Exercises — Forced oscillations — driving frequency

KYA: Steady-state frequency identify karo. KYUN: ${\omega }_d$ frequency ki ek periodic force sirf usi frequency par response sustain kar sakti hai — transient $e^{-\gamma t}$ khatam hone ke baad system driver ka "ghulam" ban jaata hai. Answer: ${\omega }_d=5$ rad/s. Natural ${\omega }_0=8$ rad/s sirf yeh govern karta hai ki response kitna bada hoga, frequency nahi. KAISA DIKHTA HAI: mass ko ek cosine wave trace karte hue socho jiske crests ke beech ki spacing pusher ki rhythm se set hoti hai, spring se nahi — jaise ek swing jise jo koi push kare uski hi pace par force kiya jaaye.

KYA: Raw constants ko standardised pair mein convert karo. KYUN: Upar ke har formula mein ${\omega }_0,\gamma$ likhe hain, $k,b$ nahi — isliye pehle translate karo.

• ${\omega }_0=\sqrt{k/m}=\sqrt{200/2}=\sqrt{100}=10$ rad/s.
• $2\gamma =b/m=8/2=4$, toh $\gamma =2$ s⁻¹. KAISA DIKHTA HAI: ${\omega }_0$ undriven spring ki crest-to-crest rhythm hai; $\gamma$ woh rate hai jis par free wobble ki height shrink hoti hai — mota $\gamma$ matlab free wiggle jaldi khatam ho jaata hai.

KYA: Amplitude formula mein ${\omega }_d\to 0$ lo. WHY: ${\omega }_d=0$ par drive essentially ek constant push hai; mass bas Hooke's law $F_0=kx$ se displaced hokar baitha rehta hai. $A\to \frac{F_0/m}{\sqrt{({\omega }_0^2{)}^2}}=\frac{F_0/m}{{\omega }_0^2}=\frac{F_0}{m{\omega }_0^2}=\frac{F_0}{k}=\frac{20}{200}=0.1\text{ m}.$ KAISA DIKHTA HAI: koi wobble nahi — mass bas ek fixed offset hold karta hai, jaise ek haath aahista spring par jhukke use rest se thoda dur pin kar de.

KYA: $A({\omega }_d)$ mein numbers plug karo. KYUN: Yeh direct substitution hai, lekin denominator ke do pieces dhyan se dekho: mismatch $({\omega }_0^2-{\omega }_d^2)$ aur damping cost $2\gamma {\omega }_d$.

• ${\omega }_0=\sqrt{200/0.5}=20$ rad/s, $2\gamma =b/m=4$.
• Mismatch: ${\omega }_0^2-{\omega }_d^2=400-100=300$.
• Damping term: $2\gamma {\omega }_d=4\times 10=40$.
• Denominator: $\sqrt{300^2+40^2}=\sqrt{90000+1600}=\sqrt{91600}\approx 302.66$.
• Numerator: $F_0/m=10/0.5=20$. $A=\frac{20}{302.66}\approx 0.0661\text{ m}.$ KAISA DIKHTA HAI: ek chhoti wobble — kyunki ${\omega }_d=10$ sirf ${\omega }_0=20$ ka aadha hai, bada mismatch (300) denominator mein dominate karta hai, isliye driven swing modest rehti hai, static offset se thodi hi zyada.

KYA: $\tan \varphi =2\gamma {\omega }_d/({\omega }_0^2-{\omega }_d^2)$ use karo. KYUN: $\varphi$ woh angle hai jiska tangent (damping term)/(mismatch term) hai. "arctan" ka jawab hai: is tan ke liye kaun sa angle?

• $\tan \varphi =40/300=0.1333$.
• Mismatch positive hai (hum resonance se neeche hain, ${\omega }_d<{\omega }_0$), toh $\varphi$ pehle quadrant mein hai — ek chhota positive lag. $\varphi =\arctan (0.1333)\approx 7.59^{\circ }\approx 0.1325\text{ rad}.$ KAISA DIKHTA HAI: mass ka peak force ke peak ke thoda baad aata hai — lagbhag ek saath.

KYA: Mismatch ko zero set karo. KYUN: Jab ${\omega }_d={\omega }_0$, term $({\omega }_0^2-{\omega }_d^2)=0$ khatam ho jaata hai; denominator mein sirf damping bachti hai, toh $A$ lagbhag apne peak par hota hai.

