Energy in SHM

Pluck the low string of a guitar and it sings for several seconds — at 110 vibrations every second, that is close to a thousand complete swings before the note fades. Nothing keeps pushing the string. What sustains the sound is a nearly lossless trade: the energy your fingertip deposited sloshes back and forth between motion (the string sweeping through its straight shape at top speed) and stretch (the string paused, taut, at the edge of its arc). The total barely leaks. A child on a swing runs the same two-column ledger — all height at the ends of the arc, all speed at the bottom — and so does every clock pendulum, every vibrating bridge, and every atom jiggling in a crystal. You already know how an oscillator moves: the previous modules gave you $a = -\omega^2 x$ as the definition of SHM and the sinusoidal timetable $x(t)$ , $v(t)$ , $a(t)$ that follows from it. This module asks the money question: where does the energy live at each instant? The answer is astonishingly clean. The total is a constant, $E = \tfrac12 kA^2$ , fixed by the stiffness and the amplitude alone — and the two accounts, kinetic and potential, trade the whole sum back and forth twice every cycle. For exams this budget is a superpower. "Find the speed at $x$ " collapses to one line. "At what displacement is the energy half kinetic?" becomes a two-second ratio. And the fraction questions that JEE, NEET and every board love — 75% kinetic at half amplitude, equal shares at $A/\sqrt2$ — all fall out of one parabola sketch you can draw in five seconds.

Key takeaways

The total mechanical energy of SHM is constant: $E = \tfrac12 kA^2 = \tfrac12 m\omega^2 A^2$ — set by stiffness and amplitude, independent of time and of where the particle happens to be.
At displacement $x$ : potential energy $U = \tfrac12 kx^2$ , kinetic energy $K = \tfrac12 k(A^2 - x^2)$ . Their graphs against $x$ are two parabolas that always add up to the same horizontal line.
Because $E \propto A^2$ , doubling the amplitude quadruples the energy — the square law is the fingerprint of SHM.
In time, $K$ and $U$ oscillate as $\sin^2$ and $\cos^2$ : each peaks twice per cycle, so energy sloshes at frequency $2f$ , double the motion's frequency.
The landmark fractions: at $x = A/2$ the motion is 75% kinetic; at $x = A/\sqrt2$ the split is exactly 50–50; averaged over a full cycle each account holds $E/2$ .
Key idea: SHM is a fixed energy budget $\tfrac12 kA^2$ sloshing between a kinetic and a potential account at twice the oscillation frequency — amplitude sets the budget, position sets the split.

The core insight

In SHM the total energy $E = \tfrac12 kA^2 = \tfrac12 m\omega^2 A^2$ never changes; at displacement $x$ it is split as $U = \tfrac12 kx^2$ stored and $K = \tfrac12 k(A^2 - x^2)$ moving, so amplitude fixes the budget and position fixes the split.

Where this lives today

The energy budget of SHM is the working currency of modern precision measurement. The mirrors of the LIGO gravitational-wave detectors hang as pendulums whose thermal energy budget — each quadratic account holding its equal average share, exactly equation (6) — sets the noise floor against which passing spacetime ripples must be read. Atomic force microscopes measure surfaces by watching the energy budget of a vibrating cantilever change as atomic forces add stiffness; the shift in $\omega$ betrays forces of piconewtons. And vibration energy harvesters — the chips that power wireless sensors on bridges and train bogies — are deliberately built as SHM systems that skim $\tfrac12 kA^2$ out of ambient shaking, cycle after cycle. Meanwhile every guitar note, struck bell and MRI-magnet quench alarm you hear is a two-account budget being slowly taxed by friction into sound and heat — which is precisely the story of the damping module two steps ahead.

