Worked examples — Resonance — physical consequences, design implications

1.6.12 · D3 · Physics › Oscillations & Waves › Resonance — physical consequences, design implications

Yeh page parent note on Resonance ka "har case cover karo" companion hai. Kuch bhi compute karne se pehle, hum har tarah ki situation lay out karte hain jo resonance formulas dे sakti hain, taaki baad mein koi bhi example tumhe surprise na kare.

Har symbol jo hum use karte hain woh parent note mein build kiya gaya hai, lekin yahan ek line ka reminder hai jis par tum constantly lean karoge:

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The scenario matrix

Resonance problems ko ek grid mein rehte hue socho. Har cell ek alag regime hai, aur neeche har worked example us cell(s) ke saath tagged hai jise woh cover karta hai.

Cell	Case class	Usmein kya special hai
C1	Resonance se bahut neeche drive karna ($\omega \ll {\omega }_0$)	Amplitude ≈ static; spring term dominate karta hai
C2	Resonance par drive karna ($\omega ={\omega }_0$)	Denominator collapse hokar $b\omega /m$ reh jaata hai; amplitude peak karta hai
C3	Resonance se bahut upar drive karna ($\omega \gg {\omega }_0$)	Inertia dominate karta hai; amplitude $1/{\omega }^2$ ki tarah die hoti hai
C4	Degenerate: zero damping ($b\to 0$)	Amplitude infinity tak blow up ho jaata hai — limiting catastrophe
C5	Degenerate: zero drive frequency ($\omega =0$, ek steady push)	Koi oscillation nahi; pure static displacement $F_0/k$
C6	Peak ${\omega }_0$ par nahi hai: exact ${\omega }_{res}$	Displacement peak ${\omega }_0$ se thoda neeche baith jaata hai
C7	Sharpness & bandwidth: $Q$, half-power width $\Delta \omega$	Woh do frequencies jahan power half ho jaati hai
C8	Real-world word problem	Bridge / soldiers / earthquake — words ko $\omega$ mein translate karo
C9	Electrical twin (LC circuit)	Same math, alag symbols: ${\omega }_0=1/\sqrt{LC}$
C10	Exam-style twist: "microwave resonance nahi hai"	Pehchano jab formula apply nahi hota

Ab hum saare das cells ko aath examples mein walk karte hain.

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Forecast: aage padhne se pehle guess karo teeno mein se kaunsa sabse bada swing dega. Zyaadatar log sahi guess karte hain (b) — lekin kitne farak se?

1. ${\omega }_0$ find karo. ${\omega }_0=\sqrt{k/m}=\sqrt{100/1}=10\text{ rad/s}$. Yeh step kyun? Har case ${\omega }_0$ ke relative measure hota hai; yeh hamara yardstick hai.

2. (a) Far below, $\omega =1$. Plug in karo: $A=\frac{10/1}{\sqrt{(100-1{)}^2+(2\cdot 1/1{)}^2}}=\frac{10}{\sqrt{99^2+2^2}}=\frac{10}{99.02}\approx 0.101\text{ m}.$ Yeh step kyun? Jab $\omega \ll {\omega }_0$, term ${\omega }_0^2-{\omega }^2\approx {\omega }_0^2=k/m$, isliye $A\approx F_0/k=10/100=0.1$ — system notice hi nahi karta ki use shake kiya ja raha hai; woh bas apni static stretch par baith jaata hai. Yeh C1 hai.

3. (b) On resonance, $\omega =10$. Ab ${\omega }_0^2-{\omega }^2=0$: $A=\frac{10}{\sqrt{0+(2\cdot 10{)}^2}}=\frac{10}{20}=0.5\text{ m}.$ Yeh step kyun? Bada square-root term vanish ho jaata hai, sirf damping term $b\omega /m$ reh jaata hai. Yeh C2 hai — amplitude static value se 5× zyaada jump karta hai.

4. (c) Far above, $\omega =100$. $A=\frac{10}{\sqrt{(100-10000{)}^2+(2\cdot 100{)}^2}}=\frac{10}{\sqrt{9900^2+200^2}}=\frac{10}{9902}\approx 1.01\times 10^{-3}\text{ m}.$ Yeh step kyun? Jab $\omega \gg {\omega }_0$, inertia term $-m{\omega }^2$ dominate karta hai; $A\approx F_0/(m{\omega }^2)$. Mass itna heavy hai ki keep up nahi kar sakta — woh barely hilta hai. Yeh C3 hai.

