Energy in SHM — KE + PE = ½kA² (constant): Worked examples

Before anything, let me re-state the toolbox in plain words so no symbol here is unearned:

The scenario matrix

Every energy question is really one of these cases. The right column names the example that covers it.

#	Case class	What makes it tricky	Covered by
C1	Positive $x$ inside the range ( $0 < x < A$ )	standard "find speed / KE / PE here"	Ex 1
C2	Negative $x$ ( $x < 0$ )	does the sign of $x$ change $U$ or $v$ ?	Ex 2
C3	Degenerate: $x = 0$ (centre)	speed is maximum, PE = 0	Ex 3
C4	Degenerate: $x = \pm A$ (turning point)	speed is zero, all energy is PE	Ex 3
C5	Fraction/ratio question ("at what $x$ is KE = 3·PE?")	solve for position from an energy split	Ex 4
C6	Limiting / scaling (change $A$ , $k$ , or $m$ )	how $E$ , $v_{ma x}$ respond	Ex 5
C7	Real-world word problem (car suspension)	translate words → symbols → numbers	Ex 6
C8	Exam twist: energy given, work backwards	you're handed $E$ and $v$ , find $A$	Ex 7
C9	Impossible / boundary check ( $x > A$ ?)	recognise a physically forbidden input	Ex 8

The figure below is the map every example lives on: the energy "bowl" $U = \frac{1}{2} k x^{2}$ (black parabola), the flat ceiling $E = \frac{1}{2} k A^{2}$ (red), and — this is the key move used in almost every example — KE is always the vertical gap between the bowl and the ceiling. Keep this picture open as you read: Ex 1 reads the gap at $x = 0.08$ , Ex 3 reads the gap at the centre and edges, Ex 4 finds where the gap is three times the bowl height.

Figure — Energy in SHM — KE + PE = ½kA² (constant)

Worked examples

Worked example Example 1 — C1: positive

$x$ inside the range A mass on a spring has $k = 200$ N/m, $m = 0.50$ kg, amplitude $A = 0.12$ m. Find the speed and the KE/PE split at $x = + 0.08$ m.

Forecast: guess — is the mass faster or slower than at the centre? (It must be slower: it's away from the middle.)

Total energy. $E = \frac{1}{2} k A^{2} = \frac{1}{2} (200) (0.12)^{2} = 1.44$ J. (the red ceiling in the figure.) Why this step? The ceiling is fixed by $A$ alone — everything else is carved out of it.
PE here. $U = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.08)^{2} = 0.64$ J. (height of the black bowl at $x = 0.08$ .) Why this step? $U$ depends only on how far out you are.
KE is the gap. $K = E - U = 1.44 - 0.64 = 0.80$ J. (this is exactly the dashed gap drawn in the figure.) Why this step? Energy is conserved, so whatever isn't stored is motion.
Speed. $ω = 200/0.5 = 400 = 20$ rad/s. $∣ v ∣ = ω A^{2} - x^{2} = 20 0.1 2^{2} - 0.0 8^{2} = 20 0.0080 = 20 (0.08944) = 1.789$ m/s. Why this step? $A^{2} - x^{2}$ shrinks as you approach $A$ — the mass slows near the edge, exactly as forecast.

Verify: $\frac{1}{2} m ∣ v ∣^{2} = \frac{1}{2} (0.5) (1.789)^{2} = 0.80$ J $= K$ . ✓ Units: N/m·m² = J. ✓

Worked example Example 2 — C2: negative displacement

Same spring as Ex 1 ( $k = 200$ , $m = 0.5$ , $A = 0.12$ ). Now find speed and PE at $x = - 0.08$ m.

Forecast: will the answers differ from Ex 1? Guess: no — because energy only cares about distance from centre, not direction.

PE. $U = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (- 0.08)^{2} = 0.64$ J. Why this step? $x$ is squared, so $(- 0.08)^{2} = (+ 0.08)^{2}$ . The minus sign vanishes. (Physically: $x < 0$ means the spring is compressed, but compression stores exactly as much energy as the same amount of stretch — that's why we call $U$ elastic potential energy, not "stretch" energy.)
Speed magnitude. $∣ v ∣ = ω A^{2} - x^{2} = 20 0.1 2^{2} - (- 0.08)^{2} = 1.789$ m/s. Why this step? Again $x$ appears squared — sign of $x$ is invisible to speed.
What the sign DOES tell you. The $\pm$ in $v = \pm ω A^{2} - x^{2}$ is the velocity direction (left vs right), not about which side of centre you're on. At $x = - 0.08$ the mass could be moving either way; energy is identical to $x = + 0.08$ . Why this step? Separating "which side" (sign of $x$ ) from "which way moving" (sign of $v$ ) is the whole point of this case.