• Denominator $=2\gamma {\omega }_d=4\times 20=80$.
• $A=\frac{20}{80}=0.25$ m. KAISA DIKHTA HAI: ek badi wobble — L2.1 ki off-resonance swing (0.066 m) se lagbhag chaar guna. Mismatch khatam hone par, sirf friction amplitude ko rok raha hai, toh swing balloon hokar apni near-maximum height tak pahunch jaati hai.

KYA: Pehle check karo ki peak exist karti hai, phir ${\omega }_{res}=\sqrt{{\omega }_0^2-2{\gamma }^2}$ use karo. KYUN: Amplitude tab sabse bada hota hai jab denominator sabse chhota ho. $({\omega }_0^2-{\omega }_d^2{)}^2+(2\gamma {\omega }_d{)}^2$ minimise karne par (parent note mein calculus step hai) yeh milta hai. $-2{\gamma }^2$ dikhata hai ki damping peak ko ${\omega }_0$ se neeche kheench laati hai — lekin sirf tab jab square root real rahe.

• Existence check: kya $\gamma <{\omega }_0/\sqrt{2}$ hai? Yahan ${\omega }_0/\sqrt{2}=20/1.414\approx 14.14$, aur $\gamma =2\ll 14.14$. Haan — light damping, real peak exist karti hai.
• ${\omega }_{res}=\sqrt{400-2(4)}=\sqrt{400-8}=\sqrt{392}\approx 19.799$ rad/s.
• Yeh ${\omega }_0=20$ se thoda neeche hai — kyunki damping motion ko resist karti hai, "best" drive bare natural frequency se thoda slow hoti hai. KAISA DIKHTA HAI: socho (aur figure s01 dekho) ek smooth curve jahan amplitude $A$ upar plot hai drive frequency ${\omega }_d$ ke against. Baayein taraf se static value se shuru hokar, yeh ek single rounded hump tak uthti hai aur phir zero ki taraf girne lagti hai. Us hump ki top ${\omega }_{res}\approx 19.8$ hai; ${\omega }_0=20$ par khींchi gayi vertical reference line hump ke top se thoda daayein hai — woh chhoti gap hi $-2{\gamma }^2$ shift visible hai.

KYA: ${\omega }_d={\omega }_{res}$ ko $A$ mein substitute karo. KYUN: Ab dono terms nonzero hain, isliye poora denominator compute karte hain.

• ${\omega }_d^2=392$, toh mismatch $=400-392=8$; squared $=64$.
• Damping term $2\gamma {\omega }_d=4\times 19.799=79.196$; squared $=6272$.
• Denominator $=\sqrt{64+6272}=\sqrt{6336}\approx 79.599$.
• $A=\frac{20}{79.599}\approx 0.2513$ m. KAISA DIKHTA HAI: yeh figure s01 par hump ki top ki height hai — L2.3 mein ${\omega }_d={\omega }_0$ par mili $A=0.25$ m ki value se thodi hi zyada. Light damping ke liye peak sirf thodi si zyada oonchi hai, jo confirm karta hai ${\omega }_{res}\approx {\omega }_0$.

KYA: $\tan \varphi$ compute karo, phir angle sahi jagah rakho. KYUN: Resonance ke upar mismatch ${\omega }_0^2-{\omega }_d^2$ negative ho jaata hai, toh raw arctan ek negative angle deta hai — lekin physically mass force se $90°$ se zyada peeche rehta hai. Hume $180°$ add karna padega.

• Mismatch: $400-900=-500$ (negative).
• Damping: $2\gamma {\omega }_d=4\times 30=120$ (hamesha positive).
• $\tan \varphi =120/(-500)=-0.24$. Bare $\arctan (-0.24)=-13.5^{\circ }$.
• Sahi physical lag hai $\varphi =180^{\circ }-13.5^{\circ }=166.5^{\circ }$ (second quadrant), kyunki lag hamesha $0$ se $180^{\circ }$ ki taraf badhta hai jab ${\omega }_d$ resonance se guzarta hai. $\varphi \approx 166.5^{\circ }\approx 2.906\text{ rad}.$ KAISA DIKHTA HAI: Figure s02 mein phase-lag curve baayein se $0^{\circ }$ ke paas se uthti hai (in phase, resonance ke neeche), exactly $90^{\circ }$ par ${\omega }_d={\omega }_0=20$ par guzarti hai (dashed vertical reference), aur high ${\omega }_d$ par $180^{\circ }$ (anti-phase) ki taraf chadh jaati hai. ${\omega }_d=30$ par hum $166^{\circ }$ ke paas value padhte hain — mass ab force ke lagbhag opposite move karta hai, uski inertia sath nahi rakh pati, toh woh flip ho jaata hai.

