AP Physics 1 · Topic 7.4

Topic 7.4: Energy of Simple Harmonic Oscillators

Q: What is the total energy of a mass on a spring?

For a spring and object system oscillating with amplitude A, the total mechanical energy is E_{total} = \tfrac{1}{2}kA^2, where k is the spring constant. That is the relevant equation the AP Physics 1 CED attaches to essential knowledge 7.4.A.4.ii. It comes from evaluating the elastic potential energy U_s = \tfrac{1}{2}k(\Delta x)^2 at the turning point, where the object is momentarily at rest so all of the energy is stored in the spring. Because the total is conserved when nothing drains the system, that same number describes the oscillator at every instant, with the split between kinetic and potential shifting as the object moves. Notice what the expression leaves out: mass and time. Two different masses pulled to the same amplitude on the same spring store the same energy, though they reach different maximum speeds and have different periods.

Q: If you double the amplitude of an oscillator, what happens to its energy?

It quadruples. The total energy of a spring oscillator is E_{total} = \tfrac{1}{2}kA^2, so the energy depends on the square of the amplitude: doubling A multiplies E by 2^2 = 4, and tripling A multiplies it by 9. Essential knowledge 7.4.A.4.ii states the qualitative version, that changing the amplitude changes the maximum potential energy and therefore the total energy of the system. Other quantities scale differently under the same change, and sorting them out is what the exam asks for. The maximum speed and the maximum acceleration are both linear in A, so both double when A doubles. The period and the frequency do not change at all, by essential knowledge 7.3.A.2. Guessing that energy scales linearly is one of the most common errors in this topic.

Q: Is the SHM energy equation on the AP Physics 1 equation sheet?

No. E_{total} = \tfrac{1}{2}kA^2 appears in the CED as a relevant equation under essential knowledge 7.4.A.4.ii, but it is not printed on the AP Physics 1 Table of Information. Neither is E_{total} = U + K. What the sheet does give you is everything needed to build them: the elastic potential energy U_s = \tfrac{1}{2}k(\Delta x)^2, the kinetic energy K = \tfrac{1}{2}mv^2, the near-surface gravitational relation \Delta U_g = mg\Delta y, and the spring force \vec{F}_s = -k\Delta\vec{x}. Evaluate U_s at the turning point, where the speed is zero, and \tfrac{1}{2}kA^2 falls straight out. The five oscillation entries that are printed are T = 1/f, T_s = 2\pi\sqrt{m/k}, T_p = 2\pi\sqrt{\ell/g}, x = A\cos(2\pi f t), and x = A\sin(2\pi f t).

Q: Does the spring energy formula work for a pendulum?

No. E_{total} = \tfrac{1}{2}kA^2 is stated in the CED specifically as the relevant equation for a spring and object system, and a pendulum has no spring constant to put into it. A pendulum stores gravitational potential energy instead, so use mgh, where h = \ell(1 - \cos\theta) is the height of the bob above the lowest point of the swing and \theta is the string's angle from the vertical, measured in degrees for that cosine. Setting mgh equal to \tfrac{1}{2}mv_{max}^2 gives v_{max} = \sqrt{2gh}, with the mass cancelling out. What does carry over is the general statement 7.4.A.1, that the total energy of a system exhibiting SHM is the sum of its kinetic and potential energies, and the pattern of maxima: potential energy peaks at the turning points, kinetic energy peaks at the lowest point.

Unit 7: Oscillations5-8% of the multiple-choice section

A spring oscillator's energy shuttles between elastic potential energy and kinetic energy. With nothing draining the system, the total stays fixed at one half k A squared, set by the amplitude alone. Potential energy peaks at the turning points and kinetic energy peaks at equilibrium.

AP Physics: Unit 7 (topics 7.4 Energy of Simple Harmonic Oscillators). Topic 7.4 carries a single CED learning objective, 7.4.A, describe the mechanical energy of a system exhibiting SHM, with four essential knowledge statements: 7.4.A.1 gives the relevant equation E_total = U + K; 7.4.A.2 states that conservation of energy indicates the total energy is constant; 7.4.A.3 and 7.4.A.4 pair maximum kinetic energy with minimum potential energy and the reverse; 7.4.A.4.i states that the minimum kinetic energy is zero; and 7.4.A.4.ii states that changing the amplitude changes the maximum potential energy and therefore the total energy, with the relevant equation for a spring and object system given as E_total = (1/2) k A squared, which is not itself printed on the AP Physics 1 equation sheet. The CED's suggested skills are 1.A, 2.B, 2.C, and 3.B, and it prints no boundary statement for this topic or for any topic in Unit 7. Unit 7 is weighted at 5 to 8 percent of the multiple-choice section and estimated at about 5 to 10 class periods.

