Visual walkthrough — SHM differential equation — solution - x = A cos(ωt + φ)

1.6.2 · D2 · Physics › Oscillations & Waves › SHM differential equation — solution - x = A cos(ωt + φ)

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Step 0 — $\dot{x}$ aur $\ddot{x}$ ke ye squiggles matlab kya hain?

KYA HAI. Kisi bhi physics se pehle, is notation se milte hain. Maano $x$ ek ball ki distance hai table ki centre line se, metres mein, har instant of time $t$ par.

• $x$ = position (ball kahan hai).
• $\dot{x}$ = velocity = "$x$ abhi kitni tezi se change ho raha hai" (metres per second). Ek dot shorthand hai "rate of change of" ke liye.
• $\ddot{x}$ = acceleration = "velocity kitni tezi se change ho rahi hai" (metres per second, per second). Do dots = "rate of change of the rate of change."

YEH notation kyun. Physicists letter ke upar dots lagate hain $\frac{dx}{dt}$ baar baar likhne ki jagah — matlab same hi hai, bas chhota. Har dot matlab hai "time ke saath graph ka slope lo."

PICTURE. Neeche: ek position curve (blue), ek marked point par uska slope (velocity) orange tangent line ke roop mein. Jahan tangent tedi ho, velocity badi hai; jahan curve flat ho, velocity zero hai.

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Step 1 — Yeh equation aslaan kya keh rahi hai

KYA HAI. Simple Harmonic Motion ka defining law hai $\underset{\text{acceleration}}{\underbrace{\ddot{x}}}=-\underset{\begin{array}{c}\text{ek fixed}\\ \text{positive number}\end{array}}{\underbrace{{\omega }^2}}\underset{\text{displacement}}{\underbrace{x}}.$

Term by term: $\ddot{x}$ hai jahan hum accelerate kar rahe hain; $x$ hai jahan hum abhi hain; ${\omega }^2$ ek constant "stiffness" number hai (hamesha positive); minus sign direction ko flip karta hai.

YEH shape kyun. Mass on a spring ke liye hum teen plain quantities use karte hain:

• $m$ = ball ki mass, kilograms mein (kitna stuff hai push karne ke liye).
• $k$ = spring stiffness, newtons per metre mein (spring har metre stretch par kitni zyada khichti hai) — dekho Hooke's law.
• $F$ = ball par force, newtons mein (total push ya pull jo use feel hoti hai).

Hooke's law kehta hai spring ki pull hai $F=-kx$ (pull stretch ke opposite hai, isliye minus). Newton's second law kehta hai $F=m\ddot{x}$ (force = mass × acceleration). Kyunki dono $F$ ke barabar hain, unhe equal set karo aur $m$ se divide karo: $m\ddot{x}=-kx⟹\ddot{x}=-\frac{k}{m}x.$ Hum positive number $\frac{k}{m}$ ko naam dete hain ${\omega }^2$ (ise square ke roop mein likhne se guarantee hoti hai ki yeh positive hai, jo baad mein kaam aayega). Toh $\omega =\sqrt{k/m}$.

PICTURE. Acceleration ka arrow hamesha $x=0$ ki taraf wapas point karta hai, aur jitna door jaate ho utna lamba hota jaata hai. Yahi hai minus sign ki poori personality.

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Step 2 — Equation ko integrable banane ki ek trick: $\dot{x}$ se multiply karo

KYA HAI. Hum $\ddot{x}=-{\omega }^{2}x$ ke dono sides ko velocity $\dot{x}$ se multiply karte hain: $\underset{\text{velocity}\times \text{accel}}{\underbrace{\dot{x}\ddot{x}}}=-{\omega }^2\underset{\text{position}\times \text{velocity}}{\underbrace{x\dot{x}}}.$

