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

WHAT: total energy sirf amplitude se fix hoti hai. WHY: parent note ne dikhaya ki $E=\frac{1}{2}k{A}^2$ constant hai. $E=\frac{1}{2}(200)(0.05{)}^2=\frac{1}{2}(200)(0.0025)=0.25\text{J}.$ Answer: $0.25$ J.

WHY: turning point par mass momentarily rest mein hota hai ($K=0$), isliye saari energy potential hai. $U=\frac{1}{2}k{A}^2=0.25\text{J}=E.$ Answer: $0.25$ J — $E$ ke barabar, kyunki wahan $K=0$ hai.

WHAT LOOKS LIKE: Figure 1 mein, mint $K$-parabola bilkul centre $x=0$ par peak karti hai. WHY: $K=\frac{1}{2}k(A^2-x^2)$ tab sabse badi hoti hai jab $x=0$ ho toh $A^2-x^2=A^2$. $K_{\max }=\frac{1}{2}k{A}^2=E=0.25\text{J}\text{at }x=0.$ Answer: equilibrium $x=0$ par, $K=0.25$ J.

WHY $\omega$ pehle: speed ke liye $\omega =\sqrt{k/m}$ chahiye. $\omega =\sqrt{32/0.5}=\sqrt{64}=8\text{rad/s}.$ Max speed $x=0$ par hai, jahan $\sqrt{A^2-x^2}=A$: $v_{\max }=\omega A=8\times 0.10=0.80\text{m/s}.$ Answer: $\omega =8$ rad/s, $v_{\max }=0.80$ m/s (speed; mass yeh magnitude $x=0$ se guzarte hue dono directions mein achieve karta hai).

WHY: speed $v=\pm \omega \sqrt{A^2-x^2}$ ki magnitude hai; hum positive root lete hain. $v=\omega \sqrt{A^2-x^2}=8\sqrt{0.10^2-0.06^2}=8\sqrt{0.0064}=8(0.08)=0.64\text{m/s}.$ WHY $v_{\max }$ se kam: $\sqrt{A^2-x^2}$ tab shrink hota hai jab $x$, $A$ ki taraf move kare — mass turning point ki taraf slow ho raha hai. Answer: $0.64$ m/s (magnitude; yeh $x=0.06$ m se baahir jaate aur wapas aate dono baar is speed se guzarta hai).

WHY: energy $v^2$ par depend karti hai, isliye direction ka $\pm$ sign fraction ke liye irrelevant hai. $\frac{K}{E}=\frac{\frac{1}{2}k(A^2-x^2)}{\frac{1}{2}k{A}^2}=\frac{A^2-x^2}{A^2}=\frac{0.0064}{0.01}=\mathrm{0.64.}$ Answer: energy ka $64\%$ kinetic hai (toh $36\%$ potential hai). Note karo yeh $1-(x/A{)}^2$ hai.

WHY: KE-in-$x$ formula ko $\frac{1}{4}E$ ke barabar set karo aur position ke liye solve karo. $\frac{1}{2}k(A^2-x^2)=\frac{1}{4}\cdot \frac{1}{2}k{A}^2\Rightarrow A^2-x^2=\frac{1}{4}{A}^2\Rightarrow x^2=\frac{3}{4}{A}^2.$ $x=\pm \frac{\sqrt{3}}{2}A\approx \pm 0.866A.$ Answer: $x=\pm \frac{\sqrt{3}}{2}A$. (Figure 1 dekho: edges ke paas zyaadatar energy PE hai, isliye KE kam hai.)

WHY: equal energies ka matlab hai dono parabolas same height par milti hain. $K=U$ set karo: $\frac{1}{2}k(A^2-x^2)=\frac{1}{2}k{x}^2\Rightarrow A^2-x^2=x^2\Rightarrow x^2=\frac{A^2}{2}$. $x=\pm \frac{A}{\sqrt{2}}\approx \pm 0.707A.$ WHAT LOOKS LIKE: Figure 1 mein, yeh exactly wahan hai jahan mint aur coral parabolas cross karti hain (do slate dots) — har ek wahan $\frac{1}{2}E$ carry karti hai. Answer: $x=\pm A/\sqrt{2}$.

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

(a) WHY: $E\propto A^2$, isliye factor $(9/3{)}^2=9$ hai. Energy $9\times$ ho jaati hai. (b) WHY: $v_{\max }=\omega A\propto A$ (kyunki $\omega =\sqrt{k/m}$ unchanged hai), isliye speed factor $=9/3=3$ hai. $v_{\max }$ triple ho jaati hai. WHY dono alag hain: energy mein amplitude ka square hai lekin speed amplitude mein linear hai. $A$ triple karne se speed triple hoti hai lekin energy nine-fold ho jaati hai.

