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

WHAT: total energy is fixed by amplitude alone. WHY: the parent note showed $E=\frac{1}{2}k{A}^2$ is constant. $E=\frac{1}{2}(200)(0.05{)}^2=\frac{1}{2}(200)(0.0025)=0.25\text{J}.$ Answer: $0.25$ J.

WHY: at the turning point the mass is momentarily at rest ($K=0$), so all energy is potential. $U=\frac{1}{2}k{A}^2=0.25\text{J}=E.$ Answer: $0.25$ J — the same as $E$, because $K=0$ there.

WHAT LOOKS LIKE: in Figure 1, the mint $K$-parabola peaks dead centre at $x=0$. WHY: $K=\frac{1}{2}k(A^2-x^2)$ is biggest when $x=0$ so $A^2-x^2=A^2$. $K_{\max }=\frac{1}{2}k{A}^2=E=0.25\text{J}\text{at }x=0.$ Answer: at equilibrium $x=0$, $K=0.25$ J.

WHY $\omega$ first: speed needs $\omega =\sqrt{k/m}$. $\omega =\sqrt{32/0.5}=\sqrt{64}=8\text{rad/s}.$ Max speed is at $x=0$, where $\sqrt{A^2-x^2}=A$: $v_{\max }=\omega A=8\times 0.10=0.80\text{m/s}.$ Answer: $\omega =8$ rad/s, $v_{\max }=0.80$ m/s (speed; the mass reaches this magnitude moving in either direction through $x=0$).

WHY: speed is the magnitude of $v=\pm \omega \sqrt{A^2-x^2}$; we take the positive root. $v=\omega \sqrt{A^2-x^2}=8\sqrt{0.10^2-0.06^2}=8\sqrt{0.0064}=8(0.08)=0.64\text{m/s}.$ WHY smaller than $v_{\max }$: $\sqrt{A^2-x^2}$ shrinks as $x$ moves toward $A$ — the mass is slowing on its way to the turning point. Answer: $0.64$ m/s (magnitude; it passes $x=0.06$ m at this speed both on the way out and on the way back).

WHY: energy depends on $v^2$, so the $\pm$ sign of direction is irrelevant to the fraction. $\frac{K}{E}=\frac{\frac{1}{2}k(A^2-x^2)}{\frac{1}{2}k{A}^2}=\frac{A^2-x^2}{A^2}=\frac{0.0064}{0.01}=\mathrm{0.64.}$ Answer: $64\%$ of the energy is kinetic (so $36\%$ is potential). Notice this is $1-(x/A{)}^2$.

WHY: set the KE-in-$x$ formula to $\frac{1}{4}E$ and solve for position. $\frac{1}{2}k(A^2-x^2)=\frac{1}{4}\cdot \frac{1}{2}k{A}^2\Rightarrow A^2-x^2=\frac{1}{4}{A}^2\Rightarrow x^2=\frac{3}{4}{A}^2.$ $x=\pm \frac{\sqrt{3}}{2}A\approx \pm 0.866A.$ Answer: $x=\pm \frac{\sqrt{3}}{2}A$. (See Figure 1: near the edges most energy is PE, so KE is small.)

WHY: equal energies means the two parabolas hit the same height. Set $K=U$: $\frac{1}{2}k(A^2-x^2)=\frac{1}{2}k{x}^2\Rightarrow A^2-x^2=x^2\Rightarrow x^2=\frac{A^2}{2}$. $x=\pm \frac{A}{\sqrt{2}}\approx \pm 0.707A.$ WHAT LOOKS LIKE: in Figure 1, this is exactly where the mint and coral parabolas cross (the two slate dots) — each carries $\frac{1}{2}E$ there. Answer: $x=\pm A/\sqrt{2}$.

(a) WHY: total energy is $\frac{1}{2}k{A}^2$. $E=\frac{1}{2}(120)(0.08{)}^2=\frac{1}{2}(120)(0.0064)=0.384\text{J}.$ (b) WHY: speed is $\omega \sqrt{A^2-x^2}$; here $A^2-x^2=A^2-\frac{A^2}{2}=\frac{A^2}{2}$. $\omega =\sqrt{120/0.30}=\sqrt{400}=20\text{rad/s},v=\omega \sqrt{\frac{A^2}{2}}=20\cdot \frac{0.08}{\sqrt{2}}=20(0.05657)=1.131\text{m/s}.$ (c) WHY: compute $K=\frac{1}{2}m{v}^2$ and compare to $\frac{1}{2}E$. $K=\frac{1}{2}(0.30)(1.131{)}^2=\frac{1}{2}(0.30)(1.280)=0.192\text{J}=\frac{1}{2}(0.384).✓$ Answers: $E=0.384$ J, $v\approx 1.131$ m/s (speed), and indeed $K=0.192$ J $=\frac{1}{2}E$.

(a) WHY: $E\propto A^2$, so the factor is $(9/3{)}^2=9$. Energy becomes $9\times$. (b) WHY: $v_{\max }=\omega A\propto A$ (since $\omega =\sqrt{k/m}$ is unchanged), so speed factor $=9/3=3$. $v_{\max }$ triples. WHY they differ: energy carries a square of amplitude but speed is linear in amplitude. Tripling $A$ triples speed but nine-folds energy.