The physical picture

Watch a child on a swing at the two special moments. At the highest point of the arc she hangs motionless for a heartbeat — all the energy is stored as height, none as motion. Sweeping through the bottom she is fastest — all motion, no storage. In between, the books are split. The total never changes; it only migrates between two accounts, and it makes the full round trip twice every swing: stored → moving → stored → moving → stored is one complete cycle of the motion. A block on a spring keeps the same books with "height" replaced by "stretch". At the turning points $x = \pm A$ the block is momentarily at rest and the spring is at maximum stretch or squeeze: the whole budget sits in the spring. At the centre the spring is relaxed and the block is at top speed: the whole budget is motion. The cleanest way to see the entire story in one picture is the energy diagram below: draw the potential-energy parabola $U = \tfrac12 kx^2$ , then rule a horizontal line at the total $E$ . The motion lives where the line is above the parabola. The line meets the curve exactly at $x = \pm A$ — that is why the particle turns around there — and at any position in between, the vertical gap between line and curve is the kinetic energy. Every energy question in this chapter is a question about that gap.

Figure

The energy diagram — one parabola and one horizontal line answer everything

One more thing the picture makes obvious: the particle hurries through the middle and lingers near the edges. Near $x = 0$ the gap (kinetic energy) is at its widest, so the particle is fast; near $\pm A$ the gap pinches to zero, so it crawls. At half amplitude the gap is still three-quarters of the total — the particle still has 87% of its top speed. That lopsidedness is why a pendulum seems to "pause" at the ends of its swing and blur through the bottom.

The energy budget

Everything in this module flows from two entries in the ledger. The potential energy is the work you must do against the restoring force to move the particle from the centre to $x$ . The force you fight grows linearly from $0$ to $kx$ , so its average is $\tfrac12 kx$ , and the work is average force times distance:

U(x) = \tfrac12\,k\,x^2 \qquad (1)

where: $k$ — the stiffness (for a spring, literally the spring constant; for any other SHM, the effective stiffness $m\omega^2$ ); $x$ — displacement measured from the mean position, not from a wall or a ceiling. Equation (1) is a parabola: storing energy at double the displacement costs four times as much, because both the force and the distance doubled. This triangle-area argument is exact for a linear force — no calculus needed, though if you have met integration you can recognise $\tfrac12 kx^2$ as the area under the $F$ – $x$ line. The kinetic energy comes from the speed law of the previous module, $v = \pm\omega\sqrt{A^2 - x^2}$ . Square it, multiply by $\tfrac12 m$ , and use $m\omega^2 = k$ :

K(x) = \tfrac12\,m\,v^2 = \tfrac12\,k\,(A^2 - x^2) \qquad (2)

where: $A$ — the amplitude. Kinetic energy is an upside-down parabola in $x$ : maximum $\tfrac12 kA^2$ at the centre, zero at the turning points. Notice that (1) and (2) are perfect complements — one grows as $x^2$ , the other shrinks as $x^2$ — so their sum has no $x$ in it at all:

E = K + U = \tfrac12\,k\,A^2 = \tfrac12\,m\,\omega^2 A^2 = \tfrac12\,m\,v_{\max}^2 \qquad (3)

where: the three forms are the same number dressed for different data. Use $\tfrac12 kA^2$ when you know the stiffness, $\tfrac12 m\omega^2 A^2$ when you know the timing, $\tfrac12 mv_{\max}^2$ when you know the top speed $v_{\max} = \omega A$ . The equality of the first two rests on $k = m\omega^2$ — checking that identity is the fastest sanity test of any SHM energy answer. And read the headline: the budget depends on amplitude squared. Double the pull before you release, and you have deposited four times the energy.

you want	use	notes
total energy from amplitude	$E = \tfrac12 kA^2$	needs the stiffness
total energy from timing	$E = \tfrac12 m\omega^2 A^2$	best when $T$ or $f$ is given
speed at position $x$	$K = E - U$ , then $v = \sqrt{2K/m}$	or directly $v = \omega\sqrt{A^2-x^2}$
the split at position $x$	$\dfrac{U}{E} = \dfrac{x^2}{A^2}$ , $\dfrac{K}{E} = 1 - \dfrac{x^2}{A^2}$	the ratio form kills most MCQs
where $K = U$	$x = \pm A/\sqrt2 \approx \pm 0.707A$	equal split, NOT at $A/2$
cycle-averaged shares	$\langle K\rangle = \langle U\rangle = E/2$	equal on time-average

Derivation AThe invariant budget — why the total cannot depend on position

We derive the constancy of $E = K + U$ for SHM using nothing but the work–energy theorem and the two parabolas, then extract the master ratio $U/E = x^2/A^2$ that solves most exam questions in one line. The strategy: build $U$ from the triangle area under the force graph, import the speed law from SHM kinematics, and watch the $x^2$ terms cancel. Watch for the one identity doing all the work: $k = m\omega^2$ .