WHY set-up: motion ko time mein $x(t)=A\cos (\omega t+\varphi )$ likho, jahan $\varphi$ (phase constant, radians mein) woh fixed number hai jo batata hai $t=0$ par mass apne swing mein kahan tha — yeh cosine ko shift karta hai lekin energy ke baare mein kuch nahi badlta. Velocity $v=-A\omega \sin (\omega t+\varphi )$ hai, isliye $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\sin }^2(\omega t+\varphi )=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi ).$

(a) WHY $2\omega$ — identity derive karna, sirf quote nahi karna. Hume ${\sin }^2$ ko single cosine wale expression mein convert karna hai taki iska frequency read off kar sakein. Do standard cosine facts se shuru karo: $\cos (2\theta )={\cos }^2\theta -{\sin }^2\theta ,{\cos }^2\theta +{\sin }^2\theta =1.$ Pehle ko doosre se subtract karo: $({\cos }^2+{\sin }^2)-({\cos }^2-{\sin }^2)=1-\cos 2\theta$, yaani $2{\sin }^2\theta =1-\cos 2\theta$, isliye ${\sin }^2\theta =\frac{1-\cos 2\theta }{2}.$ $\theta =\omega t+\varphi$ rakhho: $K=\frac{1}{2}k{A}^2\cdot \frac{1-\cos (2\omega t+2\varphi )}{2}=\frac{1}{4}k{A}^2-\frac{1}{4}k{A}^2\cos (2\omega t+2\varphi ).$ Sirf time-varying piece $\cos (2\omega t+2\varphi )$ hai, jiska angular frequency $\boxed{2\omega }$ hai (units rad/s). Kyunki ordinary frequency $f$ (units Hz = cycles per second) angular frequency se $f=\frac{\text{angular frequency}}{2\pi }$ relation se judi hai, KE $f_K=\frac{2\omega }{2\pi }=\frac{\omega }{\pi }$ Hz par cycle karti hai — displacement ki frequency $\omega /2\pi$ se exactly double. $\varphi$ sirf argument ke andar ride karta hai; yeh kabhi rate $2\omega$ nahi badlta.

(b) WHY $⟨K⟩=\frac{1}{2}E$ — same rewrite se. Upar boxed line mein, $K$ ek constant $\frac{1}{4}k{A}^2$ plus ek pure cosine hai. Cosine ek complete cycle mein zero ke upar aur neeche equal time bitaati hai, isliye iska average $0$ hai. 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, yeh kehta hai $⟨{\sin }^2⟩=\frac{1}{2}$, jo humne assume karne ki jagah derive kiya.) Mirror argument se $⟨U⟩=\frac{1}{2}E$ bhi. Answer: angular frequency $2\omega$ rad/s (yaani $\omega /\pi$ Hz, displacement frequency se double); $⟨K⟩=\frac{1}{2}E=\frac{1}{4}k{A}^2$.

Energy — WHY: $E=\frac{1}{2}k{A}^2\propto k$, isliye $E_B/E_A=4$. B $4\times$ energy store karta hai. Frequency — WHY: $\omega =\sqrt{k/m}\propto \sqrt{k}$, isliye ${\omega }_B/{\omega }_A=\sqrt{4}=2$. B do guna tezi se oscillate karta hai. WHY dono alag hain: stiffer spring zyada tezi se kheeenchti bhi hai (same stretch ke liye zyada stored PE) aur tezi se wapas bhi snapback karti hai. Answer: $E_B=4E_A$, ${\omega }_B=2{\omega }_A$.

WHAT: do speed readings $k$ ya $m$ jaane bina $A$ pin down karte hain. WHY: $v=\omega \sqrt{A^2-x^2}$ aur $v_{\max }=\omega A$; ratio lene se $\omega$ cancel ho jaata hai (aur kyunki hum speeds use karte hain, $\pm$ direction sign irrelevant hai). $\frac{v}{v_{\max }}=\frac{\sqrt{A^2-x^2}}{A}\Rightarrow {\left(\frac{1.6}{2.0}\right)}^2=\frac{A^2-x^2}{A^2}=1-\frac{x^2}{A^2}.$ $0.64=1-\frac{x^2}{A^2}\Rightarrow \frac{x^2}{A^2}=0.36\Rightarrow A=\frac{x}{0.6}=\frac{0.03}{0.6}=0.05\text{m}.$ Answer: $A=0.05$ m.

WHY: $\omega =v_{\max }/A$ ($v_{\max }=\omega A$ se), phir $k=m{\omega }^2$, aur $E=\frac{1}{2}m{v}_{\max }^2$ kyunki $x=0$ par saari energy kinetic hoti hai. $\omega =v_{\max }/A=2.0/0.05=40\text{rad/s},k=m{\omega }^2=0.8(40{)}^2=0.8(1600)=1280\text{N/m}.$ $E=\frac{1}{2}m{v}_{\max }^2=\frac{1}{2}(0.8)(2.0{)}^2=1.6\text{J}.$ (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 har limit: har formula mein $A\to 0$ substitute karo. Energy: $E=\frac{1}{2}k{A}^2\to 0$ — koi stretch kabhi diya hi nahi gaya, isliye swap karne ke liye kuch hai hi nahi. Max speed: $v_{\max }=\omega A\to 0$ — mass kabhi hilta hi nahi. Trade-off: $A=0$ ke saath Figure 1 mein parabolas origin par ek single point tak collapse ho jaati hain: $K=0$ aur $U=0$ har jagah. Oscillator equilibrium par frozen baitha rehta hai. WHY yeh matter karta hai: yeh poori family ke degenerate endpoint ko confirm karta hai — energy genuinely amplitude se set hoti hai, aur zero amplitude ek valid (trivial) SHM state hai, koi paradox nahi. Answer: $E\to 0$, $v_{\max }\to 0$, mass $x=0$ par rest mein rehta hai.

Energy fixed hai ($\frac{1}{2}k{A}^2$), split $x^2$ follow karta hai, speed $\sqrt{A^2-x^2}$ follow karta hai (direction ke liye $\pm$ sign ke saath), energy $A^2$ aur $k$ (linearly) ke saath scale karti hai, frequency $\sqrt{k}$ ke saath — aur speeds ke ratios $\omega$ cancel kar dete hain.