KYA: $A$ at ${\omega }_d={\omega }_0$ se $b$ solve karne ke liye backwards kaam karo. KYUN: Resonance par mismatch khatam ho jaata hai, toh $A=\frac{F_0/m}{2\gamma {\omega }_0}$. Ek unknown ($\gamma$, hence $b$), ek equation.

• ${\omega }_0=\sqrt{100/1}=10$ rad/s.
• $A=\frac{F_0/m}{2\gamma {\omega }_0}\Rightarrow 0.5=\frac{5/1}{2\gamma \cdot 10}=\frac{5}{20\gamma }$.
• $20\gamma \times 0.5=5\Rightarrow 10\gamma =5\Rightarrow \gamma =0.5$ s⁻¹.
• $b=2\gamma m=2(0.5)(1)=1$ kg/s. Check: $b=1$ ke saath, $2\gamma {\omega }_0=2(0.5)(10)=10$, toh $A=5/10=0.5$ m. ✓ (Aur $\gamma =0.5<{\omega }_0/\sqrt{2}\approx 7.07$, toh real peak exist karti hai — design physical hai.) KAISA DIKHTA HAI: hum friction ka knob dial kar rahe hain jabtab resonant hump exactly aadhe metre par land kare — kam friction hump ko taller banata, zyada friction use flatten karta; $b=1$ kg/s woh setting hai jo target height hit karta hai.

KYA: $|{\omega }_0^2-{\omega }_d^2|=2\gamma {\omega }_d$ set karo aur ${\omega }_d$ solve karo (below-resonance root lo, mismatch positive). KYUN: Jahan Pythagorean denominator ki dono "legs" barabar hain, phase lag exactly $45^{\circ }$ hota hai (kyunki $\tan \varphi =1$) — ek physically meaningful marker.

• ${\omega }_0^2-{\omega }_d^2=2\gamma {\omega }_d\Rightarrow 100-{\omega }_d^2=6{\omega }_d$.
• ${\omega }_d^2+6{\omega }_d-100=0$.
• ${\omega }_d=\frac{-6+\sqrt{36+400}}{2}=\frac{-6+\sqrt{436}}{2}=\frac{-6+20.881}{2}\approx 7.44$ rad/s. Check phase: $\tan \varphi =2\gamma {\omega }_d/({\omega }_0^2-{\omega }_d^2)=1\Rightarrow \varphi =45^{\circ }$. ✓ (Hum positive root rakhte hain kyunki ${\omega }_d>0$.) KAISA DIKHTA HAI: s02 jaise phase curve par, yeh ${\omega }_d\approx 7.44$ chadhne ka halfway point hai — bilkul jahan mass ka peak force se exactly cycle ka one-eighth peeche hota hai ($45^{\circ }$), spring-controlled aur damping-controlled behaviour ke beech ka balance point.

KYA: Force times velocity se time-averaged power compute karo, sirf already defined ek phase $\varphi$ use karke. KYUN: Instantaneous power hai $P=F\dot{x}$ (force times velocity). Hume jaanna hai ki velocity force ke saath kaise align hai, kyunki agar dono ek hi direction mein hain toh push positive work karta hai.

Do waves set up karna (sirf $\varphi$ use karke):

• Force: $F(t)=F_0\cos ({\omega }_{d}t)$ — hamaara reference, $t=0$ par peak.
• Displacement force se $\varphi$ lag karta hai: $x(t)=A\cos ({\omega }_{d}t-\varphi )$.
• Velocity time-derivative hai: $\dot{x}(t)=-A{\omega }_d\sin ({\omega }_{d}t-\varphi )=A{\omega }_d\cos ({\omega }_{d}t-\varphi +90^{\circ })$. Cosine differentiate karne par peak $90^{\circ }$ aage shift ho jaata hai, toh velocity displacement se exactly $90^{\circ }$ lead karta hai — yeh $\cos$ ke baare mein ek fact hai, kisi naye symbol ki zaroorat nahi.
• Isliye velocity peak force ke relative phase angle $-\varphi +90^{\circ }$ par hai. Force ($0^{\circ }$) aur velocity ($-\varphi +90^{\circ }$) ke beech ka gap $(90^{\circ }-\varphi )$ hai, puri tarah ek phase $\varphi$ ke through express kiya gaya.

