Visual walkthrough — Forced oscillations — driving frequency

1.6.11 · D2 · Physics › Oscillations & Waves › Forced oscillations — driving frequency

Hum vault ki kuch ideas use karenge: spring ka picture Simple Harmonic Motion se, drag ka idea Damped Oscillations se, phasor shortcut Complex Exponential Method se, aur — Step 7 mein peak par — sharpness ki language Resonance and Quality Factor se.

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Step 1 — Mass ko push karne wali har force draw karo

KYA. Mass $m$ ka ek block spring par rakho. Teen arrows us par act karte hain. Hum unhe add karte hain — bas itna hi Newton's second law kehta hai: (mass) × (acceleration) = (arrows ka sum).

KYUN. Kisi bhi symbol ka matlab samajhne se pehle, hume pata hona chahiye ki physically block ko kya dhak raha hai. Theek teen pushes hain aur koi nahi, isliye is poore page ka baaki hissa in teen arrows ka bookkeeping hai.

PICTURE. Figure dekho — block position $x$ par baith a hai jo apni rest jagah se measure ki gayi hai (vertical dashed line).

Teen arrows, picture se padhke:

• Spring pull $-kx$. Spring hamesha block ko dashed line ki taraf wapas kheenchta hai, isliye jab $x$ positive hai (block daayein) to yeh arrow baayen point karta hai → minus sign. $k$ spring ki stiffness hai (from Simple Harmonic Motion).
• Drag $-b\dot{x}$. Yahan $\dot{x}$ (padho "$x$-dot") block ki velocity hai — $x$ kitni tezi se change ho rahi hai. Drag motion ko oppose karta hai, isliye velocity $\dot{x}$ se ladta hai → ek aur minus. $b$ drag strength hai (from Damped Oscillations).
• Haath $F_0\cos ({\omega }_{d}t)$. Ek bahari agent aage-peechhe dhakelta hai. $F_0$ dhakele ki strength hai, ${\omega }_d$ hai kitni baar haath cycle karta hai, aur $\cos ({\omega }_{d}t)$ use $+F_0$ se $-F_0$ tak smoothly swing karata hai.

Inhe add karne par:

$\underset{\text{mass}\times \text{accel.}}{\underbrace{m\ddot{x}}}=\underset{\text{spring back}}{\underbrace{-kx}}\underset{\text{drag fights }\dot{x}}{\underbrace{-b\dot{x}}}+\underset{\text{the hand}}{\underbrace{F_0\cos ({\omega }_{d}t)}}$

Yahan $\ddot{x}$ ("$x$-double-dot") acceleration hai — velocity khud kitni tezi se change ho rahi hai.

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Step 2 — Equation ko pure physics constants mein tidy karo

KYA. Poori line ko $m$ se divide karo. Phir bachne wale constants ke clumps ko rename karo.

KYUN. Raw numbers $k,b,m$ un do cheezon ko chhupaate hain jo actually behaviour control karti hain: spring kitni tezi se bounce karna chahta hai aur friction kitni tezi se energy khaata hai. Hum unhe repackage karte hain taaki woh do ideas saamne aa jayein.

PICTURE. Figure mein, wohi teen arrows tidy constants se relabel kiye gaye hain — physically kuch nahi badla, sirf naam.

Do naye symbols define karo dekhkar ki $m$ se divide karne ke baad kya bachta hai:

• ${\omega }_0^2=\frac{k}{m}$, isliye ${\omega }_0=\sqrt{k/m}$ — natural frequency. Kyunki $k$ aur $m$ dono positive hain, ${\omega }_0$ ek real positive number hai (hum hamesha positive root lete hain). ${\omega }_0$ (omega-zero) woh rhythm hai jisme block koi friction aur koi haath nahi hone par bounce karta. Stiffer spring ya hafla mass → tezi natural rhythm.
• $2\gamma =\frac{b}{m}$ — damping rate. $\gamma$ (gamma) batata hai ki free wobble $e^{-\gamma t}$ ki tarah kitni tezi se marti hai. "$2$" ek convenience hai jo baad ki algebra ko cleanly land karaata hai.

