

   Velocity and acceleration — the quarter-cycle offsets

   Content

   Velocity is the slope of the position graph, and acceleration is the slope of the velocity graph. Taking derivatives of x(t) = A cos(ωt) gives v(t) = −Aω sin(ωt) and a(t) = −Aω² cos(ωt) = −ω²x(t). Three consequences follow directly. First, v leads x by T/4: when x is at its peak +A, v = 0; a quarter cycle later v reaches its extreme −Aω while x = 0. Second, a is exactly opposite in sign to x at every instant, because a = −ω²x — the two graphs are mirror images across the time axis. Third, the peak magnitudes are vₘₐₓ = ωA and aₘₐₓ = ω²A. Worked example continued: with ω = π rad/s and A = 0.10 m, vₘₐₓ = 0.31 m/s (at every equilibrium crossing) and aₘₐₓ = 0.99 m/s² (at every turning point). At t = 0.50 s the mass is at x = 0, so a = 0 and |v| is maximum. At t = 1.0 s the mass is at x = −0.10 m, so v = 0 and a = +0.99 m/s² pointing back toward equilibrium.

   Delivery

   Walk the stacked x-t, v-t, a-t graphs on the slide left to right and narrate what is happening on the spring at each vertical gridline. Hammer on two claims: (1) speed is MAX at equilibrium, not at the endpoints — this is the biggest misconception; (2) acceleration and position have OPPOSITE signs at every instant because the force always pulls back toward equilibrium. Have students point at the a-t curve and predict where the mass physically is at that moment. If a student says 'the mass stops at the turning point so the force is zero,' correct on the spot: stopped does not mean force-free — a = −ω²x is largest there.

3. 6m

   Reading and constructing the graphs — the procedure

   Content

   Given any one of the three graphs, you can construct the other two by applying three rules in order. Rule 1 — Amplitudes scale: if the position amplitude is A, the velocity amplitude is ωA, and the acceleration amplitude is ω²A. Rule 2 — Zeros and extrema swap: wherever x has an extremum, v = 0; wherever x = 0, v has an extremum. The same rule applies from v to a. Rule 3 — Signs of a and x are opposite everywhere, because a = −ω²x. Worked example: suppose you are handed a v-t graph that is a positive sine curve of amplitude 0.20 m/s with period 1.0 s. Then ω = 2π rad/s, so A = vₘₐₓ/ω = 0.20/(2π) ≈ 0.032 m and aₘₐₓ = ωvₘₐₓ = 2π·0.20 ≈ 1.26 m/s². Because v = +vₘₐₓ sin(ωt) came from x = −A cos(ωt) (slope of that x gives +Aω sin), the position graph is an inverted cosine: it starts at x = −0.032 m and rises. Acceleration a = +ω²A cos(ωt) starts at its positive maximum. Check with a = −ω²x: at t = 0, x = −A so a = +ω²A. Consistent.

   Delivery

   Give students the three-rule procedure as a checklist they can literally run down on the AP exam. Model the rules once with the v-t → x-t, a-t example above, thinking aloud. Then flip the direction: give them 20 seconds to sketch, in their notes, the v-t and a-t for a position graph you describe verbally (positive cosine, A = 0.05 m, T = 4.0 s), and pick two students to compare answers. Reinforce the slope error: the slope of the x-t graph gives velocity, NOT acceleration — reading the wrong slope is one of the most common graph-analysis errors on this standard.