YEH bizarre step kyun? Kyunki ek hidden calculus identity hai: $\dot{x}$ se multiply karne par dono sides "kisi cheez ka slope" ban jaate hain, aur jo bhi slope ho use integrate (un-slope) kiya ja sakta hai cleanly. Yahan do identities hain, chain rule se sidhe dikhaye gaye hain (kisi squared cheez ka slope 2×cheez×uska slope hai, aur $\frac{1}{2}$ 2 ko cancel karta hai): $\frac{d}{dt}\left(\frac{1}{2}{\dot{x}}^2\right)=\frac{1}{2}\cdot 2\dot{x}\cdot \ddot{x}=\dot{x}\ddot{x},\frac{d}{dt}\left(\frac{1}{2}{x}^2\right)=\frac{1}{2}\cdot 2x\cdot \dot{x}=x\dot{x}.$ Toh hamare multiplied equation ka left side hai $\frac{d}{dt}(\frac{1}{2}{\dot{x}}^2)$ aur right side hai $-{\omega }^2\frac{d}{dt}(\frac{1}{2}{x}^2)$. Yeh secretly energy conservation apne aap saamne aa rahi hai.

PICTURE. Do "bump" curves. Blue wali hai $y=\frac{1}{2}{x}^2$ position $x$ ke against plot ki gayi hai (horizontal axis) — kisi bhi point par uska tangent slope $x$ ke barabar hai. Orange wali hai $y=\frac{1}{2}{\dot{x}}^2$ velocity $\dot{x}$ ke against same horizontal axis par plot ki gayi — uska slope $\dot{x}$ ke barabar hai. Picture ek claim visible karne ke liye hai: kisi squared quantity ka slope (woh quantity) khud hoti hai, jo exactly woh chain-rule identity hai jo humne upar use ki.

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Step 3 — Integrate karo taki speed–position relation mile

KYA HAI. Dono sides ko un-slope karo (matlab integrate karo: slopes ko add karo taaki original recover ho): $\frac{1}{2}{\dot{x}}^2=-\frac{1}{2}{\omega }^2{x}^2+C.$ Yahan $C$ constant of integration hai — ek leftover number jo humne abhi tak pin down nahi kiya. Rearrange karo aur $C=\frac{1}{2}{\omega }^2{A}^2$ rename karo: $\boxed{{\dot{x}}^2={\omega }^2(A^2-x^2)}.$

Term by term: ${\dot{x}}^2$ speed-squared hai; $A$ constant ka ek naya naam hai, is liye choose kiya taaki equation tidy lage.

$C$ ko $\frac{1}{2}{\omega }^2{A}^2$ kyun likha? Boxed line dekho: jab $x=\pm A$ ho, right side zero hai, toh $\dot{x}=0$. Jahan speed zero ho woh exactly ek turning point hai — ball apni furthest reach par momentarily ruk gayi. Toh $A$ automatically amplitude ban jaata hai (maximum displacement). Humne yeh assume nahi kiya; algebra ne khud dediya.

PICTURE. Speed vs position ek ellipse-jaisi arch hai: centre $x=0$ par fastest, edges $x=\pm A$ par zero.

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Step 4 — Variables separate karo taki sinusoid forced ho ke aaye

KYA HAI. Box ka square root lo aur position ek side par, time doosri side par rakhdo ($\dot{x}=dx/dt$ use karke): $\frac{dx}{\sqrt{A^2-x^2}}=\pm \omega dt.$

Term by term: left side par, jahan hain usse related sab kuch; right side par, time se related sab kuch. $\pm$ carry karta hai ki hum bahar ja rahe hain ($+$) ya wapas aa rahe hain ($-$).

YEH exact rearrangement kyun? Kyunki left-hand side ek standard integral from circle geometry hai. Pucho "kis function ka slope $\frac{1}{\sqrt{A^2-x^2}}$ hai?" aur seedha jawab hai $\int \frac{dx}{\sqrt{A^2-x^2}}=\arcsin \left(\frac{x}{A}\right)+\text{const}=-\arccos \left(\frac{x}{A}\right)+\text{const}.$ Arcsine aur arccosine dono kaam karte hain — ye sirf ek constant se differ karte hain (actually $\arcsin u=\frac{\pi }{2}-\arccos u$), aur koi bhi constant us phase mein absorb ho jaata hai jo hum introduce karne wale hain. Hum cosine by hand choose nahi kar rahe — equation khud circle-function ko apni logic se kheench laati hai.

PICTURE. Chhoti si slice $dx$ shrinking length $\sqrt{A^2-x^2}$ ke upar (radius $A$ ke circle ke andar ek right triangle ki vertical side) — yeh circle par ek angle trace karti hai — reference circle ka janm.

