1.6.11 · D3 · Physics › Oscillations & Waves › Forced oscillations — driving frequency
Intuition Yeh page kis liye hai
Parent note ne tumhe do master formulas diye the. Yeh page har tarah ke questions throw karta hai jo un formulas ko face karne padte hain — slow drives, fast drives, exact resonance, dono directions mein phase lags, zero damping, aur ek real-world word problem — taaki jab exam koi scenario de, tum us jaisa kuch pehle dekh chuke ho.
Shuru karne se pehle, yahan sirf do tools hain jo humein chahiye, copy karke likhe hain taaki kuch assume na ho:
Do symbols ko abhi plain words mein samjhana zaroori hai use karne se pehle:
Common mistake Units ka convention:
ω radians-per-second mein, lekin ϕ hum degrees mein report karte hain
Log yahan confuse kyun hote hain: is page par har frequency (ω 0 , ω d , γ ) radians per second mein hai, aur tan ϕ unhi raw SI numbers se compute hoti hai. Lekin answer ϕ hum degrees mein quote karte hain, kyunki "7. 6 ∘ " "0.132 rad" se zyada samajh aata hai. Rule yeh hai: formula mein pure SI numbers (rad/s) daalo, tan ϕ milega (ek plain unitless ratio), phir result angle ko sirf end mein degrees mein convert karo — ϕ deg = ϕ rad × π 180 . Calculation ke dauran kabhi degrees ko tan mein mat daalo. Landmarks ke taur par: ϕ = 9 0 ∘ = 2 π rad, ϕ = 18 0 ∘ = π rad.
Ek aur quantity Example 2 mein aati hai, toh chaliye usse abhi plain words mein pin down karte hain:
Definition Quality factor
Q
Q = 2 γ ω 0 ek single unitless number hai jo measure karta hai ki resonance kitni sharp aur tall hai. Isse aise padho: "resonance shake kitne times bada hai uss lazy static push se". Bada Q = light damping = tall, narrow peak; chota Q = heavy damping = short, wide bump. Hamare base system ke liye Q = 20/4 = 5 . Poori kahani ke liye Resonance and Quality Factor dekho.
Is topic ka har question inhi cells mein se kisi ek mein fit hota hai. Neeche har worked example ka tag us cell ke saath hai jo woh fill karta hai.
Cell
Kya vary karta hai
Kya tricky hai
Example
A. Slow drive (ω d ≪ ω 0 )
limiting value
Hooke's law A → F 0 / k tak reduce hona chahiye
Ex 1
B. Exact resonance denominator (ω d = ω 0 )
mismatch term = 0
sirf damping bachti hai; ϕ = 9 0 ∘
Ex 2
C. True amplitude peak (ω d = ω r es )
peak ω 0 se neeche hai
ω 0 2 − 2 γ 2 use karna padega
Ex 3
D. Phase below resonance (ω d < ω 0 )
ω 0 2 − ω d 2 > 0
tan ϕ > 0 , ϕ acute
Ex 4
E. Phase above resonance (ω d > ω 0 )
ω 0 2 − ω d 2 < 0
tan ϕ < 0 — quadrant fix zaroori
Ex 5
F. Fast drive (ω d ≫ ω 0 )
limiting value
inertia jeet jaata hai, A → F 0 / ( m ω d 2 ) , ϕ → 18 0 ∘
Ex 6
G. Zero damping (γ = 0 )
degenerate input
ω 0 par A → ∞ ; ϕ 0 → 18 0 ∘ jump karta hai
Ex 7
H. Real-world word problem
modelling
words ko symbols mein translate karo
Ex 8
I. Exam twist (ω d nikalo diye gaye A se)
inverse problem
root ke andar chhupe unknown ke liye solve karo
Ex 9
Hum number-crunching ke liye ek base system use karte hain taaki results comparable rahein:
m = 0.5 kg , k = 200 N/m , b = 2 kg/s , F 0 = 10 N .
Toh ω 0 = 200/0.5 = 20 rad/s aur 2 γ = b / m = 4 ⇒ γ = 2 s − 1 .