Verify: teeno amplitudes hain $0.101>$ … nahi — size ke hisaab se order: $0.5\text{ (resonance)}\gg 0.101\text{ (low)}\gg 0.00101\text{ (high)}$. Resonance jeetta hai, exactly jaisa figure ke peak mein predict hota hai. Units: $F_0/k=\text{N}/(\text{N/m})=\text{m}$. ✓

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Forecast: finite hai ya infinite? Guess karo.

1. Formula mein $b=0$ set karo, phir $\omega ={\omega }_0$. $A=\frac{F_0/m}{\sqrt{({\omega }_0^2-{\omega }^2{)}^2+0}}=\frac{F_0/m}{|{\omega }_0^2-{\omega }^2|}.$ Yeh step kyun? Koi damping term nahi hai, toh denominator sirf spring-minus-inertia part hai.

2. Ab $\omega \to {\omega }_0$ push karo. Denominator $\to 0$, isliye $A\to \infty$. Yeh step kyun? Yeh "resonance disaster" ka mathematical statement hai — energy bleed karne ke liye kuch nahi hai, toh har push coherently forever add hota rehta hai. Yeh C4 hai.

Verify: Example 1(b) se compare karo: wahan, $b=2$ ne answer ko $0.5\text{ m}$ par cap kiya tha. $b\to 0$ lo: $A_{max}=F_0/(b{\omega }_0)=10/(b\cdot 10)$ blow up karta hai jab $b\to 0$. Toh finite $0.5\text{ m}$ entirely damping ki wajah se tha. ✓ Reality mein hamesha kuch $b>0$ hota hai, isliye true infinity kabhi occur nahi hoti.

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Forecast: ek $100\text{ N/m}$ spring par steady $10\text{ N}$ pull se kya displacement expect karte ho?

1. $\omega =0$ set karo. $A=\frac{F_0/m}{\sqrt{({\omega }_0^2-0{)}^2+0}}=\frac{F_0/m}{{\omega }_0^2}=\frac{F_0/m}{k/m}=\frac{F_0}{k}.$ Yeh step kyun? $\omega =0$ par driver kabhi reverse nahi karta; damping ($b\omega$) aur inertia dono kuch contribute nahi karte. Hum Hooke's law par aa jaate hain: force = stiffness × stretch.

2. Number: $A=10/100=0.1\text{ m}$. Yeh step kyun? Yeh C5 degenerate case hai aur yeh poori curve ka anchor hai — graph ke ekdum left par amplitude. Notice karo ki yeh Example 1(a) se closely match karta hai (isliye $\omega \ll {\omega }_0$ "static" lagta hai).

Verify: Hooke's law rearrange karo: $F=kx\Rightarrow x=F/k=10/100=0.1\text{ m}$. Units: $\text{N}/(\text{N/m})=\text{m}$. ✓

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Forecast: $10$ se kitna neeche — bahut zyaada ya thoda sa?

1. Exact peak formula use karo. ${\omega }_{res}=\sqrt{{\omega }_0^2-\frac{b^2}{2m^2}}$. Yeh step kyun? Yeh $dA/d\omega =0$ set karne se aata hai; damping term peak of $A$ ko thoda left shift kar deta hai kyunki friction higher $\omega$ par zyaada "khata" hai.

2. Plug in karo: ${\omega }_{res}=\sqrt{100-\frac{2^2}{2\cdot 1^2}}=\sqrt{100-2}=\sqrt{98}\approx 9.899\text{ rad/s}$. Yeh step kyun? Yeh C6 hai: peak $9.899$ par baith jaata hai, $10$ par nahi — yahan ek tiny $\sim 0.1\text{ rad/s}$ shift hai kyunki damping light hai.

3. Direction par sanity check: $9.899<10={\omega }_0$. ✓ Neeche, jaisa promise kiya tha.

Verify: velocity/power resonance exactly ${\omega }_0=10$ par hai; displacement peak $\sqrt{98}\approx 9.899$ par hai. Limiting rule bhi check karo: jab $b\to 0$, ${\omega }_{res}\to \sqrt{100-0}=10={\omega }_0$. ✓ Light damping ⇒ shift almost vanish ho jaata hai.