1. 27m

   Motion-sensor SHM: discovering the quarter-cycle phase shiftsLab

   Groups of 3. Set up a cart between two springs on a track with a motion sensor at one end. Students record ~5 s of oscillation, then use the three stacked graphs to extract T, A, vₘₐₓ, and aₘₐₓ, verify vₘₐₓ = ωA and aₘₐₓ = ω²A, and confirm the quarter-cycle phase relationships. Targets SP 1 Creating Representations (reading and annotating three time-aligned graphs), SP 2 Mathematical Routines (computing ω, vₘₐₓ, aₘₐₓ and comparing to measured peaks), and SP 3 Argumentation (justifying the phase rule from the data and from ΣF = −kx). Circulate and check that groups (i) use the same time axis on all three graphs, (ii) do not confuse ω with f, and (iii) actually mark the equilibrium crossings on the position graph. About 5 minutes before end, cold-call two groups to share their measured phase shift and their calculated vs. measured aₘₐₓ. Student handout (below) is the full document — copy directly. Lab: Reading SHM off Three Graphs Setup. Cart between two springs on a level track. Motion sensor at one end, aimed at the cart. Displace the cart about 10 cm from equilibrium and release. Record 5 s of data. Turn on position, velocity, AND acceleration graphs, stacked with a shared time axis. Part 1 — Read the position graph. - Period T = ______ s - Amplitude A = ______ m - Angular frequency ω = 2π/T = ______ rad/s - Times of the first two equilibrium crossings: t = ______ s and t = ______ s - Times of the first two turning points: t = ______ s and t = ______ s Part 2 — Predict, then measure. Using ONLY A and ω from Part 1, predict: - Predicted maximum speed vₘₐₓ = ωA = ______ m/s - Predicted maximum acceleration aₘₐₓ = ω²A = ______ m/s² Now read directly off the graphs: - Measured peak |v| = ______ m/s - Measured peak |a| = ______ m/s² Percent difference for vₘₐₓ: ______ %. Percent difference for aₘₐₓ: ______ %. Part 3 — Phase shifts. Pick one full cycle in the middle of your data. On the printed sketch grid below, sketch all three curves stacked with a shared time axis. Mark: - Every point where x = 0 (label them E for equilibrium) - Every point where x = ±A (label them T for turning point) Answer: - At each E, what is v? ________________ - At each E, what is a? ________________ - At each T, what is v? ________________ - At each T, what is a? ________________ - Phase difference between x and v, in fractions of T: ______ - Phase difference between x and a, in fractions of T: ______ Part 4 — Argue from the physics. In 2–3 sentences, use the restoring force ΣF = −kx to explain WHY the acceleration graph is always the mirror image of the position graph across the time axis. Reference the sign of the force at x = +A and at x = −A in your answer. Part 5 — Prediction check. Without touching the equipment, predict what will happen to T, vₘₐₓ, and aₘₐₓ if you (a) double the amplitude, (b) double the cart's mass. Then test one of the two predictions and report the result. - (a) Doubling A: T ______, vₘₐₓ ______, aₘₐₓ ______ - (b) Doubling m: T ______, vₘₐₓ ______, aₘₐₓ ______ - Which did you test? ______ Result: ______

   Materials

   • Vernier or PASCO motion sensor with interface
   • Data-collection software (LoggerPro, Capstone, or Graphical Analysis) set to display three stacked graphs: position, velocity, acceleration vs. time
   • Low-friction dynamics cart on a track
   • Two springs (spring constant roughly 5–15 N/m) to attach cart between track endpoints
   • Meter stick
   • Ruler and pencil for student sketches
   • Printed lab handout (one per student)

   Example outputs

   • Part 1: T = 1.10 s, A = 0.095 m, ω = 5.71 rad/s. First equilibrium crossings at t ≈ 0.28 s and t ≈ 0.83 s (separated by T/2). First turning points at t ≈ 0.55 s and t ≈ 1.10 s.
   • Part 2: Predicted vₘₐₓ = 5.71 × 0.095 = 0.54 m/s, measured 0.52 m/s, 3.8% difference. Predicted aₘₐₓ = 5.71² × 0.095 = 3.10 m/s², measured 2.95 m/s², 4.8% difference — within the noise of the sensor's numerical derivative.
   • Part 3: At every E, v = ±vₘₐₓ and a = 0. At every T, v = 0 and a = ±aₘₐₓ. Phase difference x-to-v = T/4; phase difference x-to-a = T/2.
   • Part 4: 'At x = +A the spring force is −kA, pointing back toward equilibrium (negative), so a is negative and maximum in magnitude. At x = −A the force is +kA (positive), so a is positive and maximum. Because a = −(k/m)x at every instant, the a-t curve is exactly x-t flipped across the time axis.'
   • Part 5: T is unchanged by doubling A (period depends only on k and m). Doubling A doubles both vₘₐₓ and aₘₐₓ. Doubling m increases T by √2 and decreases vₘₐₓ (= ωA) and aₘₐₓ (= ω²A) accordingly.