Tidy equation:

$\ddot{x}+\underset{\text{friction rate}}{\underbrace{2\gamma }}\dot{x}+\underset{{\text{natural rhythm}}^2}{\underbrace{{\omega }_0^2}}x=\underset{\text{drive per kg}}{\underbrace{\frac{F_0}{m}}}\cos ({\omega }_{d}t)$

Har term ab ek acceleration hai (units of ${\text{m/s}}^2$). Left side oscillator ki apni personality hai; right side bahari duniya hai jo use dhakela maar rahi hai.

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Step 3 — Transient fade ho jaata hai; sirf driven part bachta hai

KYA. Full motion do pieces ka sum hai: ek transient (block ki apni damped wobble) aur ek steady state (haath ki response). Hum transient ko marte dekhte hain aur use fenk dete hain.

KYUN. Ek periodic haath apni frequency ke alawa kisi aur frequency par response support nahi kar sakta. Natural frequency ${\omega }_0$ par jo kuch bhi hai usse koi feed nahi kar raha, isliye friction use $e^{-\gamma t}$ ke zariye drain kar deta hai. Kaafi der tak intezaar karo aur sirf fed motion bachti hai.

PICTURE. Figure teen curves stack karta hai: dying transient (flat line mein fade ho jaata hai), never-dying steady state, aur unka sum (shuru mein messy, phir clean repeat mein settle ho jaata hai).

Common underdamped case ke liye (halka friction, $\gamma <{\omega }_0$) transient ek decaying wobble hai:

$x(t)=\underset{\text{transient}\to 0}{\underbrace{C{e}^{-\gamma t}\cos ({\omega }_{1}t+\psi )}}+\underset{\text{steady state, survives}}{\underbrace{x_{ss}(t)}}$

Transient ke symbols padhkar:

• $C$ — transient ka starting size, aur $\psi$ (psi) uska starting phase. Yeh do haath se set nahi hote; yeh is baat se fix hote hain ki block kaise release kiya gaya (uski initial position aur velocity). Yeh sirf shuru ke messy moments ko affect karte hain, final motion ko kabhi nahi.
• ${\omega }_1=\sqrt{{\omega }_0^2-{\gamma }^2}$ — damped wobble frequency. Friction free bounce ko ${\omega }_0$ se thoda slow kar deta hai; woh shifted rhythm ${\omega }_1$ hai (yeh seedha Damped Oscillations se aata hai). Yeh real hai sirf jab $\gamma <{\omega }_0$, jo exactly underdamped assumption hai.

Envelope $e^{-\gamma t}$ (figure mein orange fading curtain) transient ko kuch nahi kar deta. Yahan se hum sirf $x_{ss}$ ka peechha karte hain.

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Step 4 — Response shape guess karo, ek lag ke saath

KYA. Propose karo ki steady state drive ki tarah dikhti hai lekin shifted hai: $x_{ss}(t)=A\cos ({\omega }_{d}t-\varphi )$

KYUN. Kyunki Step 3 frequency ${\omega }_d$ fix kar deta hai, baaki sirf do freedoms hain — swing kitni badi hai ($A$) aur haath se kitni peeche hai ($\varphi$). ${\omega }_d$ par do dials wala pure cosine aise sab se general motion hai.

PICTURE. Figure haath ki push aur block ki response ko overlay karta hai. Block ka peak baad mein aata hai — woh time gap, angle mein convert karke, $\varphi$ hai.

• $A$ — amplitude, response ki height (jo hum solve karna chahte hain).
• $\varphi$ — phase lag (phi). Haath peak karta hai; thodi der baad block peak karta hai. $({\omega }_{d}t-\varphi )$ mein minus ka matlab hai block peeche chalta hai. Damping is delay ki wajah hai: friction block ko turant jawab dene se rokta hai. Hum $\varphi$ ko $0\le \varphi \le \pi$ range mein rakhte hain (response "in step" se "opposite" tak lag kar sakta hai, kabhi lead nahi karta).

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Step 5 — Substitute karo, term-by-term match karo, ek right triangle padho

KYA. Guess ko tidy equation mein daalo. Hume ${\dot{x}}_{ss}$ aur ${\ddot{x}}_{ss}$ chahiye: ${\dot{x}}_{ss}=-A{\omega }_d\sin ({\omega }_{d}t-\varphi ),{\ddot{x}}_{ss}=-A{\omega }_d^2\cos ({\omega }_{d}t-\varphi )$ Har dot ek ${\omega }_d$ ka factor kheench laata hai aur cosine ko rotate karta hai → sine → cosine (signs ke saath).

