Topic 7.3: Representing and Analyzing SHM

What Topic 7.3 requires

Topic 7.3 carries one learning objective, 7.3.A: describe the displacement, velocity, and acceleration of an object exhibiting SHM. Three essential knowledge statements sit under it, and two sub-statements sit under the first.

• 7.3.A.1 For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations $x = A\cos(2\pi f t)$ or $x = A\sin(2\pi f t)$.
• 7.3.A.1.i Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion.
• 7.3.A.1.ii Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion.
• 7.3.A.2 Changing the amplitude of a system exhibiting SHM will not change the period of that system.
• 7.3.A.3 Properties of SHM can be determined and analyzed using graphical representations.

Read that list again and notice what is missing. The CED hands you an equation for displacement and nothing else: no velocity equation, no acceleration equation. What the statements ask for instead is where those quantities peak and where they vanish, which is a reading skill before it is a computing skill.

The suggested skills say the same thing. The CED lists four: 1.C, create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; and 3.C, justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

The CED prints no boundary statement for Topic 7.3. It prints none for any of the four topics in Unit 7, which is weighted at 5 to 8 percent of the multiple-choice section and estimated at about 5 to 10 class periods.

Cosine or sine is decided by where the object starts

Both forms are on the AP Physics 1 equation sheet, side by side:

They describe the same motion. They differ only in what is happening at $t = 0$, so the initial condition picks one for you.

• Use cosine when the clock starts at maximum displacement. At $t = 0$, $\cos(0) = 1$, so $x = A$. This is the released-from-rest case: a block pulled aside and let go, a pendulum bob held at its highest point and dropped.
• Use sine when the clock starts at equilibrium moving in the positive direction. At $t = 0$, $\sin(0) = 0$, so $x = 0$, and $x$ grows positive immediately after. This is the struck or pushed case: a glider given a shove as it sits at the equilibrium position.

Two consequences follow, and both get tested.

First, $A$ is always the amplitude, always measured from the equilibrium position, never across the full swing. A block that travels 24 cm from one turning point to the other has an amplitude of 12 cm. That is the same displacement convention Topic 2.8 uses for $\vec{F}_s = -k\Delta\vec{x}$.

Second, the argument of the trig function is in radians, not degrees. Geometric angles elsewhere in AP Physics 1, the $\theta$ in $W = Fd\cos\theta$ or in $\tau = rF\sin\theta$, are given in degrees, so the switch is easy to miss. But $2\pi f t$ is an angle in radians by construction: after one full period, $t = T = 1/f$, the argument is $2\pi$, one complete cycle. In degree mode every SHM value you compute will be wrong and none of them will look obviously wrong. Rotational angles in Units 5 and 6 are in radians too, so this is a switch between two conventions the course already uses, not a one-off.

Throughout this page, take the positive $x$ direction to be the direction of the initial displacement and measure $x$ from the equilibrium position. Every sign below follows from that one choice.

The three graphs and how they line up

Position, velocity, and acceleration are the same three quantities you graphed for straight-line motion in Topic 1.3, and the same two slope rules apply. Velocity is the slope of the position graph; acceleration is the slope of the velocity graph. Nothing new is being asked, the curves are just periodic now.

At the turning points, where $x = \pm A$, the object has momentarily stopped. The position graph is flat there, so $v = 0$. But the restoring force is largest there, because $|F_s| = k|\Delta x|$, so the acceleration is largest too, pointing back toward equilibrium.

At the equilibrium position, where $x = 0$, the net force is zero by the definition in essential knowledge 7.1.A.2.ii, so $a = 0$. The position graph is steepest there, so the speed is at its maximum.

That is the whole phase relationship, and it answers most Topic 7.3 multiple-choice questions. Here is one full cycle of the cosine case:

Three facts from that table are standard exam distractors on their own.

Velocity and position are a quarter of a period out of step. Whenever one is at an extreme, the other is zero. They are never both zero and never both at a maximum.