   No-equipment fallback

   Provide each group a printed, labeled x-t graph of a cart oscillating with A = 0.080 m and T = 1.20 s. Students carry out Parts 1–4 above using slope estimation on the printed graph in place of sensor data, then compare their sketched v-t and a-t curves to a printed 'answer key' curve at the 20-minute mark.

1. A mass on a spring oscillates with x(t) = (0.12 m) cos((4.0 rad/s) t). At the instant t = π/8 s ≈ 0.39 s, which statement best describes the motion? (A) x is maximum positive, v = 0, a is maximum positive. (B) x = 0, |v| is maximum, a = 0. (C) x is maximum negative, v = 0, a is maximum positive. (D) x = 0, v = 0, a is maximum negative.

   multiple choice(B). At t = π/8 s, ωt = π/2, so cos(ωt) = 0 → x = 0 (equilibrium crossing). Since a = −ω²x, a = 0. Speed is maximum at equilibrium: |v| = ωA = 4.0 × 0.12 = 0.48 m/s. Targets SP 2 Mathematical Routines (evaluating sinusoids at a specified phase) and SP 3 Argumentation (justifying which quantities are extreme where).

2. The velocity-versus-time graph of a mass on a spring is a positive sine curve with amplitude 0.30 m/s and period 0.50 s. Determine (a) the amplitude of the position-time graph, (b) the maximum acceleration, and (c) the value of the acceleration at t = 0. Justify (c) using the relation between a and x.

   calculationω = 2π/T = 2π/0.50 = 4π rad/s ≈ 12.57 rad/s. (a) A = vₘₐₓ/ω = 0.30 / (4π) ≈ 0.024 m. (b) aₘₐₓ = ω·vₘₐₓ = 4π × 0.30 ≈ 3.77 m/s². (c) v(t) = +vₘₐₓ sin(ωt) implies (by integrating or by the quarter-cycle rule) x(t) = −A cos(ωt), so at t = 0, x = −A. Then a = −ω²x = +ω²A = +3.77 m/s². The acceleration is at its maximum positive value at t = 0 because the mass is at the negative turning point and the spring is pulling it back toward equilibrium. Targets SP 1 (constructing the x-t and a-t representations from the v-t graph) and SP 2 (numerical routines).

3. A student claims: 'The mass is moving fastest at the endpoints because the spring is stretched the most there, so it pushes back hardest.' In 2–3 sentences, identify what is correct and what is wrong in this claim, and state where in the cycle the speed is actually maximum and why.

   short answerThe student is correct that the spring force (and thus the acceleration) is largest at the endpoints, since ΣF = −kx has maximum magnitude at x = ±A. But 'largest force' does not mean 'largest speed' — at the endpoints the mass is momentarily at rest (v = 0) as it reverses direction. Speed is maximum at the equilibrium crossing (x = 0), where the mass has been accelerated over the full amplitude and the net force is instantaneously zero; |v|ₘₐₓ = ωA. Targets SP 3 Argumentation (distinguishing force from velocity and justifying from ΣF = −kx).

• ☐Vertical spring and 200 g hanging mass for the hook demo
• ☐Ring stand or support to hang the demo spring
• ☐One motion sensor per group (4–6 groups) with interface and cables
• ☐Laptops or lab computers running LoggerPro / Capstone / Graphical Analysis, one per group
• ☐One low-friction dynamics track per group
• ☐Two springs (k ≈ 5–15 N/m) per group and one dynamics cart per group
• ☐Meter sticks (one per group)
• ☐Printed lab handout (one per student)
• ☐Slide deck cued to display: stacked x-t/v-t/a-t graphs, annotated single sine with equilibrium and turning-point labels, and the five-snapshot oscillator strip with x, v, a arrows