LC circuit — oscillations (electrical analog of SHM): Visual walkthrough

1.8.30 · D2 · Physics › Electromagnetism › LC circuit — oscillations (electrical analog of SHM)

Step 1 — Do dabbe aur unke andar kya rehta hai

KYA HAI: Hum do cheezein naam karte hain jo time ke saath change hoti hain: $q$ (bucket kitna bhara hai) aur $i$ (charge kitni tezi se move ho raha hai).

KYUN: Jo kuch bhi aage aata hai woh in do numbers ki ek-doosre ki jagah lene ki kahani hai. Agar hum jaanein ki inhe kya push karta hai, toh hum poori motion jaante hain.

PICTURE: Figure mein, upar wali plate $+ q$ carry karti hai, neeche wali $- q$ . Coil daayein taraf hai. Abhi koi current flow nahi ho raha — switch open hai.

Figure — LC circuit — oscillations (electrical analog of SHM)

Step 2 — Har daabba kisse push karta hai (do voltages)

KYA HAI: Hum likhte hain ki har daabba kitna push karta hai.

YEH DO FORMS KYUN, AUR KOI NAHI?

$V_{C} = q / C$ isliye kyunki zyada bhara capacitor zyada push karta hai — charge double karo, push double ho jaata hai. Number $C$ (capacitance, dekho Capacitance and Energy in Capacitors) yeh batata hai ki ek volt banane ke liye kitna charge chahiye. Bada $C$ = bharna easy = charge ke per push weak.
$V_{L} = L \frac{d i}{d t}$ isliye kyunki inductor current ki parwah nahi karta, sirf current ke change ki karta hai (dekho Inductance and Self-Induction). Yeh $\frac{d i}{d t}$ hai "current har second mein kitni tezi se change ho raha hai." Number $L$ (inductance) batata hai ki yeh us change ko kitni zidd se resist karta hai.

PICTURE: Capacitor ka push (magenta arrow) current ko top plate se bahar drive karta hai. Inductor ka back-push (violet arrow) current jo bhi karna chahta hai uske opposite point karta hai.

Step 3 — Loop walk karo (Kirchhoff ka rule)

KYA HAI: Hum loop mein do pushes add karte hain aur sum ko zero set karte hain.

KYUN: Koi battery energy feed nahi kar rahi, loop mein aur kuch nahi hai. Capacitor ka outward push aur inductor ka back-push sirf do voltages hain, aur kisi bhi complete loop mein voltages cancel hone chahiye — warna charge ko har trip mein kahin se bhi energy milti rahegi.

PICTURE: Green arrow loop trace karta hai. Jab hum capacitor se guzarte hain toh $+ q / C$ count karte hain; jab inductor se guzarte hain toh $+ L \frac{d i}{d t}$ count karte hain. Shuru ki jagah wapas pahuncho → net push $= 0$ .

Step 4 — Isko sirf $q$ ki ek equation mein badlna

KYA HAI: Hum $i$ ko eliminate karte hain taaki equation sirf $q$ ki baat kare.

KYUN: Symbol $\frac{d ^{2} q}{d t ^{2}}$ (kaho "d-two-q-d-t-squared") ka matlab hai rate of change ka bhi rate of change — charge ki acceleration. Ek unknown mein ek equation kuch aisa hai jise hum actually solve kar sakte hain.

Boxed result ko term by term padhna:

$\frac{d ^{2} q}{d t ^{2}}$ — current kitni tezi se change ho raha hai (charge ki "acceleration").
$- \frac{1}{L C}$ — ek positive number jiske aage minus sign hai. Minus hi poori kahaani hai.
$q$ — current charge.

PICTURE: Plot dikhata hai: jab bhi $q$ positive hai (bucket mein positive charge hai), acceleration arrow zero ki taraf neeche point karta hai. Jab bhi $q$ negative hai, woh upar point karta hai. Push hamesha waapas empty ki taraf aim karta hai — yehi ek restoring force karta hai.

Step 5 — Ek puraane dost ko pehchanana: yeh SHM equation hi hai

KYA HAI: Hum apni boxed equation ko SHM equation se, symbol by symbol, compare karte hain.

KYUN: Agar do equations ki shape same hai, toh unka solution bhi same hoga. Humein kabhi bhi kisi differential equation ko scratch se solve nahi karna — bas match read off karo. Jo cheez $q$ (ya $x$ ) ko minus sign ke saath multiply kar rahi hai woh hamesha $ω^{2}$ hoti hai.

Match force karta hai: $ω^{2} = \frac{1}{L C} ⟹ ω = \frac{1}{L C}$

Square root kyun? Kyunki equation humein $ω^{2}$ deti hai, $ω$ nahi. $ω$ paane ke liye squaring ko undo karna padega — exactly yehi karta hai.

PICTURE: Do panels side by side — ek mass spring par bounce kar raha hai, aur hamaara LC loop — dictionary $q \leftrightarrow x$ , $L \leftrightarrow m$ , $\frac{1}{C} \leftrightarrow k$ matching arrows ki tarah draw ki gayi hai.

Step 6 — Solution aur $q$ aur $i$ ka 90° dance

Isse padhna:

$q_{0}$ — starting charge (cosine ki height).
$cos (ω t)$ — $1$ par shuru hota hai, toh $q$ full shuru hota hai. ✔ hamaari setup se match.
$- q_{0} ω sin (ω t)$ — $0$ par shuru hota hai, toh $i$ rest par shuru hota hai. ✔
$i_{0} = q_{0} ω$ — peak current, tab milta hai jab cosine zero cross karta hai.