What Topic 7.4 requires

Topic 7.4 carries one learning objective, 7.4.A: describe the mechanical energy of a system exhibiting SHM. Four essential knowledge statements sit under it, with two sub-statements under the last.

7.4.A.1 The total energy of a system exhibiting SHM is the sum of the system's kinetic and potential energies. The CED gives the relevant equation $E_{total} = U + K$ .
7.4.A.2 Conservation of energy indicates that the total energy of a system exhibiting SHM is constant.
7.4.A.3 The kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum.
7.4.A.4 The potential energy of a system exhibiting SHM is at a maximum when the system's kinetic energy is at a minimum.
7.4.A.4.i The minimum kinetic energy of a system exhibiting SHM is zero.
7.4.A.4.ii Changing the amplitude of a system exhibiting SHM will change the maximum potential energy of the system and, therefore, the total energy of the system. The CED gives, as the relevant equation for a spring and object system, $E_{total} = \tfrac{1}{2}kA^2$ .

Two things about that list are worth noticing.

The whole topic is Unit 3 applied to repeating motion. Nothing here is a new principle. Conservation of energy is Topic 3.4, elastic potential energy belongs to Topic 3.3, and the accounting is the same accounting. What Unit 7 adds is that the two forms swap places twice per cycle, on a schedule you can predict.

Only two equations are named, and only one is system-specific. $E_{total} = U + K$ applies to any oscillator; $E_{total} = \tfrac{1}{2}kA^2$ is stated for a spring and object system, and the CED says so in as many words. Statements 7.4.A.3 and 7.4.A.4 are comparisons rather than calculations.

The CED lists four suggested skills for this topic: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Skill 1.A is what makes the energy bar chart a Topic 7.4 skill rather than a nice extra, and 2.C appears here and nowhere else in Unit 7.

The CED prints no boundary statement for Topic 7.4, or for any of the four topics in Unit 7. Unit 7 is weighted at 5 to 8 percent of the multiple-choice section, with an estimate of about 5 to 10 class periods.

The total energy is set by the amplitude alone

Take a block on a horizontal spring. At the turning point, $x = \pm A$ , the block is momentarily at rest, so $K = 0$ and every joule sits in the spring. Evaluating the elastic potential energy there:

E_{total} = U_{s,max} = \tfrac{1}{2}k A^2

That is the relevant equation the CED attaches to 7.4.A.4.ii. Because the total is conserved, this one number describes the system at every instant, not just at the turning point.

Three features of that expression are each a test question waiting to happen.

Mass does not appear. Two blocks of different mass pulled to the same amplitude on the same spring store the same energy. They reach different maximum speeds, because $\tfrac{1}{2}mv_{max}^2$ must equal the same total, and they have different periods, because $T_s = 2\pi\sqrt{m/k}$ does contain the mass.

Time does not appear. The total is the same at every instant. What changes with time is the split between the two forms.

The amplitude appears squared. Increase $A$ by a factor of 3 and the total energy grows by a factor of 9. Getting that linear instead of quadratic is a standard wrong answer.

Once you have the total, everything else follows by subtraction. At any position $x$ the spring holds $U_s = \tfrac{1}{2}k(\Delta x)^2$ , so

K = \tfrac{1}{2}kA^2 - \tfrac{1}{2}kx^2

and the speed comes from $K = \tfrac{1}{2}mv^2$ . That is the whole computational content of the topic, and it is skill 2.B in one line.

Where each form peaks, and the table that answers most questions

Statements 7.4.A.3 and 7.4.A.4 are a matched pair: kinetic energy is at a maximum when potential energy is at a minimum, and potential energy is at a maximum when kinetic energy is at a minimum. Statement 7.4.A.4.i pins down the second half of that: the minimum kinetic energy of a system exhibiting SHM is zero.

That zero matters. It says the block genuinely stops at the turning point, so the full total sits in potential energy there, which is what makes $E_{total} = \tfrac{1}{2}kA^2$ exact rather than approximate.

