An oscillator has position-time graph with peaks at 0.030 m and one full cycle every 0.80 s. State its amplitude and period. [2 points]

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How do the position, velocity and acceleration of an SHM oscillator vary with time, and how are their graphs related?

Topic 7.3 Representing and Analyzing SHM: describe the position, velocity and acceleration of an oscillator using sinusoidal functions and graphs.

A focused answer to AP Physics 1 Topic 7.3, covering the sinusoidal position function x = A cos(2 pi f t), the phase relationships between position, velocity and acceleration, reading amplitude and period from a graph, and where each quantity reaches its extremes, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The sinusoidal position function
The phase relationships
Reading and using the graphs
Try this

What this topic is asking

The College Board (Topic 7.3) wants you to represent and analyze simple harmonic motion using sinusoidal functions and graphs of position, velocity and acceleration against time, and to read amplitude and period off a graph. The key skill is knowing the phase relationships: where each quantity is at a maximum, a minimum or zero through the cycle.

The sinusoidal position function

Because SHM has a restoring force proportional to displacement, its solution is a smooth sine or cosine curve in time. Which function you use depends only on where the oscillator starts: cosine if it begins at a turning point (maximum displacement), sine if it begins at equilibrium moving outward. Both describe the same kind of motion, just started at different points in the cycle.

The phase relationships

These relationships are the analytical core of the topic. The reliable way to keep them straight is to track one full cycle starting at a turning point: there the oscillator is momentarily at rest (velocity zero) with maximum restoring force (acceleration maximum); it accelerates toward equilibrium, where it has maximum speed but zero acceleration; then it decelerates to the far turning point, where again velocity is zero and acceleration is maximum the other way.

Reading and using the graphs

A position-time graph of SHM is a sinusoid from which you read two quantities directly: the amplitude $A$ is the maximum displacement from the center line, and the period $T$ is the time for one full cycle (then $f = 1/T$ ). The velocity-time graph is the slope of the position graph, peaking where position crosses zero and vanishing at the position peaks; the acceleration-time graph is the slope of the velocity graph and is the negative mirror image of the position graph. This stacked relationship, each graph being the slope of the one above, is the same calculus-of-graphs logic you used for translational motion in Topic 1.3, now applied to oscillation. The strategic payoff is that once you can locate the extremes and zeros of all three quantities, you can answer almost any representational question in the unit: where the speed is greatest, where the force is largest, when the acceleration changes sign, and how the graphs shift if you change the amplitude (taller curves, same period) or the period (wider curves, same height). This graphical fluency is exactly what the representational free-response question rewards.

Locating the extremes from a position graph

An oscillator follows $x = 0.050\cos(4\pi t)$ (meters, seconds). Identify its amplitude and period, and state the velocity and acceleration at $t = 0$ and at the first equilibrium crossing.

step 1 Read amplitude and frequency

Comparing with $x = A\cos(2\pi f t)$ : amplitude $A = 0.050$ m, and $2\pi f = 4\pi$ so $f = 2.0$ Hz. The period is $T = 1/f = 0.50$ s.

step 2 At t = 0

The cosine is at its maximum, so $x = +A = 0.050$ m, a turning point. There the velocity is zero and the acceleration is maximum in magnitude, directed back toward equilibrium (negative).

step 3 First equilibrium crossing

Equilibrium ( $x = 0$ ) is reached a quarter period later, at $t = T/4 = 0.125$ s. There the velocity is maximum in magnitude and the acceleration is zero.

step 4 Interpret

In a quarter cycle the oscillator goes from rest with maximum acceleration (at the extreme) to maximum speed with zero acceleration (at equilibrium). This quarter-cycle phase shift between position and velocity is the signature of SHM.

Try this

Q1. An oscillator has position-time graph with peaks at $\pm 0.030$ m and one full cycle every $0.80$ s. State its amplitude and period. [2 points]

Cue. Amplitude $0.030$ m; period $0.80$ s.

Q2. State where in the cycle an SHM oscillator's acceleration has its greatest magnitude. [1 point]

Cue. At the turning points (extremes), where the displacement is largest.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)6 marksSection II (representational FRQ). An oscillator starts at maximum positive displacement and has amplitude

0.080

m and period

0.40

s. (a) Write an equation for its position as a function of time. (b) Sketch and describe the velocity-time graph, stating where the velocity is maximum and zero. (c) State the phase relationship between the acceleration and the position, and explain it physically.

Show worked answer →

A 6-point representational FRQ on the sinusoidal description of SHM.

(a) Position (2 points): starting at maximum displacement, use cosine. $x = A\cos(2\pi f t)$ with $A = 0.080$ m and $f = 1/T = 1/0.40 = 2.5$ Hz: $x = 0.080\cos(2\pi(2.5)t) = 0.080\cos(5\pi t)$ (meters, seconds).
(b) Velocity graph (2 points): the velocity is a negative sine curve, zero at the turning points (when $x = \pm A$ ) and maximum in magnitude as the oscillator passes through equilibrium ( $x = 0$ ). It is one quarter period out of step with the position.
(c) Acceleration and position (2 points): the acceleration is exactly out of phase with the position (180 degrees): when $x$ is at its maximum positive value, $a$ is at its maximum negative value. This is because $a = -\tfrac{k}{m}x$ , so acceleration is always opposite the displacement.

Markers reward the cosine form with correct amplitude and frequency, the velocity graph with extremes located, and the opposite-phase relation $a = -\tfrac{k}{m}x$ .

AP 2022 (style)1 marksSection I (multiple choice). At the instant an SHM oscillator passes through its equilibrium position, which statement is true? (A) velocity zero, acceleration maximum (B) velocity maximum, acceleration zero (C) both zero (D) both maximum. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the phase relationships. The answer is (B).

At equilibrium the displacement is zero, so the restoring force and acceleration ( $a = -\tfrac{k}{m}x$ ) are zero, while the oscillator has been accelerated all the way in and so moves at maximum speed. The trap is (A), which describes the turning points, not equilibrium.

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Sources & how we know this

AP Physics 1: Algebra-Based Course and Exam Description — College Board (2024)