Figure s01 — Amplitude resonance curve (alt-text). Horizontal axis driving frequency ω d hai 0 se 45 rad/s tak; vertical axis steady-state amplitude A hai 0 se lagbhag 0.28 m tak. Ek blue curve static value 0.05 m ke paas se shuru hoti hai (dashed gray line, labelled A → F 0 / k ), ek sharp red peak tak ≈ 0.251 m par ω r es ≈ 19.8 rad/s par uthti hai (ω 0 = 20 ko mark karti dotted orange line se thodi left), phir fast drives ke liye zero ki taraf girti hai. Coloured dots mark karte hain kahan har cell curve par hai: green "A slow" left rise par neeche, orange "B ω 0 " peak ke thoda right mein, red "C peak" bilkul top par, gray "F fast" right tail mein kaafi neeche.
Upar wali figure is base system ke liye amplitude-vs-frequency curve hai, jisme cells A–G mark kiye hain jahan woh curve par hain. Baar-baar isko dekhte raho: har example bas "is curve ka ek point read karo" ya "phase graph ka ek point read karo" hai.
Worked example Ex 1 · Cell A ·
ω d = 2 rad/s (bahut slow)
Base system ko ω d = 2 rad/s par slowly drive karo. A nikalo, aur confirm karo ki yeh Hooke's-law value F 0 / k ki taraf ja raha hai.
Forecast: compute karne se pehle guess karo — kya A F 0 / k = 10/200 = 0.05 m ke paas hoga, ya kaafi zyada?
Mismatch term ω 0 2 − ω d 2 = 400 − 4 = 396 .
Yeh step kyun? Yeh "resonance se kitni door" wala number hai; yahan bada hai kyunki 2 ≪ 20 .
Damping term 2 γ ω d = 4 × 2 = 8 .
Yeh step kyun? Denominator ka doosra piece; chota hai kyunki ω d chota hai.
Denominator 39 6 2 + 8 2 = 156816 + 64 = 156880 ≈ 396.08 .
Yeh step kyun? Dono pieces ko ( ⋯ ) 2 + ( ⋯ ) 2 mein plug kar rahe hain.
Amplitude A = 396.08 F 0 / m = 396.08 20 ≈ 0.0505 m.
Yeh step kyun? F 0 / m = 10/0.5 = 20 ; denominator se divide karo.
Verify: pure static value F 0 / k = 0.05 m hai. Hamara answer 0.0505 m thoda usse upar hai — exactly wahi jo "slow drive ≈ static" predict karta hai. Units: kg N ÷ s 2 1 = kg N ⋅ s 2 = m . ✓
Worked example Ex 2 · Cell B ·
ω d = 20 rad/s
Base system ko exactly ω d = ω 0 = 20 rad/s par drive karo. A aur phase lag ϕ nikalo.
Forecast: mismatch term ab zero hai. Kya tumhara expect hai ki yeh possible sabse bada A hoga, ya bas close? Aur jab frequencies match karen toh ϕ exactly kya hona chahiye?
Mismatch term ω 0 2 − ω d 2 = 400 − 400 = 0 .
Yeh step kyun? Jab ω d = ω 0 frequencies match karti hain, toh yeh contribution vanish ho jaata hai.
Denominator 0 2 + ( 2 γ ω d ) 2 = 2 γ ω d = 4 × 20 = 80 .
Yeh step kyun? Mismatch khatam ho jane ke baad, sirf damping cost response ko limit karti hai.
Amplitude A = 80 20 = 0.25 m.
Yeh step kyun? F 0 / m = 20 , 80 se divide karo.
Phase lag tan ϕ = ω 0 2 − ω d 2 2 γ ω d = 0 80 → + ∞ , toh ϕ = 9 0 ∘ = 2 π rad.
Yeh step kyun? Jab tan ϕ ka denominator upar se zero hit karta hai, tan ϕ blow up karta hai, aur woh angle jiska tangent + ∞ hai exactly 9 0 ∘ hai. Yeh physically special case hai: mass force se quarter-cycle peeche run karta hai, jo force ko velocity ke in phase rakhta hai → maximum power input. Units rule yaad karo: SI numbers se compute karo, phir angle degrees mein quote karo.
Verify: 0.25 m static 0.05 m se kaafi zyada hai — resonance se 5 ka boost. Yeh factor exactly quality factor Q = ω 0 / ( 2 γ ) = 20/4 = 5 hai (ω 0 par amplitude, static value F 0 / k ka Q times hota hai). Aur ϕ = 9 0 ∘ exactly Ex 4 ke 7. 6 ∘ (below) aur Ex 5 ke 166. 5 ∘ (above) ke beech hai — phase graph 9 0 ∘ cross karta hai exactly ω 0 par. ✓ (Yeh point amplitude peak bilkul nahi hai — Ex 3 dekho.)