Step 1. Potential energy from the force graph. The restoring force has magnitude $F = kx$ . To park the particle at displacement $x$ you must do work against a force that climbs linearly from $0$ to $kx$ ; a linear climb averages to $\tfrac12 kx$ . Work $=$ average force $\times$ distance $= \tfrac12 kx \cdot x = \tfrac12 kx^2$ . This is equation (1). Why we may use the average: for a straight-line graph the area under it is exactly the triangle $\tfrac12 \times \text{base} \times \text{height}$ — the same rule that gave the work stored in a stretched spring in the Hooke's-law module.

Step 2. Kinetic energy from the speed law. SHM kinematics (previous module) established $v^2 = \omega^2(A^2 - x^2)$ . Multiply by $\tfrac12 m$ : $K = \tfrac12 m\omega^2 (A^2 - x^2)$ .

Step 3. Trade $m\omega^2$ for $k$ . The definition of SHM is $a = -\omega^2 x$ ; Newton's second law reads $F = ma = -m\omega^2 x$ ; comparing with $F = -kx$ forces $k = m\omega^2$ . So $K = \tfrac12 k(A^2 - x^2)$ , equation (2).

Step 4. Add the accounts. $E = K + U = \tfrac12 k(A^2 - x^2) + \tfrac12 kx^2 = \tfrac12 kA^2$ . The $x^2$ terms cancel identically — not approximately, not on average, but at every single instant. The total is a constant fixed by $k$ and $A$ alone: equation (3).

Step 5. The master ratio. Divide (1) by (3): $\dfrac{U}{E} = \dfrac{x^2}{A^2}$ , and hence $\dfrac{K}{E} = 1 - \dfrac{x^2}{A^2}$ . Every "what fraction of the energy is kinetic at $x = \ldots$ " question is this one line. At $x = A/2$ : $U/E = 1/4$ , $K/E = 3/4$ . At $x = A/\sqrt2$ : both $1/2$ . Verified: at $x = \pm A$ the ratio gives $U = E$ , $K = 0$ — the particle is at rest at the turning points ✓; at $x = 0$ it gives $K = E = \tfrac12 mv_{\max}^2$ with $v_{\max} = \omega A$ , matching SHM kinematics ✓; dimensions: $kA^2$ is $\mathrm{(N/m)\,m^2 = N\,m = J}$ ✓.

The energy movie — double time

The diagram above freezes the budget against position. Run it against time and a famous subtlety appears. Take $x = A\cos\omega t$ (released from rest at $+A$ ). Then $v = -\omega A\sin\omega t$ , and the two accounts read

K(t) = \tfrac12 kA^2\sin^2\omega t, \qquad U(t) = \tfrac12 kA^2\cos^2\omega t \qquad (4)

where: the two shares are $\sin^2$ and $\cos^2$ of the same angle, so they sum to $\tfrac12 kA^2$ by the Pythagorean identity — conservation made visible. But look at the tempo. A squared sinusoid never goes negative and repeats every half period of the original: $\sin^2$ has hills at $\omega t = \pi/2$ and $3\pi/2$ . Kinetic energy peaks every time the particle crosses the centre — moving right and moving left — which happens twice per cycle. So:

f_{\text{energy}} = 2f_{\text{motion}} \qquad (5)

where: if a block bounces at 2 Hz, its energy sloshes between the accounts at 4 Hz. Exam setters love this: "the frequency of oscillation of the kinetic energy of a particle executing SHM with frequency $f$ is…" — answer $2f$ , and the graph of $K(t)$ they show you touches zero twice per period, always staying above the axis. A related freebie: since $\sin^2$ and $\cos^2$ are mirror images sharing the interval symmetrically, each account averages to exactly half the budget over a full cycle:

\langle K\rangle = \langle U\rangle = \tfrac12 E \qquad (6)

where: $\langle\ \rangle$ means the time average over one period. On average the oscillator keeps equal balances in both accounts — a fact worth a mark on its own, and the seed of deep physics (in thermal physics each such "quadratic account" claims an equal share of thermal energy).