KYUN. Guess solution hone ke liye, left side ko right side ke barabar hona chahiye har ek instant $t$ par. Ek cosine aur ek sine usi argument ke same independent shapes hain — woh kabhi ek doosre ko cancel nahi karte — isliye equation sirf tab sabhi $t$ ke liye hold ho sakti hai jab cosine coefficient left par right ke cosine coefficient ke barabar ho, AUR sine coefficient left par right ke sine coefficient ke barabar ho. Woh "term-by-term match" rule ek equation-jo-hamesha-hold-hoti-hai ko do ordinary equations mein badal deta hai, ek shape ke hisaab se, jo humari do unknowns $A$ aur $\varphi$ pin down karne ke liye exactly kaafi hai.

Sab kuch argument $u={\omega }_{d}t-\varphi$ mein likhte hue, left side collect hoke banti hai $[({\omega }_0^2-{\omega }_d^2)A]\cos u-[(2\gamma {\omega }_d)A]\sin u,$ aur right side $\frac{F_0}{m}\cos ({\omega }_{d}t)$ expand hoti hai (with ${\omega }_{d}t=u+\varphi$) ek cosine-part aur ek sine-part mein. Do shapes match karne par milta hai: $\underset{\text{cosine coefficients}}{\underbrace{({\omega }_0^2-{\omega }_d^2)A=\frac{F_0}{m}\cos \varphi }},\underset{\text{sine coefficients}}{\underbrace{(2\gamma {\omega }_d)A=\frac{F_0}{m}\sin \varphi }}.$

PICTURE — kyun woh do equations ek right triangle ki legs hain. Yeh Complex Exponential Method phasor idea hai, jo hum yahan seedha bolte hain taaki isse faith par na lena pade. Phasor bas ek arrow hai jiska length term ka amplitude hai aur jiska direction uska phase record karta hai; $\cos$ term horizontal axis ke along point karta hai, aur $\sin$ term vertical axis ke along (ek quarter-turn away), kyunki sine ek cosine hai jo 90° se shift hai. Isliye upar diye gaye do matched equations literally ek arrow-sum ke horizontal aur vertical components hain, aur drive $F_0/m$ woh arrow hai. Figure dekho:

• Horizontal leg cosine equation hai: $({\omega }_0^2-{\omega }_d^2)A$ — in-step part, spring-vs-inertia frequency mismatch.
• Vertical leg sine equation hai: $(2\gamma {\omega }_d)A$ — quarter-turn damping cost, right angles par kyunki drag velocity follow karta hai, jo displacement se 90° lead karta hai.
• Hypotenuse drive strength $F_0/m$ hai.

Do matched equations ko square-and-add karo (${\cos }^2\varphi +{\sin }^2\varphi =1$ use karke) — woh triangle par Pythagoras hai:

${\left(\frac{F_0}{m}\right)}^2=[({\omega }_0^2-{\omega }_d^2)A{]}^2+[(2\gamma {\omega }_d)A{]}^2$

$A^2$ factor out karo aur square root lo:

$\boxed{A({\omega }_d)=\frac{F_0/m}{\sqrt{({\omega }_0^2-{\omega }_d^2{)}^2+(2\gamma {\omega }_d{)}^2}}}$

Sine equation ko cosine equation se divide karo ($\tan \varphi =\sin \varphi /\cos \varphi$) — woh triangle ka corner angle hai, "opposite over adjacent":

$\tan \varphi =\frac{\text{vertical (damping)}}{\text{horizontal (mismatch)}}=\frac{2\gamma {\omega }_d}{{\omega }_0^2-{\omega }_d^2}$

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Step 6 — Dial ${\omega }_d$ ke teen regimes mein walk karo

KYA. Driving-frequency dial ko bahut dheere se, match tak, bahut tezi tak ghoom aur $A$ aur $\varphi$ dekho.

KYUN. Formula tabhi trustworthy hai jab har setting par — extremes sameta — sensible pictures de. Har ek check karo aur confirm karo ki kuch toot nahi raha.

PICTURE. Figure amplitude curve hai jisme teen flags lage hain; har flag ke neeche, ek chhoti clock phase lag $\varphi$ dikhati hai.