Working in fractions of the total is faster than working in joules, because the fractions are the same for every spring oscillator:

Position	$U_s$ as a fraction of $E$	$K$ as a fraction of $E$	Speed
$x = 0$	$0$	$1$	$v_{max}$
$x = \pm A/2$	$1/4$	$3/4$	$0.87\,v_{max}$
$x = \pm A/\sqrt{2}$	$1/2$	$1/2$	$0.71\,v_{max}$
$x = \pm A$	$1$	$0$	$0$

Read the second row carefully. At half the amplitude the spring holds only a quarter of the energy, because $U_s$ goes as $x^2$ , so three quarters is still kinetic and the block moves at $\sqrt{3}/2$ of its top speed, about 87 percent. Half is the tempting wrong answer. The third row is worth the same attention: the energy splits evenly not at half the amplitude but at $A/\sqrt{2}$ , roughly 71 percent of the way out.

One timing fact rounds this out. The energy swaps completely twice per period, once on each side of equilibrium, so a graph of $K$ against time repeats at twice the frequency of the position graph in Topic 7.3.

Constant total energy assumes nothing is draining the system

Statement 7.4.A.2 says conservation of energy indicates that the total energy of a system exhibiting SHM is constant. Be exact about the condition hiding in that sentence.

The total is constant when no energy leaves the system. On a frictionless surface, with the spring treated as ideal and massless and air resistance ignored, energy shuttles between the spring and the block forever and $\tfrac{1}{2}kA^2$ never changes. Those idealisations are the model, and the CED's statements describe the model.

A real oscillator is damped. Friction and air resistance do negative work on the block, mechanical energy leaves the system as thermal energy, and the total falls cycle after cycle. Because $E = \tfrac{1}{2}kA^2$ ties the total to the amplitude, a falling total means a shrinking amplitude: this is exactly why a real pendulum's swings get narrower and eventually stop. The period, though, does not visibly change as the amplitude decays, which is 7.3.A.2 doing its work.

So when a question says a block "eventually comes to rest", it is telling you the system is not isolated and $\tfrac{1}{2}kA^2$ is not conserved. When it says "frictionless" or "ideal spring", it is licensing the conserved total. Read for that word before you write an energy equation. The general framework, including how to account for the energy that leaves, is in the conservation of energy guide. There is no damped SHM equation in AP Physics 1: energy loss is handled with the Unit 3 machinery, not by adjusting the SHM equations.

What the half k A squared formula does and does not cover

The CED is careful with its wording, and copying that care will save you marks. $E_{total} = \tfrac{1}{2}kA^2$ is introduced as the relevant equation for a spring and object system. It is not a general SHM formula, and it is not on the equation sheet at all.

It does not apply to a pendulum. A pendulum has no spring constant. Writing $\tfrac{1}{2}kA^2$ for a swinging bob is a physics error, not a slip.

It does apply to a vertical spring, with one adjustment. Hang a block from a spring and it settles at a new equilibrium where the spring is already stretched by $mg/k$ . Measure displacement from that hanging equilibrium, not from the relaxed length. The elastic and gravitational potential energies then combine into $\tfrac{1}{2}k(\Delta y)^2$ plus a constant, because their linear terms cancel exactly, so the oscillation energy is $\tfrac{1}{2}kA^2$ with $A$ measured from the hanging equilibrium. The gravitational term has not disappeared; it has been absorbed. This goes beyond what the CED states, but it is why vertical and horizontal spring problems are solved identically.

The general statement is the one to lean on. $E_{total} = U + K$ , from 7.4.A.1, covers every oscillator. Identify what kind of potential energy the system stores, add the kinetic energy, and set the sum equal at two instants. That is skill 3.B, and it works whether the restoring agent is a spring or gravity.

The equation sheet supplies the pieces for either case: $U_s = \tfrac{1}{2}k(\Delta x)^2$ , $K = \tfrac{1}{2}mv^2$ , $\Delta U_g = mg\Delta y$ , and $\vec{F}_s = -k\Delta\vec{x}$ . The whole sheet is on the AP Physics 1 formulas page.

The pendulum trades gravitational potential energy instead

Essential knowledge 7.1.A.2.iii establishes that the motion of a pendulum with a small angular displacement can be modelled as simple harmonic motion, because the restoring torque is proportional to the angular displacement, so a pendulum is inside the scope of Topic 7.4. But Topic 7.4's own statements never mention a pendulum, and the only relevant equation they name for a specific system is the spring one. The pendulum version has to be assembled out of 7.4.A.1 and the Unit 3 material.