Figure

Kinetic and potential energy in time — the budget sloshes at twice the frequency

A caution about the vertical spring, because it wrecks more energy solutions than any other setup: hang the block, let it sag to its new equilibrium, and measure $x$ from that sagged point. Then $U = \tfrac12 kx^2$ with the same $k$ , and gravity has vanished from the books — the gravitational and elastic contributions repackage themselves into one clean parabola about the new centre. If you insist on measuring stretch from the natural length you must carry $mgh$ terms separately, and the two bookkeeping styles must never be mixed in one equation. Measure from equilibrium; it is the whole reason the vertical oscillator behaves identically to the horizontal one.

Limiting cases

case	result	meaning
$x = 0$	$K = E$ , $U = 0$ , $v = \omega A$	centre: all motion — the speed peak
$x = \pm A$	$U = E$ , $K = 0$	turning points: all storage — the pause
$x = \pm A/2$	$K = \tfrac34 E$ , $U = \tfrac14 E$	half the distance out ≠ half the energy — only a quarter is stored
$x = \pm A/\sqrt2$	$K = U = \tfrac12 E$	the true 50–50 point, at 70.7% of the amplitude
$A \to 2A$	$E \to 4E$	the square law: amplitude is the budget's only knob (for fixed $k$ )
$k \to 4k$ , same $A$	$E \to 4E$ , $f \to 2f$	stiffer well: more energy stored and faster sloshing

Common trap“At half the amplitude, the energy is half kinetic, half potential”

It feels symmetric and it is wrong. Energy goes as the square of displacement, so at $x = A/2$ the stored share is $(1/2)^2 = 1/4$ of the budget — the motion is still 75% kinetic. The genuine 50–50 point sits at $x = A/\sqrt2 \approx 0.707A$ , less than a third of the way in from the turning point. Every fraction question in this chapter is a squared ratio: $U/E = x^2/A^2$ , never $x/A$ . The same square bites in reverse: to double a pendulum's swing you must inject four times the energy, and a string plucked twice as far sounds four times as loud in energy terms. When an MCQ offers you both $A/2$ and $A/\sqrt2$ as options, the examiner is checking exactly this square.

Worked examples

1 · Board level — the full audit (CBSE/ISC pattern)

A block of mass $0.50\ \mathrm{kg}$ on a smooth table is attached to a spring of stiffness $k = 50\ \mathrm{N/m}$ , pulled $10\ \mathrm{cm}$ from equilibrium and released. Find (a) the total energy, (b) the maximum speed, (c) the kinetic and potential energies at $x = 6.0\ \mathrm{cm}$ , and (d) the speed there.

Step 1. Total budget from the release point: $E = \tfrac12 kA^2 = \tfrac12 \times 50 \times (0.10)^2 = \boxed{0.25\ \mathrm{J}}$ . Deposited once, at release; conserved forever after.

Step 2. Maximum speed, where the budget is all kinetic: $\tfrac12 mv_{\max}^2 = 0.25$ , so $v_{\max} = \sqrt{2 \times 0.25/0.50} = \boxed{1.0\ \mathrm{m/s}}$ . Cross-check via kinematics: $\omega = \sqrt{k/m} = \sqrt{100} = 10\ \mathrm{rad/s}$ and $v_{\max} = \omega A = 10 \times 0.10 = 1.0$ m/s ✓.

Step 3. The split at $x = 6.0$ cm: $U = \tfrac12 kx^2 = \tfrac12 \times 50 \times (0.060)^2 = \boxed{0.09\ \mathrm{J}}$ , and $K = E - U = 0.25 - 0.09 = \boxed{0.16\ \mathrm{J}}$ .

Step 4. Speed there: $v = \sqrt{2K/m} = \sqrt{2 \times 0.16/0.50} = \sqrt{0.64} = \boxed{0.80\ \mathrm{m/s}}$ . Direct route as a check: $v = \omega\sqrt{A^2 - x^2} = 10\sqrt{0.01 - 0.0036} = 10 \times 0.080 = 0.80$ m/s ✓. The marks-scheme skeleton: budget → split → convert. Write $E$ first in every energy problem; the examiner awards a mark for the budget line alone.