90° offset kyun? Kisi bhi cosine ko dekho: apne peak par uski slope flat hoti hai (zero). Apne zero crossing par uski slope steepest hoti hai. Kyunki current charge ka slope hai, current zero hota hai jab charge maximum ho aur maximum hota hai jab charge zero ho. Yeh quarter-cycle gap hi $9 0^{\circ}$ phase difference hai.

PICTURE: Ek axis par dono curves. Vertical dashed lines chaar moments mark karti hain: (a) $q$ max, $i = 0$ ; (b) $q = 0$ , $i$ sabse zyada negative; (c) $q$ sabse zyada negative, $i = 0$ ; (d) $q = 0$ , $i$ sabse zyada positive.

Step 7 — Energy: total kabhi nahi hilti (degenerate/limiting check)

KYA HAI: Hum do energies add karte hain aur wobble ko cancel hote dekhte hain.

KYUN: Identity $cos^{2} θ + sin^{2} θ = 1$ time-dependence ko gayab kar deti hai. Koi resistor nahi toh energy leak hone ki jagah hi nahi, isliye sum flat hona hi chahiye. Yeh ultimate sanity check hai — agar constant na hota, toh humse galti hui hoti. (Resistor add karo aur Damped Oscillations / LCR Circuit mil jaata hai.)

Woh limiting cases jo picture mein dikhaaye jaane chahiye:

$t = 0$ par: saari energy electric hai ( $U_{E}$ full, $U_{B} = 0$ ) — bucket bhara, flywheel still.
$ω t = π /2$ par: saari energy magnetic hai ( $U_{E} = 0$ , $U_{B}$ full) — bucket khaali, flywheel sabse tez spin kar raha hai.
Jahan woh cross karte hain: $cos^{2} = sin^{2} \Rightarrow ω t = π /4 \Rightarrow t = T /8$ . Energy 50/50 split.

PICTURE: $U_{E}$ (magenta) aur $U_{B}$ (violet) dono wobble karte hain, lekin orange line $U_{E} + U_{B}$ bilkul flat hai. Crossings $T /8, 3 T /8, \dots$ par mark hain.

Ek-picture summary

Recall Poore walkthrough ki Feynman-style retelling

Ek bucket of water ko ek bhaari paddle-wheel se ek pipe se joda hua imagine karo, kahin bhi friction nahi. Bucket bhara aur wheel still se shuru karo (Step 1–2). Bhara bucket paani bahar push karta hai — woh push hai $V_{C} = q / C$ (bhara = zyada push). Lekin bhaari wheel suddenly spin up hone se resist karta hai; uski change se resistance hai $V_{L} = L d i / d t$ (Step 2). Kyunki yeh ek closed loop hai aur kuch nahi, woh do pushes loop ke chaaron taraf exactly cancel hone chahiye (Step 3, Kirchhoff). "Flow" ko "bucket kitni tezi se khaali hota hai" ki tarah likhne par ek clean law milta hai: charge ki acceleration minus ek constant times charge itself hai (Step 4). Woh minus sign ka matlab hai ki circuit charge ko hamesha khaali ki taraf wapas kheenchta hai — disguise mein spring. Humne woh exact equation pehle dekhi hai — yeh mass on spring hai (Step 5). Term by term match karne par, frequency $1/ L C$ honi chahiye: bhaari wheel ( $L$ ) ya bada bucket ( $C$ ) dheema sway karta hai. Solve karo aur charge ek cosine hai, current uski slope — ek sine (Step 6). Toh paani tab sabse tez bhagta hai jab bucket khaali hota hai (steepest slope) aur bilkul band ho jaata hai jab bucket bhara hota hai (flat top). Yahi $9 0^{\circ}$ dance hai. Ant mein, bucket ki stored energy ko wheel ki spinning energy mein add karo (Step 7): $cos^{2} + sin^{2} = 1$ ki wajah se, total kabhi nahi badalta. Koi friction nahi toh sway hamesha ke liye chalta hai, energy bucket se wheel mein aur wapas poure hoti rehti hai.

Recall Quick self-check

Boxed equation of motion kya hai? ::: $\frac{d ^{2} q}{d t ^{2}} = - \frac{1}{L C} q$ $ω$ mein square root kyun hona chahiye? ::: Matched equation $ω^{2} = 1/ L C$ deti hai; lena square ko undo karta hai. Current maximum kab hota hai? ::: Jab charge zero se guzarta hai (cosine ka steepest slope). Total energy constant kya rakhta hai? ::: Identity $cos^{2} + sin^{2} = 1$ aur kisi resistor ki absence. Energy 50/50 share kab hoti hai? ::: $t = T /8$ par (jahan $ω t = π /4$ ).

Connections

Simple Harmonic Motion — woh equation jisse hum Step 5 mein match karte hain.
Capacitance and Energy in Capacitors — $V_{C} = q / C$ aur $U_{E} = q^{2} /2 C$ ka source.
Inductance and Self-Induction — $V_{L} = L d i / d t$ ki origin.
Kirchhoff's Voltage Law — Step 3 ka loop rule.
Damped Oscillations / LCR Circuit — Step 7 ki flat line kya ban jaati hai jab $R \neq = 0$ ho.
Resonance and AC Circuits — $ω = 1/ L C$ resonant frequency ki tarah dobara aata hai.