Ek cycle par average karna: $\cos ({\omega }_{d}t)\cdot \cos ({\omega }_{d}t+\theta )$ ka time-average ek poore period par $\frac{1}{2}\cos \theta$ hota hai. Yahan $\theta =(90^{\circ }-\varphi )$ force-to-velocity gap hai, toh $\bar{P}=\frac{1}{2}{F}_0(A{\omega }_d)\cos (90^{\circ }-\varphi )=\frac{1}{2}{F}_{0}A{\omega }_d\sin \varphi ,$ $\cos (90^{\circ }-\varphi )=\sin \varphi$ use karke. Sab kuch $\varphi$ mein hai — koi doosra phase symbol nahi chahiye tha.

Resonance par $\varphi =90^{\circ }$, toh $\sin \varphi =1$ (force aur velocity perfectly aligned):

• L2.3 se, $A=0.25$ m, toh peak velocity $A{\omega }_d=0.25\times 20=5$ m/s.
• $\bar{P}=\frac{1}{2}{F}_{0}A{\omega }_d\sin 90^{\circ }=\frac{1}{2}(10)(5)(1)=\boxed{25\text{ W}}.$ KAISA DIKHTA HAI: har push exactly tab lagta hai jab mass already usi direction mein move kar raha hota hai — jaise swing ke arc ke bilkul bottom par har baar dhakka dena. Koi effort motion se ladte hue waste nahi hota, isliye har second mein pump ki jaane wali energy apni ceiling par hoti hai: maximum power transfer.

KYA: Teen alag-alag frequencies par $A$ aur $\varphi$ evaluate karo. KYUN: Poori curve dekhne ke liye — spring-controlled, resonant, inertia-controlled — aur limiting behaviour verify karne ke liye.

Slow (${\omega }_d=2$):

• Mismatch $=400-4=396$; damping $=4\times 2=8$.
• $A=20/\sqrt{396^2+8^2}=20/\sqrt{156816+64}=20/396.08\approx 0.0505$ m. ($\approx F_0/k=0.05$ m ✓, spring-dominated.)
• $\tan \varphi =8/396=0.0202\Rightarrow \varphi \approx 1.16^{\circ }\approx 0$ → in phase (mass force follow karta hai).

Matched (${\omega }_d=20$):

• $A=0.25$ m (L2.3 se) — peak. $\varphi =90°$ → force velocity ke saath in phase, max power.

Fast (${\omega }_d=200$):

• Mismatch $=400-40000=-39600$; damping $=4\times 200=800$.
• $A=20/\sqrt{39600^2+800^2}=20/\sqrt{1568160000+640000}=20/39608.1\approx 0.000505$ m.
• Compare $F_0/(m{\omega }_d^2)=10/(0.5\times 40000)=0.0005$ m ✓ — inertia-dominated, tiny.
• $\tan \varphi =800/(-39600)=-0.0202\Rightarrow \varphi =180°-1.16°\approx 178.8°$ → lagbhag anti-phase (flipped).

Verdict: Slow → spring, in-phase, $A\approx F_0/k$. Matched → maximum $A$, $\varphi =90°$. Fast → faint aur flipped, $A\to 0$, $\varphi \to 180°$. Mnemonic har count par sahi hai. ✓ KAISA DIKHTA HAI: yeh teen points poori s01/s02 ki kahani trace karte hain — chhota left-end offset, tall central hump $\varphi =90°$ par, aur vanishing right tail jahan curve axis par flat ho jaata hai aur phase $180^{\circ }$ par pin ho jaata hai.

Ek driven oscillator ki steady-state frequency ::: driving frequency ${\omega }_d$ hoti hai, kabhi ${\omega }_0$ nahi. Denominator of $A$ apne do terms kaise combine karta hai ::: Pythagorean — $\sqrt{{\text{mismatch}}^2+{\text{damping}}^2}$, plain sum nahi. Resonance ke upar $\varphi$ ke liye sign fix ::: calculator ke arctan mein $180^{\circ }$ add karo (wahan mismatch negative hai). Real resonance peak ke liye condition ::: $\gamma <{\omega }_0/\sqrt{2}$; zyada heavy damping mein koi hump nahi (largest $A$ ${\omega }_d=0$ par). Resonance par $A$ finite kyun hai ::: damping term $2\gamma {\omega }_d\ne 0$ denominator ko zero se door rakhta hai. Average power formula $\varphi$ ke terms mein ::: $\bar{P}=\frac{1}{2}{F}_{0}A{\omega }_d\sin \varphi$; resonance par $\sin \varphi =1$ toh $\bar{P}=25$ W for L5.1.