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

1.6.5 · D2 · Physics › Oscillations & Waves › Energy in SHM — KE + PE = ½kA² (constant)

Kisi bhi symbol se pehle, cast ko simple words mein jaante hain.

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Step 1 — The stage: a block, a spring, a home position

KYA. Hum woh ek physical picture set up karte hain jis par baaki sab kuch tikta hai: ek block ek spring par, frictionless floor par slide karta hua. "Home" ($x=0$) woh jagah hai jahan spring na stretch hui hai na squeeze.

KYUN. Neeche har energy formula isi picture ke baare mein ek statement hai. Agar hum geometrically yeh nail down nahi karte ki $x$, $A$, aur "home" ka matlab kya hai, toh baad ka koi bhi symbol meaningful nahi hoga.

PICTURE. Teen snapshots: left ki taraf $x=-A$ tak khaincha, home $x=0$ par baitha, right ki taraf $x=+A$ tak dhakela. Floor marked hai taki tum literally $x$ padh sako.

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Step 2 — Spring ka law: stretch ke saath force badhti hai

KYA. Jab block position $x$ par hota hai, spring use $F=-kx$ force ke saath home ki taraf push karta hai. Yeh Hooke's Law hai.

KYUN. Energy ko is force ke against kiye gaye kaam ke roop mein define kiya jayega, isliye hume pehle force dekhni hogi. Minus sign ek spring ki poori personality hai: woh hamesha home ki taraf point karta hai.

PICTURE. Kai positions par force arrows. Home se door = lamba arrow; home ke paas = chhota arrow; home par = koi arrow nahi.

$F=\underset{\begin{array}{c}\text{points back}\\ \text{toward home}\end{array}}{\underbrace{-}}\underset{\begin{array}{c}\text{stiffness:}\\ \text{steeper}=\text{stronger}\end{array}}{\underbrace{k}}\underset{\begin{array}{c}\text{how far}\\ \text{from home}\end{array}}{\underbrace{x}}$

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Step 3 — Potential energy: force line ke neeche ka AREA

KYA. Potential energy $U$ = woh kaam jo tum block ko home se bahar $x$ tak khainchne mein karte ho, poore raaste spring se ladte hue. Spring se larne ke liye tum $F_{\text{you}}=+kx$ apply karte ho (spring ke opposite). Kaam = force $\times$ distance, lekin force badhti rehti hai, isliye hum patli slices add karte hain.

YEH TOOL KYUN — the integral. Kaam force times distance hota hai sirf tab jab force constant ho. Yahan force badhti hai jaise tum jaate ho. Toh hum safar ko itne patle slivers mein kaatate hain ki har sliver ke andar force muskil se badalti hai, har ek ke liye force$\times$width compute karte hain, aur unhe stack karte hain. "Infinitely many thin slices stack karo" wali operation exactly wahi hai jo $\int$ ka matlab hai. Hum ise isliye chunte hain kyunki force constant nahi hai — koi bhi simpler cheez honest nahi hogi.

PICTURE. Line $F_{\text{you}}=k{x}^{\prime }$; uske neeche $0$ se $x$ tak shaded triangle stored energy hai. Base $x$ aur height $kx$ wale triangle ka area $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}x\cdot kx=\frac{1}{2}k{x}^2$ hota hai.

$U(x)={\int }_0^x\underset{\text{force at slice }{x}^{\prime }}{\underbrace{k{x}^{\prime }}}\underset{\text{width of slice}}{\underbrace{d{x}^{\prime }}}=\frac{1}{2}k{x}^2$

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Step 4 — Kinetic energy: woh AREA jo motion carry karti hai

KYA. Kinetic energy $K$ = woh energy jo block sirf isliye rakhta hai kyunki woh chal raha hai: $K=\frac{1}{2}m{v}^2$.

KYUN. Block ko push karo, uski speed $v$ badhti hai; invest ki gayi energy yeh $\frac{1}{2}m{v}^2$ ban jaati hai. Yeh doosra piggy bank hai. Hum ise $U$ jaisi bhasha mein chahte hain taaki baad mein add kar sakein.

PICTURE. Teen moments par do banks ka bar chart — edge par (sab $U$ mein), halfway, aur home par (sab $K$ mein). Dekhte hain paisa ek bar se doosre mein kaise jata hai.

$K=\frac{1}{2}\underset{\text{mass}}{\underbrace{m}}\underset{\text{speed, squared}}{\underbrace{v^2}}$

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Step 5 — Motion ko words mein daalte hain: $x(t)$ aur $v(t)$

KYA. Block ki position over time (SHM se) ek cosine wave hai; uski speed (Velocity & Acceleration in SHM se) ek sine wave hai.

$x(t)=A\cos (\omega t+\varphi ),v(t)=-A\omega \sin (\omega t+\varphi )$

KYUN. Yeh prove karne ke liye ki total kabhi nahi badlata, hum ise har instant $t$ par dekhte hain. Hume $x$ aur $v$ time ke functions ke roop mein chahiye taaki unhe $U$ aur $K$ mein daal sakein.

PICTURE. Do stacked curves: $x(t)$ ek cosine (upar se shuru, $A$ par), $v(t)$ ek flipped sine (zero se shuru). Jahan ek apne peak par hai doosra zero par hai — woh "90° out of step" hain.