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Step 5 — Dono sides integrate karo → solution saamne aata hai (aur $\pm$ kahan jaata hai)

KYA HAI. Integrate karo. Arccosine form use karte hue, aur sign ka dhyan rakhte hue: $-\arccos \left(\frac{x}{A}\right)=\pm \omega t+(\text{const}).$ $-1$ se multiply karo aur saare constants (integration se bhi aur $\pm$ se bhi) ek naye letter $\varphi$ mein ikatha karo: $\arccos \left(\frac{x}{A}\right)=\mp \omega t+\varphi .$ Ab arccos ko "undo" karo dono sides ka cosine lekar ($\arccos$ ke inverse function hone ka yahan kaam aata hai — $\cos (\arccos u)=u$): $\boxed{x(t)=A\cos (\omega t+\varphi )}$

$\pm$ kyun gayab ho jaata hai. Kyunki $\cos$ ek even function hai, $\cos (-\theta )=\cos (\theta )$, toh $\omega t$ ke aage leftover $\mp$ sign koi fark nahi karta — cosine ke andar gayab ho jaata hai. Step 4 se woh $\pm$ do tarahon se absorb hua: kuch $\varphi$ mein (ek shift), aur baki cosine ki evenness ne maar diya. Yahi reason hai ki $\varphi$, jo poore $[0,2\pi )$ mein range karta hai, "left ki taraf move karna shuru kiya" vs "right ki taraf move karna shuru kiya" — dono capture karne ke liye kaafi hai — tumhe alag sign ki zaroorat nahi.

Term by term, boxed answer:

• $A$ — amplitude, maximum reach (Step 3 se).
• $\omega$ — angular frequency rad/s mein, phase kitni tezi se spin karta hai (Step 1 se).
• $t$ — time.
• $\varphi$ — phase constant, swing mein kahan se shuru kiya, aur kis direction mein (Step 5 se).

PICTURE. Completed cosine wave, $A$ uski height mark karta hai aur $\varphi$ uska horizontal shift.

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Step 6 — Exactly DO leftover constants ($A$ aur $\varphi$) kyun?

KYA HAI. Ek constant mila jab Step 3 mein integrate kiya ($C$, jo $A$ ban gaya) aur doosra jab Step 5 mein integrate kiya ($\varphi$). Do integrations → do constants.

Exactly do, physically kyun? Kisi aise motion ka prediction karne ke liye jiska second derivative fixed ho, tumhe do starting facts batane padenge: $t=0$ par ball kahan hai, aur kitni tezi se ja rahi hai. $t=0$ par solution aur uski velocity mein plug karo: $x(0)=A\cos \varphi ,v(0)=-A\omega \sin \varphi .$ Inhe $A$ aur $\varphi$ ke liye solve karne par: $A=\sqrt{x_0^2+\frac{v_0^2}{{\omega }^2}},\tan \varphi =-\frac{v_0}{\omega x_0}.$

$\frac{v_0}{\omega }$ isliye aata hai kyunki velocity ki units hain (position $\times \omega$); $\omega$ se divide karne par speed wapas length mein convert hoti hai taki woh $x_0$ ke saath same square root ke neeche baith sake.

PICTURE. Do dials — ek height set karta hai ($A$), ek starting angle ($\varphi$) — milkar ek unique wave lock karte hain.

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Step 7 — Har starting case, taki koi scenario surprise na kare

KYA HAI & PICTURE. $\varphi$ sahi choose karne ka matlab hai dono $x_0$ aur $v_0$ ka sign check karna, kyunki $\tan \varphi$ akela do opposite situations mein farq nahi kar sakta (tangent har $180°$ par repeat karta hai). General rule ke liye dekho phase. Chaar honest cases:

Start condition	$x_0$	$v_0$	Correct $\varphi$	Resulting wave
Right edge se rest mein release	$+A$	$0$	$0$	$A\cos \omega t$
Left edge se rest mein release	$-A$	$0$	$\pi$	$-A\cos \omega t$
Centre se right ki taraf move karte hue	$0$	$+$	$-\frac{\pi }{2}$	$A\sin \omega t$
Centre se left ki taraf move karte hue	$0$	$-$	$+\frac{\pi }{2}$	$-A\sin \omega t$