Verify: identical to Ex 1 by symmetry: $U = 0.64$ J, $∣ v ∣ = 1.789$ m/s. ✓ The energy bowl in the figure is a perfect parabola — symmetric about $x = 0$ . ✓

Worked example Example 3 — C3 & C4: the two degenerate points (

$x = 0$ and $x = \pm A$ ) Same spring. Find the energy split and speed at (a) the centre $x = 0$ and (b) a turning point $x = + A = 0.12$ m.

Forecast: where is the mass fastest? Where is it momentarily frozen? Picture the swing at the bottom vs the top. In the figure: where is the gap widest, and where does it close to zero?

(a) Centre, $x = 0$ : PE. $U = \frac{1}{2} k (0)^{2} = 0$ J. (bowl touches zero — the gap is at its widest.) Why this step? No deformation, no stored energy.
(a) Centre: KE = all of it. $K = E - U = 1.44 - 0 = 1.44$ J, so $v_{ma x} = ω A = 20 (0.12) = 2.40$ m/s. Why this step? At $x = 0$ the whole budget is motion — this is the fastest the mass ever goes.
(b) Turning point, $x = A$ : PE = all of it. $U = \frac{1}{2} k A^{2} = 1.44$ J $= E$ . (bowl rises to meet the red ceiling — gap closes to zero.) Why this step? Fully stretched → every joule is stored.
(b) Turning point: KE and speed. $K = E - U = 0$ , so $∣ v ∣ = ω A^{2} - A^{2} = 0$ . The mass is momentarily stopped. Why this step? This is the exact instant of turning around — like the top of a swing.

Verify: $v_{ma x} = ω A = 20 \times 0.12 = 2.40$ m/s and $\frac{1}{2} m v_{ma x}^{2} = \frac{1}{2} (0.5) (2.4)^{2} = 1.44$ J $= E$ . ✓ At $x = A$ , $A^{2} - x^{2} = 0 \Rightarrow v = 0$ . ✓

Worked example Example 4 — C5: solve for position from an energy split

Same spring ( $A = 0.12$ m). At what displacement is the kinetic energy three times the potential energy ( $K = 3 U$ )?

Forecast: in the figure, we want the point where the gap (KE) is three times the bowl height (PE). Should this $x$ be closer to the centre or the edge? If KE dominates, we're near the fast centre — so expect small $x$ .

Write the condition using $E$ . $K = 3 U$ and $K + U = E$ give $3 U + U = E \Rightarrow U = \frac{1}{4} E$ . Why this step? Turn a ratio into a fraction of the fixed total — much easier to solve.
Sub in the formulas. $\frac{1}{2} k x^{2} = \frac{\frac{1}{2} k A ^{2}}{4} = \frac{1}{8} k A^{2}$ . Multiply both sides by $\frac{2}{k}$ (cancelling the common $\frac{1}{2} k$ ): $x^{2} = \frac{1}{4} A^{2}$ . Why this step? Cancelling the common $\frac{1}{2} k$ shows the answer depends only on $A$ , not on $k$ or $m$ .
Solve. $x = \pm \frac{A}{2} = \pm 0.06$ m. Why this step? Both signs are valid — the condition is symmetric, matching Ex 2's lesson.

Verify: at $x = 0.06$ : $U = \frac{1}{2} (200) (0.06)^{2} = 0.36$ J, $K = E - U = 1.44 - 0.36 = 1.08$ J, and $1.08 = 3 (0.36)$ . ✓

Worked example Example 5 — C6: limiting & scaling behaviour

A spring oscillator has energy $E_{1} = 1.44$ J. You now (a) double the amplitude, and separately (b) quadruple the spring constant $k$ . What happens to $E$ and to $v_{ma x}$ ?

Forecast: $E \propto A^{2}$ , so doubling $A$ should more than double $E$ — guess the factor.

(a) Double $A$ . $E \propto A^{2} \Rightarrow E$ multiplies by $2^{2} = 4$ : new $E = 4 \times 1.44 = 5.76$ J. Why this step? $A$ is squared in $E = \frac{1}{2} k A^{2}$ , so scaling is quadratic.
(a) Effect on $v_{ma x}$ . $v_{ma x} = ω A$ ; $ω = k / m$ is unchanged, so $v_{ma x}$ just doubles. Why this step? $v_{ma x}$ is linear in $A$ even though energy is quadratic — a common exam trap.
(b) Quadruple $k$ (fix $A$ , $m$ ). $E = \frac{1}{2} k A^{2} \Rightarrow E$ multiplies by 4: new $E = 5.76$ J. But $ω = k / m$ multiplies by $4 = 2$ , so $v_{ma x} = ω A$ doubles too. Why this step? Stiffness raises energy linearly but speed only through $k$ .