• Dheera, ${\omega }_d\ll {\omega }_0$. Mismatch term $\to {\omega }_0^2$, isliye $A\to \frac{F_0/m}{{\omega }_0^2}=\frac{F_0}{k}$. Block bas wahan baithta hai jahan steady force use kheenchti hai — pure Hooke's law. Lag $\varphi \to 0$: block haath ke saath move karta hai.
• Matched, ${\omega }_d\approx {\omega }_0$. Triangle ki horizontal leg gayab ho jaati hai, isliye denominator sirf damping leg $2\gamma {\omega }_d$ hai — chhota — aur $A$ peak karta hai. Triangle ab purely vertical hai, isliye $\varphi \to \pi /2$: block quarter-cycle peeche hai, matlab woh apni khud ki velocity ke saath in step move karta hai → force har instant power feed karti hai (dekho Energy in Oscillations).
• Tez, ${\omega }_d\gg {\omega }_0$. Mismatch term $\to -{\omega }_d^2$ (bahut bada), isliye $A\to \frac{F_0/m}{{\omega }_d^2}\to 0$. Inertia jeetta hai; block pace nahi rakh sakta. Horizontal leg sign flip karta hai, isliye $\varphi \to \pi$: block haath ke ulta move karta hai.

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Step 7 — Peak exactly kahan hai? (degenerate-damping check)

KYA. ${\omega }_d$ dhundo jo $A$ maximise kare. Sabse bada $A$ matlab sabse chhota denominator, isliye root ke neeche wali cheez minimise karo: $D({\omega }_d)=({\omega }_0^2-{\omega }_d^2{)}^2+(2\gamma {\omega }_d{)}^2$ Iska slope zero set karo, $\frac{dD}{d{\omega }_d}=0$, aur solve karo.

KYUN. Hum derivative use karte hain kyunki yeh woh tool hai jo jawaab deta hai "curve kahan rise karna band karta hai aur fall karna shuru karta hai?" — bilkul top par flat point. Wahi peak hai jo hum chahte hain.

PICTURE. Figure peak par zoom karta hai aur uski sahi jagah ${\omega }_{res}$ mark karta hai jo ${\omega }_0$ se thodi si baayen baith ti hai; ek dashed light-damping curve lagbhag ${\omega }_0$ ke upar peak karti hai.

Derivative work karte hue: $\frac{dD}{d{\omega }_d}=2({\omega }_0^2-{\omega }_d^2)(-2{\omega }_d)+2(2\gamma {)}^2{\omega }_d=4{\omega }_d[{\omega }_d^2-{\omega }_0^2+2{\gamma }^2]=0$

Yeh product exactly do tareekon se zero hai, aur hume dono examine karne chahiye — $4{\omega }_d$ se divide karna tab hi allowed hai jab hum alag se us root ka hisaab kar lein jo woh throw away kar deta:

• Root 1: ${\omega }_d=0$. Yeh $D$ ka ek genuine extremum hai, throwaway nahi. ${\omega }_d=0$ par drive static hai aur $D={\omega }_0^4$, jisse $A=F_0/k$ milta hai (Step 6 ka slow-drive value). Check karo ki yeh max hai ya min: bahut halke damping ke liye $D$ ${\omega }_d={\omega }_0$ ke paas ${\omega }_0^4$ se neeche jaata hai, isliye ${\omega }_d=0$ $D$ ka local maximum hai — yaani $A$ ka local minimum resonant bump ki taraf jaate waqt. Isliye yeh ek real stationary point hai, lekin amplitude peak nahi jo hum chahte hain.
• Root 2: bracket $=0$. Ab, ${\omega }_d=0$ pehle handle hone ke baad, hum $4{\omega }_d$ se divide kar sakte hain aur bracket zero set kar sakte hain: ${\omega }_d^2-{\omega }_0^2+2{\gamma }^2=0$, jisse milta hai

$\boxed{{\omega }_{res}=\sqrt{{\omega }_0^2-2{\gamma }^2}}$

Yeh doosra root sahi amplitude peak hai (yahan $D$ bottom out karta hai). Isko padhkar:

• Light damping ($\gamma \ll {\omega }_0$): ${\omega }_{res}\approx {\omega }_0$. Peak essentially natural frequency par hai — textbook slogan.
• Heavy damping ($2{\gamma }^2\ge {\omega }_0^2$): square root toot jaata hai (imaginary ho jaata hai). Tab Root 2 exist nahi karta aur sirf Root 1 bachta hai, isliye $A$ ka koi peak hi nahi hota — woh bass monotonically apne slow-drive value $F_0/k$ se fall off karta hai. System resonate karne ke liye bahut sluggish hai.

