AP Physics 1: Graphs of SHM

Understanding graphs of simple harmonic motion is not just a test requirement; it's a fundamental skill that unlocks how oscillating systems behave in physics and engineering. From car suspensions to quantum oscillators, visualizing position, velocity, and acceleration over time allows you to predict motion, diagnose systems, and solve complex problems. Mastering these graphs will solidify your grasp of SHM for the AP Physics 1 exam and beyond.

Defining Simple Harmonic Motion and Its Key Quantities

Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that displacement. Think of a mass on a frictionless horizontal spring: when you pull it and let go, it oscillates back and forth. This motion is described by several key quantities. The amplitude (A) is the maximum displacement from equilibrium. The period (T) is the time for one complete cycle of motion, and the frequency (f) is the number of cycles per second, where $f=1/T$. The angular frequency ($\omega$) relates to period by $\omega =2\pi f=2\pi /T$. These quantities are the building blocks for creating and interpreting the graphs you'll study.

Graphing Position Versus Time: x(t)

The position of an object in SHM as a function of time is typically modeled by a cosine or sine function. For a system starting from maximum displacement, the equation is $x(t)=A\cos (\omega t+\varphi )$, where $\varphi$ is the phase constant that sets the initial position. The graph of x(t) is a smooth, periodic cosine wave. The crests represent the maximum positive displacement (+A), the troughs represent the maximum negative displacement (-A), and the points where the graph crosses the time axis are the equilibrium positions (x=0). For example, if a spring-mass system has an amplitude of 0.2 meters and a period of 2 seconds, its angular frequency is $\omega =2\pi /2=\pi$ rad/s. The position graph would be $x(t)=0.2\cos (\pi t)$, oscillating between +0.2 m and -0.2 m every 2 seconds.

Graphing Velocity Versus Time: v(t)

Velocity is the rate of change of position. For SHM, you find it by taking the derivative of x(t): $v(t)=\frac{dx}{dt}=-A\omega \sin (\omega t+\varphi )$. This graph is a sine wave shifted relative to the position graph. The key insight is the phase relationship: velocity is 90 degrees or $\pi /2$ radians ahead of position. When position is at its maximum (or minimum), the velocity is zero because the object momentarily stops to reverse direction. Conversely, when the object passes through equilibrium (x=0), its speed is at a maximum. The maximum velocity is $v_{max}=A\omega$. In our spring-mass example, $v_{max}=0.2\times \pi \approx 0.628$ m/s. The v(t) graph would cross zero when x(t) is at a peak and reach its own peaks when x(t) is zero.

Graphing Acceleration Versus Time: a(t)

Acceleration is the rate of change of velocity. Taking the derivative of v(t) gives $a(t)=\frac{dv}{dt}=-A{\omega }^2\cos (\omega t+\varphi )$. Notice this is proportional to negative position: $a(t)=-{\omega }^{2}x(t)$, which directly reflects Hooke's Law for a spring ($F=-kx$) and Newton's second law. The acceleration graph is a cosine wave like position, but inverted and with a different amplitude. It has a phase relationship of being 180 degrees or $\pi$ radians out of phase with position, and 90 degrees ahead of velocity. Acceleration is maximum (and negative) when displacement is maximum positive, because the restoring force pulls it back toward equilibrium. The maximum acceleration is $a_{max}=A{\omega }^2$. For our example, $a_{max}=0.2\times {\pi }^2\approx 1.97$ m/s².

Extracting Physical Information from the Graphs

You can determine all key SHM parameters directly from well-labeled graphs. From the x(t) graph, the amplitude is the vertical distance from the equilibrium line to a peak. The period is the horizontal time interval for one complete cycle, which is identical on all three graphs. Once you have the period T, you can calculate angular frequency $\omega =2\pi /T$. To find maximum velocity, you can read it directly from the peaks of the v(t) graph, or compute it using $v_{max}=A\omega$ if you've extracted A and T. Similarly, maximum acceleration comes from the a(t) graph or $a_{max}=A{\omega }^2$. For instance, if a v(t) graph shows peaks at ±0.5 m/s and the x(t) graph shows an amplitude of 0.1 m, you can verify consistency: $v_{max}=A\omega$ implies $\omega =0.5/0.1=5$ rad/s, and then period $T=2\pi /\omega \approx 1.26$ s.

Common Pitfalls

Confusing the phase relationships is a frequent error. Remember: velocity leads position by 90°, and acceleration is 180° out of phase with position. A common mistake is to think acceleration and velocity are in phase. Correction: When velocity is maximum, acceleration is zero because the object is at equilibrium where the restoring force is zero.

Misinterpreting slope on x(t) and v(t) graphs. The slope of an x(t) graph at any point gives instantaneous velocity, not acceleration. Similarly, the slope of a v(t) graph gives instantaneous acceleration. Students often swap these. Correction: Practice by picking a point on x(t), drawing a tangent line, and calculating its slope to find v at that time.

Incorrectly reading amplitude or period from mislabeled axes. Always check the units and scale. If the x(t) graph oscillates between +5 cm and -5 cm, the amplitude is 5 cm, not 10 cm. The period is the time from one peak to the next, not from a peak to a trough. Correction: Highlight one full cycle on the graph before taking measurements.

Forgetting that graphs are based on calculus relationships. When asked to sketch v(t) from x(t), some students draw a mirrored copy instead of the derivative. Correction: Recall that where x(t) has a maximum (zero slope), v(t) must be zero. Where x(t) crosses zero with steepest slope, v(t) has a maximum.

Summary

• Position x(t) in SHM is a cosine wave: $x(t)=A\cos (\omega t+\varphi )$. Its graph oscillates between +A and -A, with the period T easily measured between peaks.
• Velocity v(t) is the derivative of position: $v(t)=-A\omega \sin (\omega t+\varphi )$. It leads position by 90°; velocity is zero at maximum displacement and maximum at equilibrium.
• Acceleration a(t) is the derivative of velocity: $a(t)=-A{\omega }^2\cos (\omega t+\varphi )$. It is 180° out of phase with position, directly proportional to negative displacement ($a=-{\omega }^{2}x$).
• Phase relationships are crucial: v leads x by 90°, and a is opposite to x (180° phase difference).
• Extract physical parameters like amplitude (from x(t) peaks), period (from cycle time on any graph), and maximum velocity/acceleration (from v(t) and a(t) peaks or via $v_{max}=A\omega$, $a_{max}=A{\omega }^2$).
• Avoid pitfalls by remembering derivative links (slope gives rate), carefully reading axes, and internalizing the phase shifts between these three graphs.