Worked example Ex 3 · Cell C ·
ω r es aur peak A nikalo
Base system ke liye, woh frequency nikalo jahan A genuinely sabse badi ho, aur wahan amplitude nikalo. Dikhao ki yeh Ex 2 se zyada hai.
Forecast: kya peak ω 0 = 20 se upar, bilkul us par, ya neeche hogi?
Peak location ω r es = ω 0 2 − 2 γ 2 = 400 − 2 ( 4 ) = 392 ≈ 19.799 rad/s.
Yeh step kyun? Parent note ne denominator ko differentiate karke minimise kiya; result yeh shifted-down peak hai. Damping, peak ko ω 0 se neeche kheenchti hai.
Wahan mismatch ω 0 2 − ω r es 2 = 400 − 392 = 8 = 2 γ 2 .
Yeh step kyun? Sanity check: construction se, peak par mismatch 2 γ 2 ke barabar hota hai.
Denominator 8 2 + ( 2 γ ω r es ) 2 = 64 + ( 4 × 19.799 ) 2 = 64 + 6272 = 6336 ≈ 79.60 .
Amplitude A p e ak = 79.60 20 ≈ 0.2513 m.
Verify: 0.2513 > 0.25 Ex 2 se thoda zyada — true peak really mein ω 0 par value se ek baal zyada hai, aur 19.80 < 20 rad/s par hai. Dono facts match karte hain "peak below ω 0 , sirf γ → 0 par equal". ✓
Worked example Ex 4 · Cell D ·
ω d = 10 rad/s, ϕ nikalo
Base system ω d = 10 rad/s par driven (ω 0 se neeche). Phase lag ϕ nikalo.
Forecast: resonance ke neeche mass roughly force follow karta hai. Kya ϕ 0 ke paas hai, 9 0 ∘ ke paas, ya 18 0 ∘ ke paas?
Mismatch ω 0 2 − ω d 2 = 400 − 100 = 300 (positive).
Yeh step kyun? Positive mismatch matlab tan ϕ ka denominator positive hai — yeh "below resonance" ka pehchaan wala sign hai.
Numerator 2 γ ω d = 4 × 10 = 40 .
tan ϕ = 40/300 = 0.1 3 , toh ϕ = arctan ( 0.1333 ) ≈ 7.5 9 ∘ .
Yeh step kyun? arctan jawaab deta hai "kis angle ka tan yeh hai?" Kyunki numerator aur denominator dono positive hain, ϕ first quadrant (0 –9 0 ∘ ) mein aata hai — koi fix zaroori nahi. (SI numbers se ratio compute karo; end mein answer ko degrees mein convert karo.)
Verify: ϕ ≈ 7. 6 ∘ bahut chota hai → mass almost in phase move karta hai force ke saath. "Slow-ish drive force follow karta hai" se consistent hai. ✓
Worked example Ex 5 · Cell E ·
ω d = 30 rad/s, ϕ nikalo
Base system ω d = 30 rad/s par driven (ω 0 se upar). Phase lag ϕ nikalo — aur sign dekho.
Forecast: resonance ke upar mass force se ladhta hai. Kya ϕ 9 0 ∘ se kam hona chahiye ya zyada?
Mismatch ω 0 2 − ω d 2 = 400 − 900 = − 500 (negative ).
Yeh step kyun? Resonance ke upar mismatch ka sign flip ho jaata hai — yeh Ex 4 se crucial difference hai.
Numerator 2 γ ω d = 4 × 30 = 120 (hamesha positive — γ , ω d > 0 ).
tan ϕ = 120/ ( − 500 ) = − 0.24 . Naive arctan ( − 0.24 ) = − 13. 5 ∘ .
Yeh step kyun? arctan sirf − 9 0 ∘ aur + 9 0 ∘ ke beech answers deta hai, kyunki tan har 18 0 ∘ par repeat karta hai. Yahan negative answer ek red flag hai, physics nahi: phase lag kabhi negative nahi hoti. Jab ω 0 2 − ω d 2 < 0 , yeh red flag hai.
Quadrant fix: physical lag ( 9 0 ∘ , 18 0 ∘ ) mein hoti hai jab bhi mismatch negative ho, toh 18 0 ∘ add karo: ϕ = − 13. 5 ∘ + 18 0 ∘ = 166. 5 ∘ .