WHY set-up: write the motion in time as $x(t)=A\cos (\omega t+\varphi )$, where $\varphi$ (the phase constant, in radians) is the fixed number saying where in its swing the mass was at $t=0$ — it shifts the cosine but changes nothing about energy. The velocity is $v=-A\omega \sin (\omega t+\varphi )$, so $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\sin }^2(\omega t+\varphi )=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi ).$

(a) WHY $2\omega$ — deriving the identity, not just quoting it. We need to turn ${\sin }^2$ into something with a single cosine so we can read off its frequency. Start from the two standard cosine facts: $\cos (2\theta )={\cos }^2\theta -{\sin }^2\theta ,{\cos }^2\theta +{\sin }^2\theta =1.$ Subtract the first from the second: $({\cos }^2+{\sin }^2)-({\cos }^2-{\sin }^2)=1-\cos 2\theta$, i.e. $2{\sin }^2\theta =1-\cos 2\theta$, so ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}.$ Put $\theta =\omega t+\varphi$: $K=\frac{1}{2}k{A}^2\cdot \frac{1-\cos (2\omega t+2\varphi )}{2}=\frac{1}{4}k{A}^2-\frac{1}{4}k{A}^2\cos (2\omega t+2\varphi ).$ The only time-varying piece is $\cos (2\omega t+2\varphi )$, whose angular frequency is $\boxed{2\omega }$ (units rad/s). Because ordinary frequency $f$ (units Hz = cycles per second) relates to angular frequency by $f=\frac{\text{angular frequency}}{2\pi }$, the KE cycles at $f_K=\frac{2\omega }{2\pi }=\frac{\omega }{\pi }$ Hz — exactly twice the displacement's frequency $\omega /2\pi$. The $\varphi$ only rides inside the argument; it never changes the rate $2\omega$.

(b) WHY $⟨K⟩=\frac{1}{2}E$ — from the same rewrite. In the boxed line above, $K$ is a constant $\frac{1}{4}k{A}^2$ plus a pure cosine. A cosine spends equal time above and below zero, so its average over a whole cycle is $0$. Hence $⟨K⟩=\frac{1}{4}k{A}^2-\frac{1}{4}k{A}^2\underset{=0}{\underbrace{⟨\cos (2\omega t+2\varphi )⟩}}=\frac{1}{4}k{A}^2=\frac{1}{2}E.$ (Equivalently, this says $⟨{\sin }^2⟩=\frac{1}{2}$, which we just derived rather than assumed.) By the mirror argument $⟨U⟩=\frac{1}{2}E$ too. Answer: angular frequency $2\omega$ rad/s (i.e. $\omega /\pi$ Hz, double the displacement frequency); $⟨K⟩=\frac{1}{2}E=\frac{1}{4}k{A}^2$.

Energy — WHY: $E=\frac{1}{2}k{A}^2\propto k$, so $E_B/E_A=4$. B stores $4\times$ the energy. Frequency — WHY: $\omega =\sqrt{k/m}\propto \sqrt{k}$, so ${\omega }_B/{\omega }_A=\sqrt{4}=2$. B oscillates twice as fast. WHY the difference: stiffer spring both pulls harder (more stored PE for the same stretch) and snaps back faster. Answer: $E_B=4E_A$, ${\omega }_B=2{\omega }_A$.

WHAT: two speed readings pin down $A$ without ever needing $k$ or $m$. WHY: $v=\omega \sqrt{A^2-x^2}$ and $v_{\max }=\omega A$; taking the ratio cancels $\omega$ (and since we use speeds, the $\pm$ direction sign is irrelevant). $\frac{v}{v_{\max }}=\frac{\sqrt{A^2-x^2}}{A}\Rightarrow {\left(\frac{1.6}{2.0}\right)}^2=\frac{A^2-x^2}{A^2}=1-\frac{x^2}{A^2}.$ $0.64=1-\frac{x^2}{A^2}\Rightarrow \frac{x^2}{A^2}=0.36\Rightarrow A=\frac{x}{0.6}=\frac{0.03}{0.6}=0.05\text{m}.$ Answer: $A=0.05$ m.

WHY: $\omega =v_{\max }/A$ (from $v_{\max }=\omega A$), then $k=m{\omega }^2$, and $E=\frac{1}{2}m{v}_{\max }^2$ since all energy is kinetic at $x=0$. $\omega =v_{\max }/A=2.0/0.05=40\text{rad/s},k=m{\omega }^2=0.8(40{)}^2=0.8(1600)=1280\text{N/m}.$ $E=\frac{1}{2}m{v}_{\max }^2=\frac{1}{2}(0.8)(2.0{)}^2=1.6\text{J}.$ (Cross-check: $E=\frac{1}{2}k{A}^2=\frac{1}{2}(1280)(0.05{)}^2=\frac{1}{2}(1280)(0.0025)=1.6$ J. ✓) Answer: $k=1280$ N/m, $E=1.6$ J.

WHY each limit: substitute $A\to 0$ into each formula. Energy: $E=\frac{1}{2}k{A}^2\to 0$ — no stretch was ever given, so there is nothing to swap. Max speed: $v_{\max }=\omega A\to 0$ — the mass never moves. Trade-off: with $A=0$ the parabolas in Figure 1 collapse to a single point at the origin: $K=0$ and $U=0$ everywhere. The oscillator sits frozen at equilibrium. WHY this matters: it confirms the degenerate endpoint of the whole family — energy is genuinely set by amplitude, and zero amplitude is a valid (trivial) SHM state, not a paradox. Answer: $E\to 0$, $v_{\max }\to 0$, mass stays at rest at $x=0$.

Energy is fixed ($\frac{1}{2}k{A}^2$), split follows $x^2$, speed follows $\sqrt{A^2-x^2}$ (with a $\pm$ sign for direction), energy scales as $A^2$ and $k$ (linearly), frequency as $\sqrt{k}$ — and ratios of speeds cancel $\omega$.