2 · Competitive level — the fraction reflex (NEET/JEE-Main pattern, ~90 seconds)

A particle executes SHM of amplitude $A$ . (a) At what displacement are the kinetic and potential energies equal? (b) What fraction of the total energy is kinetic at $x = A/2$ ? (c) Starting from the mean position, how much time passes before the energies are first equal? Express in terms of the period $T$ .

Step 1. Equal split: $U = E/2$ means $\tfrac12 kx^2 = \tfrac12 \cdot \tfrac12 kA^2$ , so $x^2 = A^2/2$ and $x = \boxed{\pm A/\sqrt2 \approx \pm 0.707A}$ .

Step 2. At $x = A/2$ : $U/E = (1/2)^2 = 1/4$ , so $K/E = \boxed{3/4}$ — the 75% landmark.

Step 3. Timing. From the mean, $x = A\sin\omega t$ . We need $x = A/\sqrt2$ : $\sin\omega t = 1/\sqrt2$ , so $\omega t = \pi/4$ and $t = \dfrac{\pi/4}{2\pi/T} = \boxed{T/8}$ . In one-eighth of a period the particle covers 70.7% of the amplitude — another glimpse of "fast through the middle." Timing note: 60–90 s total. Parts (a) and (b) should be pure reflex — the ratio $U/E = x^2/A^2$ applied twice. Part (c) is where the unprepared lose a minute by re-deriving the sine timetable; the prepared quote it. Distractor watch: $A/2$ offered for part (a), and $T/4$ (the time to reach full amplitude) offered for part (c).

3 · Competitive level — running the budget backwards (JEE-Main pattern, ~2.5 minutes)

A particle of mass $0.10\ \mathrm{kg}$ executes SHM with period $T = \pi/10\ \mathrm{s}$ and total energy $8.0 \times 10^{-3}\ \mathrm{J}$ . Find (a) the amplitude, (b) the speed at a displacement of $1.0\ \mathrm{cm}$ from the centre.

Step 1. Timing first: $\omega = 2\pi/T = 2\pi \times \dfrac{10}{\pi} = 20\ \mathrm{rad/s}$ .

Step 2. Invert the budget: $E = \tfrac12 m\omega^2 A^2$ gives $A^2 = \dfrac{2E}{m\omega^2} = \dfrac{2 \times 8.0\times10^{-3}}{0.10 \times 400} = 4.0\times10^{-4}\ \mathrm{m^2}$ , so $A = \boxed{2.0\ \mathrm{cm}}$ .

Step 3. Speed at $x = 1.0$ cm $= A/2$ : the fraction reflex says $K = \tfrac34 E = 6.0\times10^{-3}$ J, so $v = \sqrt{2K/m} = \sqrt{0.12} = \boxed{0.35\ \mathrm{m/s}}$ (i.e. $20\sqrt3$ cm/s).

Step 4. Audit: $U = \tfrac12 m\omega^2 x^2 = \tfrac12 \times 0.10 \times 400 \times (0.010)^2 = 2.0\times10^{-3}$ J, and $K + U = 6.0\times10^{-3} + 2.0\times10^{-3} = 8.0\times10^{-3}$ J $= E$ ✓. Timing note: ~2–2.5 min. The entire problem is the three costumes of equation (3) worn in sequence: $\omega$ from $T$ , $A$ from $E$ , $v$ from the split. If your $A^2$ comes out negative or absurd, you almost certainly used $f$ where $\omega$ belongs — the $2\pi$ tax, collected every year.

Exam traps

The square, always the square. $E \propto A^2$ and $U/E = x^2/A^2$ . "Amplitude doubles → energy doubles" and "half displacement → half stored" are both wrong for the same reason, and both appear as tempting MCQ options every session.

$\omega$ versus $f$ . The budget is $\tfrac12 m\omega^2 A^2$ with $\omega$ in rad/s. Given a frequency in Hz, convert: $\omega = 2\pi f$ . Forgetting the $2\pi$ scales the energy by $4\pi^2 \approx 39.5$ — an error too large to slip past the options, which is precisely why setters plant the un-converted answer among them.