$v=\underset{\begin{array}{c}\text{moving}\\ \text{left first}\end{array}}{\underbrace{-}}\underset{\text{amplitude}}{\underbrace{A}}\underset{\begin{array}{c}\text{angular}\\ \text{speed}\end{array}}{\underbrace{\omega }}\underset{\begin{array}{c}\text{zero at edges,}\\ \text{max at home}\end{array}}{\underbrace{\sin (\omega t+\varphi )}}$

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Step 6 — Substitute karo aur do mirror-image humps reveal karo

KYA. Time-functions ko har bank mein plug karo. $m{\omega }^2=k$ use karte hue (kyunki ${\omega }^2=k/m$):

$K=\frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi )$ $U=\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2{\cos }^2(\omega t+\varphi )$

KYUN. $m{\omega }^2\to k$ swap karna dono banks ko same coat pehna deta hai — ab front mein har symbol $\frac{1}{2}k{A}^2$ hai. Sirf trig wiggle alag hai. Yahi magic cancellation ka setup hai.

PICTURE. $K$ (ek ${\sin }^2$ hump) aur $U$ (ek ${\cos }^2$ hump) saath plotted. Jab ek upar hota hai doosra neeche hota hai — perfect see-saw — aur unki heights hamesha same ceiling tak add hoti hain.

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Step 7 — Magic cancellation: humps add karo

KYA. Dono banks add karo:

$E=K+U=\frac{1}{2}k{A}^2\left[\underset{=1\text{always}}{\underbrace{{\sin }^2(\cdot )+{\cos }^2(\cdot )}}\right]=\frac{1}{2}k{A}^2$

YEH TOOL KYUN — the Pythagorean identity. ${\sin }^2\theta +{\cos }^2\theta =1$ ek coincidence nahi hai: unit circle par, $\cos \theta$ aur $\sin \theta$ ek right triangle ki do legs hain jiska hypotenuse 1 hai, aur legs² ka sum hypotenuse² hota hai (Pythagoras). Hum ise isliye use karte hain kyunki hamare do humps exactly ek ${\sin }^2$ aur ek ${\cos }^2$ hain jinke front factors identical hain — woh ek tool jo unhe ek constant mein collapse kar deta hai.

PICTURE. Do humps total height $\frac{1}{2}k{A}^2$ ki ek bar mein stack hue — poore time ke across ek flat ceiling. See-saw tips karta hai, lekin chhath kabhi nahi hilti.

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Step 8 — Edge cases: corners check karo

KYA / KYUN. Ek derivation tab tak khatam nahi hoti jab tak hum extreme moments test na karein — reader ko koi bhi aisa case nahi milna chahiye jo humne skip kiya ho.

PICTURE. Energy bar par chaar labelled moments: do turning points, home, aur equal-split point.

• $x=+A$ par (right edge): $v=0$ (block frozen). $K=0$, $U=\frac{1}{2}k{A}^2$. Sab stretch bank mein. ✓
• $x=-A$ par (left edge): same, kyunki sign chahe kuch bhi ho $x^2=A^2$ hota hai. Step 1 ki mirror-symmetry pay off hoti hai — dono edges same energy hain. ✓
• $x=0$ par (home): $U=0$, toh $K=\frac{1}{2}k{A}^2$ aur $v=v_{\max }=\omega A$. Sab motion bank mein. ✓
• Degenerate $A=0$ (kabhi khaincha nahi): $E=\frac{1}{2}k(0{)}^2=0$. Koi pull nahi, koi energy nahi, block kabhi nahi hilta. ✓
• Equal split $K=U$: har ek ceiling ka aadha, $\frac{1}{2}k{x}^2=\frac{1}{4}k{A}^2\Rightarrow x=\pm A/\sqrt{2}\approx 0.707A$. ✓

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Ek-picture summary

Sab ek saath: parabola bowl $U=\frac{1}{2}k{x}^2$, ulta $K=\frac{1}{2}k(A^2-x^2)$, aur flat ceiling $E=\frac{1}{2}k{A}^2$ jis tak woh hamesha add hote hain — block ko ek marble ki tarah bowl mein rolling karte draw kiya gaya hai, height aur speed trade karte hue.

Recall Feynman retelling — plain words mein poora walkthrough

Humne ek block ek spring par rakha aur "home" mark kiya. Spring hamesha use wapis khainchta hai, jitna door utna hi harder — ek straight-line law. Use bahar khainchne ke liye tum us badhti hui pull se ladte ho, aur jo kaam tum kharach karte ho woh ek triangle of area ke roop mein pile up hota hai, jo $\frac{1}{2}k{x}^2$ hai — yeh stretch bank hai. Jab block chalta hai toh woh $\frac{1}{2}m{v}^2$ bhi rakhta hai — motion bank. Humne dono ko time ke upar likha: ek ${\sin }^2$ hump nikla, doosra ${\cos }^2$ hump, dono same coat $\frac{1}{2}k{A}^2$ pehne. Aur yahi punchline hai: ek ${\sin }^2$ aur ek ${\cos }^2$ hamesha exactly 1 tak add hote hain (yeh sirf circle par Pythagoras hai), toh do humps hamesha ek flat ceiling tak add hote hain, $\frac{1}{2}k{A}^2$. See-saw endlessly tips karta hai, chhath kabhi nahi hilti. Edges par sab stretch hai aur block frozen hai; home par sab motion hai aur block sabse fast hai; aur ise koi parwah nahi ki tum home ke left ho ya right, kyunki energy sirf $x^2$ dekhti hai. Dobla door khincho aur, kyunki sab kuch square hai, tum chaar guna energy paate ho.