Yeh step kyun? Hum jaante hain ϕ ko 9 0 ∘ se aage badhna chahiye jab hum resonance cross karein (yeh 0 → 9 0 ∘ → 18 0 ∘ slide karta hai). Numerator positive batata hai sin ϕ > 0 ; denominator negative batata hai cos ϕ < 0 ; dono milke ϕ ko second quadrant (9 0 ∘ –18 0 ∘ ) mein pin karte hain.
Verify: 166. 5 ∘ 18 0 ∘ ke close hai — expected, kyunki ω d = 30 ω 0 = 20 se kaafi upar hai, "moves opposite the force" limit ki taraf ja raha hai. ✓ Ex 4 ke 7. 6 ∘ (below) ko is 166. 5 ∘ (above) se compare karo: resonance cross karne par lag near-0 se near-18 0 ∘ swing ho jaata hai, exactly ω 0 par 9 0 ∘ se guzarta hua. Neeche figure dekho.
Figure s02 — Phase lag vs frequency (alt-text). Horizontal axis driving frequency ω d hai 0 se 45 rad/s tak; vertical axis phase lag ϕ degrees mein hai 0 se 180 tak. Ek blue S-shaped curve slow drives ke liye 0 ∘ ke paas se shuru hoti hai, dashed gray 9 0 ∘ line se steeply rise karti hai exactly dotted orange ω 0 = 20 par, phir fast drives ke liye 18 0 ∘ ki taraf level off karti hai. Lower-left band (9 0 ∘ se neeche) green tinted hai aur labelled hai "below res: ϕ in (0,90)"; upper band red tinted hai aur labelled hai "above res: ϕ in (90,180)". Ek green dot Ex 4 mark karta hai (ω d = 10 , ϕ = 7. 6 ∘ ); ek red dot Ex 5 mark karta hai (ω d = 30 , ϕ = 166. 5 ∘ , annotated "add 180!").
Common mistake Resonance ke upar calculator ke
arctan par trust karna
Kyun sahi lagta hai: tumne 120/ ( − 500 ) type kiya aur screen ne − 13. 5 ∘ bola, toh tum ϕ = − 13. 5 ∘ likh dete ho. Fix: phase lag kabhi negative nahi hoti. Jab ω 0 2 − ω d 2 < 0 , 18 0 ∘ add karo. Rule of thumb: resonance ke neeche ϕ ∈ ( 0 , 9 0 ∘ ) , resonance ke upar ϕ ∈ ( 9 0 ∘ , 18 0 ∘ ) .
Worked example Ex 6 · Cell F ·
ω d = 200 rad/s (bahut fast)
Base system ko ω d = 200 rad/s par hammer karo. A nikalo aur confirm karo ki yeh F 0 / ( m ω d 2 ) approach karta hai.
Forecast: ω 0 se das guna upar — kya tumhara expect hai large amplitude ya tiny?
Mismatch ω 0 2 − ω d 2 = 400 − 40000 = − 39600 .
Yeh step kyun? Enormous negative mismatch — drive natural frequency ke kahin paas nahi hai.
Damping term 2 γ ω d = 4 × 200 = 800 .
Denominator ( − 39600 ) 2 + 80 0 2 = 1 , 568 , 160 , 000 + 640 , 000 ≈ 39608.1 .
Amplitude A = 39608.1 20 ≈ 5.049 × 1 0 − 4 m.
Verify: inertia-limit prediction hai A → m ω d 2 F 0 = 0.5 × 40000 10 = 5.0 × 1 0 − 4 m. Hamara exact answer 5.049 × 1 0 − 4 m teen figures tak match karta hai. ✓ Aur yeh tiny hai — mass simply keep up nahi kar sakta. Iska phase lag ϕ (numerator + , denominator − ) 18 0 ∘ ke paas hai: motion almost opposite force ke.
Worked example Ex 7 · Cell G ·
γ = 0 set karo
Saari damping remove karo (b = 0 ⇒ γ = 0 ), m , k , F 0 rakhno. ω d = ω 0 ke paas A ka kya hota hai? ω 0 se thoda neeche aur thoda upar ϕ kya hai?
Forecast: kuch bhi energy bleed karne ke liye nahi hai, toh resonance par amplitude exactly kya hai?