Verify: (a) $4 \times 1.44 = 5.76$ J and $v_{ma x}$ new $= 2 \times 2.40 = 4.80$ m/s. (b) $4 \times 1.44 = 5.76$ J; $ω_{n e w} = 800/0.5 = 40$ , $v_{ma x} = 40 \times 0.12 = 4.80$ m/s. ✓

Worked example Example 6 — C7: real-world word problem (car suspension)

One corner of a car sits on a spring with $k = 1.5 \times 1 0^{4}$ N/m carrying $m = 300$ kg. After a bump the corner oscillates with amplitude $A = 0.04$ m (before the shock absorbers matter). Find the total oscillation energy and the maximum speed of that corner.

Forecast: small amplitude but big stiffness — guess whether $E$ is more like 1 J or 10 J.

Translate words to symbols. "carries $m$ " → mass, "spring $k$ ", "amplitude $A$ ". No new physics — same equations. Why this step? The skill in word problems is naming variables correctly, not new formulas.
Total energy. $E = \frac{1}{2} k A^{2} = \frac{1}{2} (1.5 \times 1 0^{4}) (0.04)^{2} = \frac{1}{2} (15000) (0.0016) = 12$ J. Why this step? Directly the fixed ceiling from $A$ and $k$ .
Max speed. $ω = k / m = 15000/300 = 50 = 7.071$ rad/s. $v_{ma x} = ω A = 7.071 \times 0.04 = 0.2828$ m/s. Why this step? Fastest point is at equilibrium, $x = 0$ .

Verify: $\frac{1}{2} m v_{ma x}^{2} = \frac{1}{2} (300) (0.2828)^{2} = 12$ J $= E$ . ✓ Units: $(N/m) / kg = 1/ s^{2} =$ rad/s. ✓

Worked example Example 7 — C8: exam twist, given energy & speed, work backwards to

$A$ A $0.25$ kg mass oscillates on a spring of $k = 100$ N/m. At the instant its speed is $∣ v ∣ = 1.2$ m/s the potential energy is measured to be $U = 0.14$ J. Find the amplitude $A$ and the displacement $x$ at that instant.

Forecast: we're handed the pieces and must rebuild the whole — reverse of Ex 1.

KE from speed. $K = \frac{1}{2} m ∣ v ∣^{2} = \frac{1}{2} (0.25) (1.2)^{2} = 0.18$ J. Why this step? Speed gives motion energy directly.
Total energy = sum. $E = K + U = 0.18 + 0.14 = 0.32$ J. Why this step? Energy is conserved, so at any instant $K + U = E$ .
Amplitude from $E$ . $E = \frac{1}{2} k A^{2} \Rightarrow A = 2 E / k = 2 (0.32) /100 = 0.0064 = 0.08$ m. Why this step? Invert the ceiling equation to get $A$ .
Displacement from $U$ . $U = \frac{1}{2} k x^{2} \Rightarrow x = \pm 2 U / k = \pm 2 (0.14) /100 = \pm 0.0028 = \pm 0.0529$ m. Why this step? $U$ pins the magnitude of $x$ ; both signs are valid (Ex 2's lesson again).

Verify: check $K = \frac{1}{2} k (A^{2} - x^{2}) = \frac{1}{2} (100) (0.0 8^{2} - 0.052 9^{2}) = \frac{1}{2} (100) (0.0064 - 0.0028) = \frac{1}{2} (100) (0.0036) = 0.18$ J $= K$ . ✓

Worked example Example 8 — C9: the impossible input (boundary check)

A student is asked for the speed at $x = 0.15$ m on the spring of Ex 1 (where $A = 0.12$ m). What goes wrong, and what does it teach us?

Forecast: try to plug in and watch the maths complain. In the figure, $x = 0.15$ lies outside the bowl's opening.

Plug in. $∣ v ∣ = ω A^{2} - x^{2} = 20 0.1 2^{2} - 0.1 5^{2} = 20 0.0144 - 0.0225 = 20 - 0.0081$ . Why this step? We get the square root of a negative number — not a real speed.
Interpret. $x = 0.15 > A = 0.12$ is outside the allowed range. The mass never reaches there: amplitude $A$ is by definition the furthest point. Why this step? A negative under the root is nature's way of saying "this position never happens."
The rule. Physically valid displacements obey $- A \leq x \leq A$ . At the boundary $x = \pm A$ the root is exactly zero ( $v = 0$ ), the turning point from Ex 3. Why this step? Recognising forbidden inputs stops you from reporting nonsense on an exam.

Verify: $A^{2} - x^{2} = 0.0144 - 0.0225 = - 0.0081 < 0$ , confirming no real speed exists for $x > A$ . ✓

Recall Quick self-test (cover the answers)

Speed at $x$ for $k = 200, m = 0.5, A = 0.12$ at $x = 0.08$ ::: $1.789$ m/s Where is $K = 3 U$ ? ::: $x = \pm A /2$ Double the amplitude changes $E$ by what factor? ::: $\times 4$ Double the amplitude changes $v_{ma x}$ by what factor? ::: $\times 2$ Why does $x = 1.25 A$ give an error? ::: it's outside $- A \leq x \leq A$ ; root goes negative, no real speed