Measuring $x$ from the wrong place. All formulas here use displacement from the mean position. In a vertical-spring problem, that is the sagged equilibrium — not the natural length, not the release point. Mixing references double-counts gravity.

Kinetic energy graphs that dip below zero. $K(t)$ and $U(t)$ are squared sinusoids: they touch zero and bounce, never crossing the axis, and they run at $2f$ . If a sketch shows energy going negative or oscillating at the motion's own frequency, it is wrong — a favourite spot-the-error question.

"Energy at the extreme is zero." At $x = \pm A$ the kinetic energy is zero; the total is at no point zero — it is $\tfrac12 kA^2$ everywhere. Statements about "the energy" must name which account.

$v_{\max} = \omega A$ , $a_{\max} = \omega^2 A$ . Energy problems constantly hand you one and ask for the other; the ratio $a_{\max}/v_{\max} = \omega$ is the clean bridge. Mixing the two powers of $\omega$ is the classic final-line slip.

Which $g$ ? None — this module's formulas contain no $g$ at all (that is the point of measuring from equilibrium). If a $g$ has crept into your spring-energy answer, revisit the reference-point trap above. For the pendulum's energy, $g$ hides inside $\omega^2 = g/L$ ; boards state $g = 9.8\ \mathrm{m/s^2}$ (CBSE/NEET often 10) — use what the paper declares.

Origins · 1 minHermann von Helmholtz and how this was discovered

Hermann von Helmholtz

1821–1894 · Berlin

Helmholtz was the physiologist-turned-physicist who made energy bookkeeping the law of the land. His 1847 memoir On the Conservation of Force, written at age 26, argued that heat, motion, electricity and chemistry are entries in a single indestructible account — the modern principle of conservation of energy — at a time when "energy" was not yet even a standard word. He then spent a career auditing oscillations: his Sensations of Tone (1863) explained musical sound as energy fed into resonating oscillators, and the brass "Helmholtz resonators" he built to pick individual frequencies out of a chord are the ancestors of every acoustic filter. The two-account budget of this module — kinetic and potential, total fixed — is Helmholtz's ledger applied to the simplest oscillator there is.

History

1581

— Galileo, timing a swinging chandelier in Pisa against his pulse, notices the swing's tempo ignores its width — a fixed budget sloshing at a fixed rate.

1673

— Huygens' Horologium Oscillatorium analyses the pendulum's speed-height trade with full rigour, decades before "energy" has a name.

1807

— Thomas Young coins the word "energy" for the quantity

mv^2

-like accounts had been tracking for a century.

1847

— Helmholtz's On the Conservation of Force elevates the ledger to a universal law: the total account never changes, only its distribution.

1877

— Rayleigh's Theory of Sound turns the SHM energy budget into a working tool: equating average kinetic and potential energy to estimate frequencies of systems too hard to solve exactly.

Going further

Texts. NCERT Physics Class 11, Ch. 14 (Oscillations, energy in SHM section) · HC Verma, Concepts of Physics Vol. 1, Ch. 12 · Resnick–Halliday–Walker, Ch. 15 · For the full UG treatment (phase-space ellipse, action, virial theorem): the Energy in Harmonic Motion deep dive in the university mechanics track.

Provenance. All results are standard. A verifier should recompute: Example 1's chain $E = 0.25$ J, $v_{\max} = 1.0$ m/s, $U(6\ \mathrm{cm}) = 0.09$ J, $v = 0.80$ m/s; Example 2's landmarks $A/\sqrt2$ , $3/4$ , $T/8$ ; Example 3's $\omega = 20$ rad/s, $A = 2.0$ cm, $v = 0.35$ m/s with the $6+2 = 8$ mJ audit.

Prerequisites. Periodic Motion & the Definition of SHM · SHM Kinematics · Kinetic Energy & the Work–Energy Theorem · Potential Energy. Leads to. The Spring–Block Oscillator · The Simple Pendulum · Energy in Harmonic Motion (UG deep dive).

If E = ½kA² has no mass in it, does the block's mass not matter at all?