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Forecast: kya yeh ek "sharp radio" system hai ya "soft shock-absorber" system? High ya low $Q$ guess karo.

1. $Q$ compute karo. $Q=\frac{{\omega }_{0}m}{b}=\frac{10\cdot 1}{2}=5$. Yeh step kyun? $Q$ peak sharpness measure karta hai: high $Q$ = tall narrow peak, low $Q$ = broad. $Q=5$ moderate hai — ek real spring, quartz crystal nahi (jiska $Q\sim 10^4$–$10^6$ hota hai).

2. $Q$ se Bandwidth. Ek resonator ke liye, $\Delta \omega =\frac{{\omega }_0}{Q}$. Yeh step kyun? $\Delta \omega$ woh do frequencies ke beech ki distance hai jahan power apne peak se half ho jaati hai (amplitude $1/\sqrt{2}$ ho jaati hai). $Q$ literally hai hi ${\omega }_0/\Delta \omega$, toh hum bas invert kar dete hain.

3. Number: $\Delta \omega =\frac{10}{5}=2\text{ rad/s}$. Yeh step kyun? Yeh C7 hai: peak $2\text{ rad/s}$ wide hai. Light-damping approximation mein, damping $b/m=2$ yahan $\Delta \omega$ ke barabar hai — ek neat check.

4. Woh do half-power frequencies roughly ${\omega }_0\pm \Delta \omega /2=10\pm 1$ par baith jaati hain, yaani roughly $9$ aur $11\text{ rad/s}$. Yeh step kyun? Yeh ek radio engineer ko batata hai ki kaunse neighbouring stations leak through karte hain.

Verify: $\Delta \omega \approx b/m$ light damping ke liye: $b/m=2/1=2$. ✓ ${\omega }_0/Q=10/5=2$ se match karta hai. ✓ $Q$ ke units: $(\text{rad/s})(\text{kg})/(\text{kg/s})$ = dimensionless. ✓

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Forecast: kya $120$ steps/min dangerous hai ya safe? Pehle convert karo, phir guess karo.

1. Marching cadence ko Hz mein convert karo. $120\text{ steps/min}=120/60=2.0\text{ steps/s}=2.0\text{ Hz}$. Yeh step kyun? Har footfall ek push cycle hai, isliye cadence in Hz hi drive frequency $f$ hai. Hume $f_0$ ke saath like-for-like compare karna hai.

2. Compare karo. Drive $f=2.0\text{ Hz}$ equals $f_0=2.0\text{ Hz}$ — yeh dead-on resonance hai (C8, aur same physics jaise C2). Turant step break karo. Yeh step kyun? Ek coherent $2\text{ Hz}$ push $2\text{ Hz}$ resonance par exactly wahi Millennium/soldier-bridge scenario hai.

3. Ek full bandwidth kitna dur hai? $\Delta f=f_0/Q=2.0/20=0.10\text{ Hz}$. Yeh step kyun? Half-power band se bahar step karna resonant amplification kam kar deta hai. Hum chahte hain ki $f$ kam se kam $\Delta f/2=0.05\text{ Hz}$ off ho — ya better, ek full $\Delta f$.

4. Ek full bandwidth ke liye percentage slow-down: chahiye $f=2.0-0.10=1.90\text{ Hz}$, ek change of $0.10/2.0=0.05=5\%$. Yeh step kyun? "Kitne percentage se" question ka answer deta hai; lekin note karo reality mein aasaan hai — step break karna force ko randomise kar deta hai isliye koi single $f$ dominate hi nahi karta.

Verify: cadence conversion $120/60=2.0$. ✓ $\Delta f=2.0/20=0.10\text{ Hz}$. ✓ Percentage $=0.10/2.0=5\%$. ✓

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Forecast: picoFarads ya Farads? Order of magnitude guess karo.

1. Electrical natural frequency use karo. ${\omega }_0=\frac{1}{\sqrt{LC}}$, aur ${\omega }_0=2\pi f_0$. Yeh step kyun? LC circuit same differential equation follow karta hai jaise mass–spring; $L$ mass ka role play karta hai, $1/C$ stiffness $k$ ka role. Same math, naya costume — yeh C9 hai.