It assembles easily. The system is the bob and the Earth, the potential energy is gravitational, and the trade is between height and speed:

mg h_{max} = \tfrac{1}{2}mv_{max}^2 \quad\Rightarrow\quad v_{max} = \sqrt{2gh_{max}}

The mass cancels, which is the same cancellation that makes $T_p = 2\pi\sqrt{\ell/g}$ independent of the bob's mass. The one piece of geometry you need is the height of the bob above its lowest point when the string makes an angle $\theta$ with the vertical:

h = \ell(1 - \cos\theta)

That is the vertical drop of the bob: the string's vertical projection is $\ell\cos\theta$ , so the bob sits $\ell - \ell\cos\theta$ above the bottom of the swing. Use $g = 9.8$ m/s $^2$ , the value printed in the AP Physics 1 Table of Information. You will also meet $g \approx 10$ m/s $^2$ , and both come from the CED: its Topic 1.3 boundary statement says that for all situations in which a numerical quantity is required for $g$ , the value $g \approx 10$ m/s $^2$ will be used, and then adds that students will not be penalised for correctly using the more precise commonly accepted values of $9.81$ m/s $^2$ or $9.8$ m/s $^2$ .

Two warnings. First, $\theta$ is measured from the vertical and is a geometric angle, so it wants your calculator in degrees, unlike the phase $2\pi f t$ in Topic 7.3, which is in radians. Second, the SHM model holds only for small angles: a bob released from 60 degrees still obeys energy conservation, so the speed calculation is fine, but its period no longer matches $T_p = 2\pi\sqrt{\ell/g}$ and it is no longer performing SHM.

Energy bar charts, the representation the CED asks for

Suggested skill 1.A for this topic is creating diagrams, tables, charts, or schematics to represent physical situations, and the CED spells out what that looks like on the exam. It states that the second free-response question on the AP Physics 1 Exam is the Translation Between Representations question, and gives this 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 with each other.

The exception clause belongs with that, and dropping it would misdescribe the exam: the CED adds that while Unit 7 content provides especially good practice for this question, content from any unit may be included in it.

A bar chart is one bar per energy form, drawn to the same vertical scale at every snapshot. For a frictionless horizontal block and spring system, the two positions the CED names look like this:

At maximum displacement. The $U_s$ bar is at full height, the $K$ bar is zero. The free-body diagram shows the spring force at its largest, pointing back toward equilibrium, with weight and normal force cancelling vertically.
At equilibrium. The $U_s$ bar is zero and the $K$ bar is at full height, the same height the $U_s$ bar had. The free-body diagram shows no horizontal force at all.

The consistency the question asks you to articulate runs between those two panels: the spring force is largest exactly where the stored energy is largest, and zero exactly where the stored energy is zero, because both are governed by the same displacement from equilibrium, through $|F_s| = k|\Delta x|$ and $U_s = \tfrac{1}{2}k(\Delta x)^2$ .

Two rules keep bar charts from losing points. Total height must match across every snapshot, unless energy is leaving, in which case add a bar for the thermal energy generated and keep the grand total matching. And label the system, because potential energy belongs to a system rather than to the block alone.

How Topic 7.4 is tested, and the errors that cost points

The CED's own sample activity for this topic previews the reasoning style. It describes a cart wiggling on a horizontal spring; a blob of clay is dropped onto the cart and sticks, either when the cart is at the centre or at one end, and students explain what happened to the period, total energy, amplitude, and maximum speed. Notice that it asks about four quantities at once and that the answer depends on where the clay lands. Dropped at the turning point, where the cart is at rest, the clay adds no kinetic energy and leaves the amplitude and total energy alone. Dropped at equilibrium, where the cart is moving, it makes an inelastic collision that removes kinetic energy and so shrinks both the total energy and the amplitude. The period lengthens either way, because $T_s = 2\pi\sqrt{m/k}$ grew.

Beyond that, the recurring errors are these.

Halving the speed at half the amplitude. At $x = A/2$ the spring holds a quarter of the energy and the speed is about 87 percent of the maximum.

Doubling the energy when the amplitude doubles. It quadruples.