The budget doesn't care about mass — pull any block 10 cm on the same spring and you have stored the same 0.25 J, because the storing was done against the spring, not against inertia. What mass controls is the tempo and the speed: $\omega = \sqrt{k/m}$ , so a heavier block circulates the same budget more slowly and peaks at a lower $v_{\max} = \sqrt{2E/m}$ . In the alternative costume $E = \tfrac12 m\omega^2A^2$ , mass appears — but only chaperoned by $\omega^2$ , and $m\omega^2$ is just $k$ again. One number, several costumes.

Why is potential energy ½kx² and not kx² — where does the half come from?

Because the force isn't constant while you stretch. The first millimetre of stretch costs almost nothing (the spring barely pulls back); the last millimetre costs the most ( $F$ is now $kx$ ). The honest cost is the average force, $\tfrac12 kx$ , times the distance $x$ — hence the half. Graphically it is the area of a triangle under the $F$ – $x$ line, not the area of a rectangle. Dropping the half is the most common energy error in board scripts, and it propagates: your $v_{\max}$ then comes out $\sqrt2$ too large.

Does the E ∝ A² law hold for a pendulum too, where the potential energy is gravitational?

Yes, within the small-angle regime where the pendulum is honest SHM. The height of the bob above its lowest point is $h = L(1-\cos\theta) \approx L\theta^2/2$ for small angles — already quadratic in the swing angle. So $U = mgh \approx \tfrac12 (mg/L)(L\theta)^2$ : the same $\tfrac12 kx^2$ parabola with effective stiffness $k = mg/L$ and arc displacement $x = L\theta$ . Gravitational storage, elastic storage, even electrical storage in a circuit — SHM flattens them all into the same parabola, which is why one budget serves every oscillator in this chapter.

Kinetic energy peaks twice per cycle — but velocity is maximum only... twice as well? I'm confused about the doubling.

Track one full cycle from the right turning point: the particle crosses the centre moving left (speed maximum), pauses at the left turning point, crosses the centre moving right (speed maximum again), pauses at the right. Velocity as a vector completes one cycle: $+v_{\max}$ once, $-v_{\max}$ once. But kinetic energy squares the velocity, erasing the sign — both crossings look identical to the energy account. So while $v(t)$ oscillates at $f$ , $K(t) = \tfrac12 mv^2$ has two identical humps per period: frequency $2f$ . The doubling is the square's doing, not new physics.

When a vertical spring-block oscillates, gravity does work on the block — why doesn't gravitational PE appear in the energy equation?

It does appear — it has just been absorbed. Measure displacement $y$ from the sagged equilibrium (where $kx_0 = mg$ ) and expand the true total: elastic PE from the natural length plus gravitational PE. The linear terms of the two accounts cancel exactly (that cancellation is the statement " $x_0$ is equilibrium"), leaving a constant plus $\tfrac12 ky^2$ . So $\tfrac12 ky^2$ is the combined gravity-plus-spring storage, provided $y$ starts at equilibrium. That is why the vertical oscillator has the same period, the same budget shape, and the same fraction landmarks as the horizontal one — and why mixing "stretch from natural length" into the same equation double-counts gravity.

Is the average kinetic energy E/2 whether I average over time or over position?

No — and the distinction is a genuine subtlety. Averaged over time, $\langle K\rangle = E/2$ exactly (the $\sin^2$ hump averages to a half). Averaged over position — weighting every $x$ between $-A$ and $+A$ equally — the answer differs, because the particle does not spend equal time at each position: it lingers near the turning points (where $K$ is small) and rushes through the centre (where $K$ is large). Time-weighting therefore favours the low- $K$ regions less than you might guess; the position-average of $K$ works out to $\tfrac23 E$ . Exams at this level always mean the time average, but the sharper JEE-Advanced items have begun to probe whether you know there is a difference.

What happens to the energy budget when friction acts?

The total stops being a constant and starts leaking — each cycle, a slice of $E$ is converted to heat in the damper and sound in the air, so the amplitude shrinks. Because $E \propto A^2$ , the bookkeeping is quick: an oscillator that has lost half its amplitude has lost three-quarters of its energy. The systematic story — how fast the leak drains the budget, why the frequency barely changes, and what happens when you pump energy back in at just the right tempo — is the damped-and-forced-oscillations module, two steps ahead.