2. $C$ ke liye solve karo. ${\omega }_0^2=1/(LC)$ se: $C=\frac{1}{{\omega }_0^{2}L}=\frac{1}{(2\pi f_0{)}^{2}L}$. Yeh step kyun? Unknown $C$ ko algebraically isolate karo.

3. Numbers plug karo. ${\omega }_0=2\pi (10^6)=6.283\times 10^6\text{ rad/s}$. $C=\frac{1}{(6.283\times 10^6{)}^2(2\times 10^{-4})}=\frac{1}{(3.948\times 10^{13})(2\times 10^{-4})}=\frac{1}{7.896\times 10^9}\approx 1.27\times 10^{-10}\text{ F}.$ Yeh step kyun? Ab hamare paas ek real component value hai: $\approx 127\text{ pF}$ (picoFarads).

Verify: back-substitute karo: ${\omega }_0=1/\sqrt{LC}=1/\sqrt{(2\times 10^{-4})(1.27\times 10^{-10})}=1/\sqrt{2.53\times 10^{-14}}\approx 6.28\times 10^6\text{ rad/s}$, aur $f_0={\omega }_0/2\pi \approx 1.0\text{ MHz}$. ✓ Order of magnitude: $\sim 100\text{ pF}$, tuning cap ke liye typical. ✓

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Forecast: reason karne se pehle true/false decide karo.

1. Resonance ke liye kya chahiye hoga? Ek specific frequency par ek sharp, high-$Q$ peak, toh tiny frequency drift heating ko khatam kar dega. Yeh step kyun? Resonance (C2) inherently frequency-selective hoti hai — yahi uski poori nature hai.

2. Actually kya hota hai? Water heating dielectric / dipole-relaxation heating hai: polar water molecules oscillating field follow karne ke liye rotate karti hain, peeche lag jaati hain, aur energy ek broad (low-$Q$) band mein dump karti hain jo many GHz span karta hai. Yeh step kyun? Yeh ek non-resonant, robust process hai — chahe field $2.4$ ho ya $2.5\text{ GHz}$, kaam karta hai; frequency engineering ki wajah se choose ki jaati hai (magnetron cost, penetration depth, regulation), resonance ki wajah se nahi.

3. Conclusion: claim FALSE hai (C10). Agar yeh resonance hoti, toh ovens ko laboratory-grade frequency stability chahiye hoti — aisa hota nahi. Yeh step kyun? Yeh recognize karna ki resonance formula kab apply nahi hota utna hi important hai jitna use apply karna.

Verify (consistency, not arithmetic): gaseous water ki ek true molecular rotational resonance kahin aur hoti hai aur extremely sharp hoti hai; $2.45\text{ GHz}$ par ek sharp resonance ovens ko drift ke liye fragile bana deti — jo is reality se contradict karta hai ki ovens cheap aur tolerant hain. Broad low-$Q$ picture observation se match karta hai. ✓

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Active recall

Recall Har answer kaunsi cell se belong karta hai?

• Amplitude $\approx F_0/k$ jab $\omega \ll {\omega }_0$? ::: C1 (low-frequency / static-like limit)
• Denominator collapse hokar $b\omega /m$ reh jaata hai? ::: C2 (on resonance)
• Amplitude $\propto 1/{\omega }^2$? ::: C3 (far above resonance, inertia-dominated)
• Amplitude $\to \infty$? ::: C4 (zero damping)
• Amplitude $=F_0/k$ exactly? ::: C5 ($\omega =0$, static push)
• Peak ${\omega }_0$ se neeche baith jaata hai? ::: C6 (${\omega }_{res}=\sqrt{{\omega }_0^2-b^2/2m^2}$)
• $\Delta \omega ={\omega }_0/Q$? ::: C7 (bandwidth / half-power)
• Microwave-oven trap? ::: C10 (dielectric heating, NOT resonance)

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Connections

• Forced Oscillations — har example mein use hone wali $A(\omega )$ machinery provide karta hai.
• Damped Oscillations — $b$ aur isliye $Q$ provide karta hai (Examples 2, 4, 5).
• Simple Harmonic Motion — ${\omega }_0=\sqrt{k/m}$ define karta hai (sab examples).
• LC Circuits & AC Resonance — Example 7 ka electrical twin.
• Standing Waves & Normal Modes — Example 6 mein bridge modes ka resonance.
• Fourier Analysis — isliye random ("break-step") force frequency par spread ho jaata hai aur peak avoid karta hai.