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Ek-picture summary

Yeh single figure poore page ko carry karta hai: block par teen force-arrows (Step 1) do clean constants ${\omega }_0,\gamma$ ban jaate hain (Step 2); transient fade ho jaata hai aur ek response chhodd jaata hai jo ${\omega }_d$ par lock hai (Steps 3–4); cosine- aur sine-parts ko term-by-term match karne se ek right triangle banta hai jiska hypotenuse $A$ deta hai aur jiska angle $\varphi$ deta hai (Step 5); dial ghoomane se amplitude curve sweep hoti hai apne peak ke saath ${\omega }_{res}=\sqrt{{\omega }_0^2-2{\gamma }^2}$ par aur sharpness $Q={\omega }_0/2\gamma$ se set hoti hai (Steps 6–7).

Recall Feynman retelling — poori walk plain words mein

Ek block spring par teen dhakele feel karta hai: spring use ghar kheenchta hai, friction uski speed par drag karta hai, aur ek haath aage-peechhe dhakelta hai. Likho "mass times acceleration equals woh teen add up." Saaf karo: spring ki eagerness ${\omega }_0$ ban jaati hai (uska favourite rhythm) aur friction $\gamma$ ban jaata hai (kitni tezi se thak ta hai). Pehle block ek saath do kaam karta hai — apni dying wobble aur haath ki driven motion — lekin wobble fade ho jaati hai (uska size aur starting phase, $C$ aur $\psi$, sirf is baat ke leftovers hain ki tumne kaise chhoda), aur hamesha ke liye block bas haath ki beat ${\omega }_d$ par march karta hai, sirf thoda peeche kyunki friction uske reflexes slow kar deta hai. March ki size dhundhne ke liye, hum marching guess wapas plug in karte hain aur insist karte hain ki yeh har instant par kaam kare; kyunki cosine-follow-karne-wali wiggles aur sine-follow-karne-wali wiggles alag shapes hain jo kabhi cancel nahi karti, hume har shape ko apne aap match karna padega — woh do tidy equations deta hai, jo exactly ek right triangle ki do legs hain. Ek leg hai kitna mismatched haath ki speed block ke favourite rhythm se hai, doosri leg damping cost hai, aur tirchi side haath ki strength hai. Us triangle par Pythagoras amplitude deta hai; triangle ka corner angle lag deta hai. Finally, haath ki speed dial ghoomao: bahut dheere aur block bass ek stretched spring ki tarah follow karta hai; matched aur block sabse badi swing karta hai (resonance); bahut tezi aur inertia jeetta hai isliye woh barely hiltata hai, haath ke ulta chalte hue. Sabse badi swing favourite rhythm se ek baal neeche baith ti hai — woh swing kitni tall aur narrow hai $Q={\omega }_0/2\gamma$ par depend karta hai — aur agar friction bahut badi hai, to koi badi-swing bump nahi hota bilkul.

Recall Quick self-check

Amplitude formula ka denominator kahan se aata hai? ::: Step 5 ke term-balance triangle par Pythagoras: horizontal leg = mismatch $({\omega }_0^2-{\omega }_d^2)$, vertical leg = damping $2\gamma {\omega }_d$. Hum cosine- aur sine-parts alag kyun match kar sakte hain? ::: Kyunki ek hi argument ke cosine aur sine independent shapes hain; sabhi $t$ ke liye hold hone wali equation har shape ke coefficient ko apne aap balance karne par majboor karti hai. Transient kyun disappear hota hai? ::: Yeh sirf decaying exponentials ($e^{-\gamma t}$ etc.) se bana hota hai, isliye yeh marr jaata hai — har damping case mein. Exact peak location, aur dD/dwd=0 ke dono roots? ::: Roots hain ${\omega }_d=0$ ($A$ ka ek minimum) aur ${\omega }_{res}=\sqrt{{\omega }_0^2-2{\gamma }^2}$ (sahi peak). Resonant peak kab nahi hota? ::: Jab $2{\gamma }^2\ge {\omega }_0^2$ (yaani $Q\le 1/\sqrt{2}$) — square root imaginary ho jaata hai. $Q={\omega }_0/2\gamma$ kya measure karta hai? ::: Resonance peak ki sharpness/height: bada $Q$ = tall narrow spike, chhota $Q$ = low broad hump.