Amplitude formula collapse ho jaata hai: γ = 0 ke saath,
A = ( ω 0 2 − ω d 2 ) 2 + 0 F 0 / m = ∣ ω 0 2 − ω d 2 ∣ F 0 / m .
Yeh step kyun? Damping term ( 2 γ ω d ) 2 ab literally 0 hai, toh root bas mismatch ki absolute value hai.
ω d = ω 0 par: denominator = ∣400 − 400∣ = 0 , toh A → ∞ .
Yeh step kyun? Koi damping nahi matlab har push energy hamesha ke liye add karta rehta hai — textbook "infinite resonance". Physically spring pehle toot jaayegi; mathematically yeh diverge karta hai.
Phase just below (ω d → ω 0 − ): mismatch → 0 + , aur γ = 0 ke saath tan ϕ ka numerator 0 hai, toh tan ϕ = 0 + / 0 + → 0 ⇒ ϕ = 0 .
Phase just above (ω d → ω 0 + ): mismatch → 0 − , toh tan ϕ = 0/ 0 − negative side se ⇒ ϕ = 18 0 ∘ .
Yeh step kyun? Koi damping nahi hone par phase 9 0 ∘ se smoothly glide nahi karta — jaise hi tum ω 0 cross karte ho, yeh 0 se 18 0 ∘ tak discontinuously jump karta hai. Damping hi us transition ko smooth banata hai (Ex 4→5).
Verify: thoda off-resonance number se check karo, ω d = 19 rad/s, γ = 0 : A = ∣400 − 361∣ 20 = 39 20 ≈ 0.513 m — finite aur large, aur jab ω d → 20 toh yeh blow up ho jaata hai. ✓ 0 → 18 0 ∘ phase jump, figure s02 mein smooth curve ka light-damping limit hai.
Worked example Ex 8 · Cell H · ek washing machine
Ek washing machine drum ka mass m = 8 kg hai, total stiffness k = 3200 N/m ke springs par mounted hai, damping b = 64 kg/s hai. Spin ke dauran, ek off-centre geela towel F 0 = 40 N amplitude ka periodic shaking force produce karta hai. Ek spin rate par jo ω d = 20 rad/s deta hai, drum kitna shake karta hai? Kya yeh spin rate resonance ke paas hai?
Forecast: compute karne se pehle guess karo ki shaking millimetres mein hogi ya centimetres mein.
Words ko symbols mein translate karo: natural frequency ω 0 = k / m = 3200/8 = 400 = 20 rad/s.
Yeh step kyun? "Springs + mass" exactly ek driven oscillator hai; ω 0 = k / m uski preferred rhythm hai.
Damping rate 2 γ = b / m = 64/8 = 8 ⇒ γ = 4 s⁻¹.
Yeh step kyun? Parent note jaisi hi standardising.
Notice karo ω d = 20 = ω 0 : spin exactly resonance par drive kar rahi hai. Mismatch ω 0 2 − ω d 2 = 400 − 400 = 0 .
Yeh step kyun? Yeh dangerous case hai — washing machines "walk" across the floor jab spin rate resonance hit kare. Iska matlab hai amplitude formula sirf damping term tak collapse ho jaata hai.
Amplitude A = 2 γ ω d F 0 / m = 8 × 20 40/8 = 160 5 = 0.03125 m ≈ 3.1 cm.
Yeh step kyun? Mismatch khatam hone par, denominator sirf 2 γ ω d tak collapse ho jaata hai; F 0 / m = 40/8 = 5 ko 8 × 20 = 160 se divide karo.
Phase check: resonance par ϕ = 9 0 ∘ (Cell B), toh shaking towel ke force se quarter-cycle peeche hai — force drum ki velocity ke in phase hai, maximum energy pump kar raha hai. Isliye shake itna bada hai.
Yeh step kyun? Resonance diagnosis confirm karta hai aur violence ko physically explain karta hai.
Verify: resonance par 3.1 cm ki shaking exactly isliye machines mid-spin mein violently vibrate karti hain, phir jab woh ω 0 se aage speed karte hain toh calm ho jaati hain. Units: s − 1 ⋅ s − 1 N / kg = m . ✓ Real machines is cheez ko shrink karne ke liye extra damping/counterweights add karte hain. Energy zyada kyun pump hoti hai, yeh energy mein dekho.