Assuming a bigger amplitude means a longer period. The energy changes, the period does not. Statements 7.3.A.2 and 7.4.A.4.ii sit next to each other saying opposite-looking things about the same change.

**Using $\tfrac{1}{2}kA^2$ for a pendulum.** There is no $k$ . Use $mgh$ with $h = \ell(1 - \cos\theta)$ .

Measuring spring displacement from the wrong place. For a hanging spring the oscillation is about the stretched equilibrium, not the relaxed length.

Crediting the energy to the object alone. It is stored in the block and spring system, or the bob and Earth system, a convention set in Topic 3.3.

Conserving energy in a problem that says friction. If the amplitude is shrinking, the total is not constant.

Total energy, maximum speed, and the speed partway in

A $0.35$ kg cart sits on a frictionless horizontal track, attached to a spring with $k = 140$ N/m. It is pulled $0.080$ m from equilibrium and released from rest. Find the total energy of the system, the maximum speed of the cart, the speed when the cart is $0.040$ m from equilibrium, and the fraction of the total energy that is kinetic at that point.

Identify the amplitude. Released from rest at $0.080$ m means the cart has no kinetic energy there, so that is a turning point and $A = 0.080$ m.
Total energy. $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(140)(0.080)^2 = \tfrac{1}{2}(140)(0.0064) = 0.448$ J, so $E = 0.45$ J to two significant figures.
Maximum speed. At equilibrium the spring is relaxed, so all $0.448$ J is kinetic: $\tfrac{1}{2}mv_{max}^2 = 0.448$ J, giving $v_{max}^2 = 2(0.448)/0.35 = 2.56$ and $v_{max} = 1.6$ m/s.
**Potential energy at $x = 0.040$ m.** $U_s = \tfrac{1}{2}k(\Delta x)^2 = \tfrac{1}{2}(140)(0.040)^2 = \tfrac{1}{2}(140)(0.0016) = 0.112$ J.
Kinetic energy there. By 7.4.A.1 the two must add to the total, so $K = 0.448 - 0.112 = 0.336$ J.
Speed there. $v^2 = 2K/m = 2(0.336)/0.35 = 1.92$ , so $v = 1.386$ m/s, which is $1.4$ m/s.
Fraction kinetic. $0.336/0.448 = 0.75$ , so three quarters of the energy is kinetic. That matches the table above: at half the amplitude, $U_s$ takes a quarter and $K$ takes three quarters.
Check the speed against the fraction. $v/v_{max}$ should be $\sqrt{0.75} = 0.866$ , and $1.386/1.6 = 0.866$ . Consistent. Note that halfway out in position is nowhere near halfway down in speed.

The total energy is $0.45$ J, the maximum speed is $1.6$ m/s, and at $0.040$ m from equilibrium the cart moves at $1.4$ m/s with $75$ percent of the energy still kinetic.

What changes when the amplitude changes

The same $0.35$ kg cart on the same $k = 140$ N/m spring is now pulled to $0.12$ m instead of $0.080$ m and released. Find the new total energy, the new maximum speed, the new maximum acceleration, and the new period. State each as a factor of change as well as a number.

Factor of change in amplitude. $0.12/0.080 = 1.5$ .
Total energy. $E \propto A^2$ , so the factor is $1.5^2 = 2.25$ . Numerically, $E = \tfrac{1}{2}(140)(0.12)^2 = \tfrac{1}{2}(140)(0.0144) = 1.008$ J, so $1.0$ J, and $1.008/0.448 = 2.25$ as predicted. This is essential knowledge 7.4.A.4.ii in action: the larger amplitude raises the maximum potential energy and therefore the total.
Maximum speed. From $\tfrac{1}{2}mv_{max}^2 = E$ , speed goes as $\sqrt{E}$ , which goes as $A$ , so the factor is $1.5$ . Numerically, $v_{max}^2 = 2(1.008)/0.35 = 5.76$ and $v_{max} = 2.4$ m/s, which is $1.5 \times 1.6$ m/s.
Maximum acceleration. At the turning point the spring force is $kA$ , so $a_{max} = kA/m$ , linear in $A$ and again a factor of $1.5$ . Numerically, $a_{max} = (140)(0.12)/0.35 = 48$ m/s $^2$ , up from $(140)(0.080)/0.35 = 32$ m/s $^2$ .
Period. $T_s = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.35/140} = 2\pi\sqrt{0.0025} = 2\pi(0.050) = 0.314$ s. The amplitude does not enter, so the factor of change is exactly $1$ , per 7.3.A.2. The period is $0.31$ s before and after.
Read the pattern. One change to the amplitude produced three different scalings: quadratic for energy, linear for both maxima, and none at all for the period. Sorting quantities by how they scale is what suggested skill 2.C is for.