Acceleration and position are in step but opposite in sign. The acceleration graph is the position graph flipped upside down and stretched, which restates the restoring force in 7.1.A.2: $ma_x = -k\Delta x$. Acceleration in SHM never points the way the object has been displaced.

Speed and acceleration are never large at the same moment. Where the object is fastest it is not accelerating at all, and where it is accelerating hardest it is not moving.

Reading a number off each graph

Essential knowledge 7.3.A.3 says properties of SHM can be determined and analyzed using graphical representations, and skill 1.C asks you to sketch those graphs yourself. In practice that means being able to point at a feature and name the quantity it gives you.

Two habits make these questions faster.

**Find $T$ first, then $f$, then everything else.** Almost every quantity here is built out of $A$ and $f$, $f$ comes from $T$, and $T$ comes straight off the horizontal axis. Reading the period wrongly poisons every later answer, so measure across several cycles and divide when the graph offers them.

Check the phase before you commit. If you have decided a graph is the velocity partner of a given position graph, confirm the velocity is zero where the position peaks. If it is not, you have picked up the acceleration graph instead.

The simple harmonic motion guide works the period and frequency calculations, and Topic 7.2 covers where the period formulas come from.

Maximum speed and maximum acceleration are derived, not given

Skill 2.A for this topic is deriving a symbolic expression by following a logical mathematical pathway, and the two maxima are the derivation this topic sets up. Neither result is printed on the equation sheet, so knowing where they come from matters more than memorising them.

Start from the spring period, which is printed:

Square both sides and rearrange, then substitute $T = 1/f$:

Now use energy. At the turning point the block is at rest and all the energy sits in the spring; at equilibrium the spring is relaxed and all of it is kinetic. Setting those equal, $\tfrac{1}{2}kA^2 = \tfrac{1}{2}mv_{max}^2$, gives $v_{max} = A\sqrt{k/m}$, and substituting the box above:

The acceleration maximum is even shorter. At $x = \pm A$ the spring force has magnitude $kA$, so by Newton's second law:

Both scale with the amplitude, which is the point of skill 2.D. Double the amplitude and you double both maxima, while the period does not move. The energy behaves differently again, quadrupling rather than doubling, which is Topic 7.4.

The same energy argument gives the speed anywhere, not just at equilibrium:

At $x = 0$ it collapses to $v_{max}$. At $x = \pm A$ it gives zero. Because of the square, the speed at half the amplitude is not half the maximum speed; it is $\sqrt{3}/2$, or about 87 percent of it.

Amplitude does not change the period (7.3.A.2)

Essential knowledge 7.3.A.2 is one sentence: changing the amplitude of a system exhibiting SHM will not change the period of that system. It is stated flatly because it is counterintuitive, and it is worth understanding rather than memorising.

Pull the block twice as far and it has twice as far to travel each quarter cycle. But the restoring force at every corresponding point is also twice as large, so the acceleration is twice as large, so it covers that doubled distance in the same time. The two effects cancel exactly, and only because the restoring force is proportional to displacement. That proportionality is the definition of SHM in 7.1.A.2, so this independence is not a coincidence of springs; it is what makes the motion simple harmonic.

The period formulas on the sheet show the same thing by omission. $T_s = 2\pi\sqrt{m/k}$ holds a mass and a spring constant and no amplitude. $T_p = 2\pi\sqrt{\ell/g}$ holds a length and a gravitational field strength, no amplitude and no mass.

Two cautions belong with this.

It is a statement about a system already exhibiting SHM. A pendulum is modelled as SHM only for small angular displacements, which is what 7.1.A.2.iii says. Swing it through a large angle and the restoring torque stops being proportional to the angular displacement, the motion stops being simple harmonic, and the period does start to depend on amplitude.

It says nothing about speed, acceleration, or energy. Those all change with amplitude. Only the period, and therefore the frequency, holds still.