Centre-crossing cases $\mp \frac{\pi }{2}$ kyun pick karte hain. Centre par $x_0=0$ hai, toh $\cos \varphi =x_0/A=0$, jo force karta hai $\varphi =\pm \frac{\pi }{2}$ (wo do angles jinka cosine zero hai). Unke beech ka tie velocity ke sign se toot ta hai: yaad karo $v(0)=-A\omega \sin \varphi$. Right ki taraf move karne ka matlab $v_0>0$, jo $\sin \varphi <0$ maangta hai, yani $\varphi =-\frac{\pi }{2}$. Left ki taraf move karne ka matlab $v_0<0$, jo $\sin \varphi >0$ maangta hai, yani $\varphi =+\frac{\pi }{2}$. Toh tangent-sign discussion aur velocity sign dono agree karte hain: same rule, "check karo kis direction mein move kar rahe ho," dono resolve karta hai.

Degenerate check — agar $\omega \to 0$ ho? Toh $k/m\to 0$: koi spring nahi, koi restoring pull nahi. "Oscillation" period $T=2\pi /\omega \to \infty$ — ball kabhi wapas nahi aati. Correct hai: jab kuch ghar ki taraf nahi kheench raha, koi wiggle hi nahi hogi. (Thodi si friction add karne par damped motion milti hai, woh alag story hai.)

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One-picture summary

Upar sab kuch teen linked panels mein compress kiya gaya hai, left se right: (1) the law — red arrows hamesha $x=0$ ki taraf wapas point karte hue, dur jaane par lambe hote hue (Step 1); (2) speed-vs-position arch aur woh circle jo ban jaata hai (Steps 3–4); (3) resulting cosine wave (Step 5). Panels ko ek sentence ki tarah padho: pull ghar ki taraf → speed law ek circle hai → position ek cosine hai.

Recall Feynman: poora walkthrough ek 12-saal ke bachche ko batao

Ek rubber band par ball ek tug feel karti hai jo jitna zyada khiche, utna strong hota jaata hai — aur tug hamesha beech ki taraf point karta hai. Woh ek fact ("pull ghar ki taraf, aur distance ke saath badhta hai") hamaari starting equation hai. Hum answer guess nahi kar sakte, toh ek trick khelte hain: equation ko ball ki speed se multiply karo. Magically dono sides "kisi pahaad ki steep ness" ban jaate hain, aur steepness-of-a-hill kuch aisa hai jo hum add back kar sakte hain (integrate). Aisa karne par speed aur position ko link karne wala rule milta hai: beech mein fastest, dono edges par frozen — aur un edges ko hum naam dete hain $A$, the amplitude. Us rule ko rearrange karne par, circle-geometry ka ek shape (arcsine/arccosine) force ho ke saamne aata hai — humne use choose nahi kiya. Use undo karo aur nikal aata hai $x=A\cos (\omega t+\varphi )$. Do mystery letters bas do facts hain jo koi bhi motion shuru karne ke liye chahiye: swing kitni badi hai ($A$) aur swing mein kahan se shuru kiya, plus kis direction mein the ($\varphi$). Check karo tum kahan se start kar rahe ho aur kis direction mein move kar rahe ho, aur tumhare paas exact wave hai.

Recall Quick self-test

Exactly do constants kyun hone chahiye? ::: Equation mein second derivative hai, toh do integrations hote hain, har ek ek constant chhod jaata hai; physically, tumhe starting position aur starting velocity batani padti hai. Cosine kahan se aata hai — kya humne assume kiya tha? ::: Nahi; integral $\int dx/\sqrt{A^2-x^2}$ equals $\arcsin (x/A)$ (ya $-\arccos (x/A)$), ek circle-function, toh sinusoid algebra ki apni logic se force ho ke nikalta hai. Do situations dono $\tan \varphi =0$ deti hain. $\varphi$ kaise choose karte ho? ::: $x_0$ ka sign check karo: $\cos \varphi$ ko us se match karna chahiye, jo $x_0>0$ ke liye $\varphi =0$ aur $x_0<0$ ke liye $\varphi =\pi$ deta hai. Step 4 se variables separate karne wala $\pm$ kahan gaya? ::: Uska kuch hissa phase $\varphi$ mein fold ho gaya; baaki maar diya gaya kyunki cosine even hai, $\cos (-\theta )=\cos \theta$.