$E = 1.0$ J (up by a factor of $2.25$ ), $v_{max} = 2.4$ m/s (up by $1.5$ ), $a_{max} = 48$ m/s $^2$ (up by $1.5$ ), and $T = 0.31$ s, unchanged.

A pendulum, where the potential energy is gravitational

A $0.60$ kg bob hangs from a light string of length $1.2$ m. It is pulled aside until the string makes an angle of $15$ degrees with the vertical and released from rest. Find the height gained, the total energy of the swing relative to the lowest point, and the maximum speed of the bob. Use $g = 9.8$ m/s $^2$ .

Height above the lowest point. $h = \ell(1 - \cos\theta) = 1.2(1 - \cos 15^\circ) = 1.2(1 - 0.9659) = 1.2(0.03407) = 0.04089$ m, about $4.1$ cm. Calculator in degree mode for this step.
Total energy. At release the bob is at rest, so all of the energy is gravitational potential: $E = mgh = (0.60)(9.8)(0.04089) = 0.2404$ J, so $0.24$ J. Note that $\tfrac{1}{2}kA^2$ is unusable here; there is no spring constant in this system.
Maximum speed. At the lowest point the potential energy is back to zero, so all $0.2404$ J is kinetic: $\tfrac{1}{2}mv_{max}^2 = 0.2404$ J, giving $v_{max}^2 = 2(0.2404)/0.60 = 0.8014$ and $v_{max} = 0.895$ m/s, so $0.90$ m/s.
Notice the mass cancelling. Combining the two steps gives $v_{max} = \sqrt{2gh} = \sqrt{2(9.8)(0.04089)} = 0.895$ m/s with no mass in sight. A bob of any mass on this string released from $15$ degrees arrives at the bottom at the same speed. The total energy does depend on the mass; the speed does not.
Cross-check against the SHM model. The period is $T_p = 2\pi\sqrt{\ell/g} = 2\pi\sqrt{1.2/9.8} = 2.199$ s. The arc from the lowest point out to the turning point is $s = \ell\theta = 1.2(0.2618) = 0.3142$ m, with $\theta$ in radians this time. Treating that arc as the amplitude, $v_{max} = 2\pi f s = (2\pi/2.199)(0.3142) = 0.898$ m/s, within $0.3$ percent of the energy answer. That is the small-angle approximation showing its accuracy at $15$ degrees; the two methods drift apart at larger angles, where the motion stops being simple harmonic.

The bob rises $0.0409$ m, the swing carries $0.24$ J of energy, and the maximum speed at the bottom is $0.90$ m/s. The speed is independent of the bob's mass; the energy is not.

Frequently asked questions

What is the total energy of a mass on a spring?

For a spring and object system oscillating with amplitude $A$ , the total mechanical energy is $E_{total} = \tfrac{1}{2}kA^2$ , where $k$ is the spring constant. That is the relevant equation the AP Physics 1 CED attaches to essential knowledge 7.4.A.4.ii. It comes from evaluating the elastic potential energy $U_s = \tfrac{1}{2}k(\Delta x)^2$ at the turning point, where the object is momentarily at rest so all of the energy is stored in the spring. Because the total is conserved when nothing drains the system, that same number describes the oscillator at every instant, with the split between kinetic and potential shifting as the object moves. Notice what the expression leaves out: mass and time. Two different masses pulled to the same amplitude on the same spring store the same energy, though they reach different maximum speeds and have different periods.

Where is kinetic energy maximum in simple harmonic motion?

At the equilibrium position, where the potential energy is at its minimum. Essential knowledge 7.4.A.3 states the rule directly: the kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum. For a horizontal spring, equilibrium is where the spring is relaxed, so $U_s = 0$ and the entire total is kinetic. For a pendulum it is the lowest point of the swing, where the bob has descended as far as it will go. The reverse holds at the turning points: the object is momentarily at rest there, so its kinetic energy is zero, which is essential knowledge 7.4.A.4.i, and the potential energy carries the whole total. The complete swap happens twice per period, so a graph of kinetic energy against time repeats twice as often as the position graph.