What the equation sheet actually prints

Claims about the equation sheet are easy to get wrong from memory, so here is the full set of entries that touch Topic 7.3, counted off the AP Physics 1 Table of Information.

• $T = \dfrac{1}{f}$
• $T_s = 2\pi\sqrt{\dfrac{m}{k}}$
• $T_p = 2\pi\sqrt{\dfrac{\ell}{g}}$
• $x = A\cos(2\pi f t)$
• $x = A\sin(2\pi f t)$

Five entries, and both trig forms are there. A student who remembers only the cosine version is not remembering the sheet.

Two entries from the translational block also do work here: $\vec{F}_s = -k\Delta\vec{x}$ for the restoring force, and $\vec{F}_{net} = \Delta\vec{p}/\Delta t = m\Delta\vec{v}/\Delta t = m\vec{a}$, which turns that force into the acceleration you graph.

What is not printed tells you what you are expected to construct:

• No velocity equation for SHM.
• No acceleration equation for SHM.
• No expression for $v_{max}$ or $a_{max}$.
• No angular frequency $\omega = 2\pi f$. On the AP Physics 1 sheet $\omega$ appears only in the rotational entries, as the angular velocity in $\omega = \omega_0 + \alpha t$, $v = r\omega$, $K = \tfrac{1}{2}I\omega^2$, and $L = I\omega$. The oscillation entries spell the argument out as $2\pi f t$, so a solution written in terms of $\omega$ has to define $\omega$ first.

The full sheet is on the AP Physics 1 formulas page.

How Topic 7.3 is tested

The CED is specific about where oscillations show up on the free-response section. It says the second free-response question on the AP Physics 1 Exam is the Translation Between Representations question, which requires students to create graphical and verbal models of the same situation and compare those models to mathematical representations. Its oscillation example: a student might be asked to sketch free-body diagrams of a block oscillating on a spring at the maximum displacement and at equilibrium, then create energy bar charts for the block and spring system at those same two positions, then explain how the two representations are consistent.

The exception clause matters and is easy to drop. The CED adds that while Unit 7 content provides especially good practice for this question, content from any unit may be included in it. Unit 7 is not guaranteed to appear there.

Notice which two positions the CED names: maximum displacement and equilibrium. Those are where every quantity in this topic takes an extreme value or a zero, which is what 7.3.A.1.i and 7.3.A.1.ii are about. If you can describe $x$, $v$, $a$, and the net force at those two positions without hesitating, you have most of what the topic asks for.

On the multiple-choice section, expect graph-matching: given one of the three curves, pick the correct partner. Expect qualitative comparisons: at the instant shown, is the speed increasing or decreasing? Expect proportional reasoning driven by skill 2.D: the amplitude is tripled, what happens to the maximum speed, the maximum acceleration, and the period?

Errors that cost points on SHM graph questions

Calculator in degree mode. $\cos(2\pi f t)$ takes radians. This produces answers that are wrong but plausible-looking, the worst kind.

Treating amplitude as the full travel. Amplitude runs from equilibrium to one extreme. The distance covered in one full period is $4A$.

Saying acceleration is zero where the object stops. The object is momentarily at rest at the turning points, and that is exactly where the acceleration is largest. Being at rest and having zero acceleration are different conditions, and SHM separates them cleanly.

Giving acceleration the same sign as displacement. If your sketch has $a$ and $x$ peaking positive at the same instant, one of the two graphs is upside down.

Reading half a period as a full period. Consecutive midline crossings are $T/2$ apart; consecutive peaks are $T$ apart. Mixing these up doubles or halves $f$ and every quantity derived from it.

Assuming a bigger push means a longer period. It does not, by 7.3.A.2, however the change is dressed up: a stronger pull on the block, a wider swing of the pendulum.

Using the kinematic equations. $v_x = v_{x0} + a_x t$ and its relatives assume constant acceleration. In SHM the acceleration changes continuously, so they do not apply anywhere in this unit. Use the trig form or use energy, which is what Topic 7.4 supplies.