Worked example Ex 9 · Cell I · diye gaye amplitude se driving frequency nikalo
Base system (m = 0.5 , ω 0 = 20 , γ = 2 , F 0 = 10 ). Resonance ke neeche kaunsi drive frequency par amplitude A = 0.10 m hogi?
Forecast: hum pehle se jaante hain A = 0.05 m jab ω d → 0 aur A = 0.25 m jab ω 0 par. Toh target 0.10 m 0 aur 20 rad/s ke beech kahin hona chahiye. Roughly guess karo kahan.
Amplitude formula ko invert karo. A = D F 0 / m se, jahan D = ( ω 0 2 − ω d 2 ) 2 + ( 2 γ ω d ) 2 :
D = A F 0 / m = 0.10 20 = 200 ⇒ D = 40000.
Yeh step kyun? Humein A diya hua hai aur ω d chahiye, toh hum formula ko peel back karte hain: root ke andar chhupe frequency ko nikalne ke liye dono sides square karte hain.
D likhte hain. Maano u = ω d 2 . Toh
( 400 − u ) 2 + ( 2 γ ω d ) 2 = ( 400 − u ) 2 + 16 u = 40000.
Yeh step kyun? ( 2 γ ω d ) 2 = ( 4 ω d ) 2 = 16 ω d 2 = 16 u . u = ω d 2 use karne se yeh u mein quadratic ban jaata hai.
Expand: 160000 − 800 u + u 2 + 16 u = 40000 ⇒ u 2 − 784 u + 120000 = 0 .
Quadratic solve karo: u = 2 784 ± 78 4 2 − 4 ( 120000 ) = 2 784 ± 614656 − 480000 = 2 784 ± 134656 = 2 784 ± 366.96 .
Toh u = 575.5 ya u = 208.5 .
Physical root chuno. u = ω d 2 , toh ω d = 575.5 ≈ 23.99 (resonance ke upar) ya ω d = 208.5 ≈ 14.44 (neeche).
Yeh step kyun? Do frequencies ek hi amplitude deti hain — peak ke ek-ek taraf ek. Hum se resonance ke neeche waali maangi gayi thi: ω d ≈ 14.44 rad/s.
Verify: ω d = 14.44 back plug karo: mismatch = 400 − 208.5 = 191.5 ; damping = 4 × 14.44 = 57.76 ; denominator = 191. 5 2 + 57.7 6 2 = 36672 + 3336 = 40008 ≈ 200.0 ; A = 20/200 = 0.10 m. ✓ Doosra root 23.99 bhi 0.10 m deta hai — resonance curve ω d 2 mein roughly symmetric hai.
Recall Matrix ke across quick self-test
A ka slow drive limit kya hai? ::: F 0 / k (Hooke's law), Cell A.
Amplitude at ω d = ω 0 ? ::: F 0 / ( 2 mγ ω 0 ) = Q ⋅ F 0 / k , Cell B.
Phase lag at ω d = ω 0 ? ::: ϕ = 9 0 ∘ = 2 π rad — force velocity ke in phase, max power, Cell B.
True peak location? ::: ω r es = ω 0 2 − 2 γ 2 , ω 0 se thoda neeche, Cell C.
Resonance ke upar tan ϕ ka sign, aur fix? ::: Negative; 18 0 ∘ add karo taaki ϕ ∈ ( 9 0 ∘ , 18 0 ∘ ) , Cell E.
ω d → ∞ par A kya hai? ::: F 0 / ( m ω d 2 ) → 0 , Cell F.
γ = 0 par ω 0 par A ka kya hota hai? ::: Infinity tak diverge karta hai; phase 0 → 18 0 ∘ jump karta hai, Cell G.
Inverse problem: ek below-peak amplitude kitni drive frequencies deta hai? ::: Do — peak ke ek-ek taraf; ω d 2 mein quadratic solve karo, Cell I.
"Ek height, do frequencies." Peak se neeche koi bhi amplitude do baar hit hoti hai — ek baar ω 0 ke neeche, ek baar upar. Inverse problems (Cell I) hamesha ω d 2 mein quadratic dete hain; woh root chuno jo question ne naam liya ho.
Yeh bhi dekho: Complex Exponential Method A aur ϕ ek saath nikalne ke slick tarike ke liye, Damped Oscillations us transient ke liye jiske upar yeh steady states baithe hain, aur Simple Harmonic Motion b = 0 , F 0 = 0 ancestor ke liye jo yahan sab ka bunyaad hai.