If you double the amplitude of an oscillator, what happens to its energy?

It quadruples. The total energy of a spring oscillator is $E_{total} = \tfrac{1}{2}kA^2$ , so the energy depends on the square of the amplitude: doubling $A$ multiplies $E$ by $2^2 = 4$ , and tripling $A$ multiplies it by $9$ . Essential knowledge 7.4.A.4.ii states the qualitative version, that changing the amplitude changes the maximum potential energy and therefore the total energy of the system. Other quantities scale differently under the same change, and sorting them out is what the exam asks for. The maximum speed and the maximum acceleration are both linear in $A$ , so both double when $A$ doubles. The period and the frequency do not change at all, by essential knowledge 7.3.A.2. Guessing that energy scales linearly is one of the most common errors in this topic.

Is the SHM energy equation on the AP Physics 1 equation sheet?

No. $E_{total} = \tfrac{1}{2}kA^2$ appears in the CED as a relevant equation under essential knowledge 7.4.A.4.ii, but it is not printed on the AP Physics 1 Table of Information. Neither is $E_{total} = U + K$ . What the sheet does give you is everything needed to build them: the elastic potential energy $U_s = \tfrac{1}{2}k(\Delta x)^2$ , the kinetic energy $K = \tfrac{1}{2}mv^2$ , the near-surface gravitational relation $\Delta U_g = mg\Delta y$ , and the spring force $\vec{F}_s = -k\Delta\vec{x}$ . Evaluate $U_s$ at the turning point, where the speed is zero, and $\tfrac{1}{2}kA^2$ falls straight out. The five oscillation entries that are printed are $T = 1/f$ , $T_s = 2\pi\sqrt{m/k}$ , $T_p = 2\pi\sqrt{\ell/g}$ , $x = A\cos(2\pi f t)$ , and $x = A\sin(2\pi f t)$ .

How do you find the speed of an oscillator at a given position?

Use conservation of energy rather than the kinematic equations, which do not apply because the acceleration in SHM is not constant. Write the total as $E = \tfrac{1}{2}kA^2$ , subtract the potential energy stored at the position you care about, $U_s = \tfrac{1}{2}kx^2$ , and set the remainder equal to $\tfrac{1}{2}mv^2$ . Solving gives $v = \sqrt{(k/m)(A^2 - x^2)}$ , which can also be written $v = 2\pi f\sqrt{A^2 - x^2}$ . Two checkpoints tell you the algebra is right: at $x = 0$ it reduces to the maximum speed, and at $x = \pm A$ it gives zero. Because of the square, positions and speeds do not scale together. At half the amplitude the object still has three quarters of its energy as kinetic energy and is moving at about 87 percent of its top speed.

Does the spring energy formula work for a pendulum?

No. $E_{total} = \tfrac{1}{2}kA^2$ is stated in the CED specifically as the relevant equation for a spring and object system, and a pendulum has no spring constant to put into it. A pendulum stores gravitational potential energy instead, so use $mgh$ , where $h = \ell(1 - \cos\theta)$ is the height of the bob above the lowest point of the swing and $\theta$ is the string's angle from the vertical, measured in degrees for that cosine. Setting $mgh$ equal to $\tfrac{1}{2}mv_{max}^2$ gives $v_{max} = \sqrt{2gh}$ , with the mass cancelling out. What does carry over is the general statement 7.4.A.1, that the total energy of a system exhibiting SHM is the sum of its kinetic and potential energies, and the pattern of maxima: potential energy peaks at the turning points, kinetic energy peaks at the lowest point.

What happens to the energy of a real oscillator that slowly stops?

It leaves the system, mostly as thermal energy generated by friction and air resistance. Essential knowledge 7.4.A.2 says conservation of energy indicates that the total energy of a system exhibiting SHM is constant, and that is true of the idealised system the model describes: frictionless surface, ideal spring, no air resistance. A real oscillator is damped, so mechanical energy drains away and the total falls with each cycle. Because the total is tied to the amplitude through $E = \tfrac{1}{2}kA^2$ , a falling total shows up directly as a shrinking amplitude, which is why a real pendulum's swings get narrower until it hangs still. The period, however, stays essentially unchanged as the amplitude decays, since the period does not depend on the amplitude. AP Physics 1 has no damped SHM equation; energy that leaves is handled with the Unit 3 energy accounting.