Exercises — Resonance — physical consequences, design implications

Kaunsa tool: ${\omega }_0=\sqrt{k/m}$ — yeh seedha Simple Harmonic Motion se aata hai, jahan spring ki restoring force $-kx$ aur Newton's $F=ma$ combine hokar $\ddot{x}=-(k/m)x$ banate hain, jiska rhythm $\sqrt{k/m}$ hai. Kyun: natural frequency woh "target" hai jiske paas resonance rehti hai, isliye yeh hamesha step one hai. ${\omega }_0=\sqrt{\frac{200}{0.50}}=\sqrt{400}=20\text{rad/s}.$

Hump ke top par amber point resonance hai — woh driving frequency jo sabse bada swing $A$ deti hai. Uski height is baat se set hoti hai ki master formula ka denominator kitna chota ho sakta hai, jo peak par damping term $b\omega /m$ se dominate hota hai. Toh peak height damping par depend karti hai: kam damping → taller, narrower peak.

Kaunsa tool: master formula. $\omega ={\omega }_0$ par kyun: term $({\omega }_0^2-{\omega }^2)$ $0$ ho jaata hai, toh messy denominator sirf $b\omega /m$ tak collapse ho jaata hai. $A=\frac{F_0/m}{\sqrt{0+(b\omega /m{)}^2}}=\frac{F_0/m}{b\omega /m}=\frac{F_0}{b\omega }.$ Numerically: $A=\frac{2.0}{0.50\times 20}=\frac{2.0}{10}=0.20\text{m}.$ Notice karo ki mass cancel ho gaya — resonance par, amplitude sirf damping se limited hai.

Kaunsa tool: $Q={\omega }_{0}m/b$, phir $\Delta \omega ={\omega }_0/Q$. Kyun: $Q$ batata hai ki kitne "saaf" oscillations damping ke energy khaane se pehle hote hain — bada $Q$ = narrow peak. $Q=\frac{{\omega }_{0}m}{b}=\frac{20\times 0.50}{0.50}=20.$ $\Delta \omega =\frac{{\omega }_0}{Q}=\frac{20}{20}=1.0\text{rad/s}.$

Kaunsa tool: ${\omega }_{res}=\sqrt{{\omega }_0^2-b^2/(2m^2)}$ — yeh Forced Oscillations mein $dA/d\omega =0$ set karke milta hai. Kyun yeh ${\omega }_0$ se neeche hai: damping term $b\omega /m$ $\omega$ ke saath badhta hai, isliye poore denominator ka minimum uss point se thoda neeche push ho jaata hai jahan $({\omega }_0^2-{\omega }^2)=0$ hota hai. $\frac{b^2}{2m^2}=\frac{0.50^2}{2\times 0.50^2}=\frac{0.25}{0.50}=0.50{\text{rad}}^2/{\text{s}}^2.$ ${\omega }_{res}=\sqrt{400-0.50}=\sqrt{399.5}\approx 19.9875\text{rad/s}.$ ${\omega }_0$ se sirf $0.06\%$ neeche — is light damping ke liye shift bahut tiny hai, yeh confirm karta hai ki ${\omega }_{res}\approx {\omega }_0$ lekin exactly equal nahi.

Kyun yeh frequency: do half-power points roughly ${\omega }_0\pm \Delta \omega /2$ par hote hain; wahan amplitude peak $0.20\text{m}$ ka roughly $1/\sqrt{2}\approx 0.707$ honi chahiye, yaani approximately $0.1414\text{m}$. $\omega =20.5$ par master formula ke har piece ko compute karo:

• ${\omega }_0^2-{\omega }^2=400-420.25=-20.25$.
• $b\omega /m=(0.50\times 20.5)/0.50=20.5$.
• denominator $=\sqrt{(-20.25{)}^2+20.5^2}=\sqrt{410.06+420.25}=\sqrt{830.31}\approx 28.82$.
• numerator $F_0/m=2.0/0.50=4.0$. $A=\frac{4.0}{28.82}\approx 0.1388\text{m}.$ Yeh peak ka $0.1388/0.20=0.694$ hai — ideal $0.707$ ke bahut close, yeh confirm karta hai ki ${\omega }_0\pm \Delta \omega /2$ (approximately) half-power edges mark karta hai.

Kyun yeh hamare formula se map hota hai: LC circuit mass–spring ka electrical mirror hai: $1/C$ stiffness $k$ ka role play karta hai, $L$ mass $m$ ka. Uski natural frequency ${\omega }_0=1/\sqrt{LC}$ hai. Dial tune karna ${\omega }_0$ ko station ki frequency ke equal set karta hai, toh sirf wahi station resonate karta hai. ${\omega }_0=1/\sqrt{LC}$ ko $C$ ke liye solve karo: $C=1/({\omega }_0^{2}L)$. ${\omega }_0=2\pi (100\times 10^6)=6.2832\times 10^8\text{rad/s}.$ $C=\frac{1}{(6.2832\times 10^8{)}^2\times 2.0\times 10^{-6}}=\frac{1}{(3.9478\times 10^{17})(2.0\times 10^{-6})}.$ $C=\frac{1}{7.8957\times 10^{11}}\approx 1.266\times 10^{-12}\text{F}=1.27\text{pF}.$

Marching rate convert karo: $120\text{steps/min}=120/60=2.0\text{steps/s}=2.0\text{Hz}$. Kyun dangerous: driving frequency $f=2.0\text{Hz}$ equals $f_0=2.0\text{Hz}$, toh har footfall bridge ke apne sway ke saath in phase push karta hai — coherent pushes stack up (resonance) hote hain aur amplitude damping-limited maximum $F_0/(b{\omega }_0)$ ki taraf badhti hai. Fix: "break step" footfall timing ko randomise karta hai. Ek random force ki koi single dominant frequency nahi hoti (dekho Fourier Analysis) — uski energy kai frequencies mein spread ho jaati hai, isliye almost kuch bhi exactly $f_0$ par nahi land karta, aur resonant build-up kabhi nahi hota.

(a) $Q={\omega }_{0}M/b=(1.2\times 6.0\times 10^6)/(4.0\times 10^5)=7.2\times 10^6/4.0\times 10^5=18.$ Ek moderate $Q$ — building kaafi sharply ring karti hai, jo exactly danger hai.

(b) Resonance par $A=F_0/(b{\omega }_0)$ ($({\omega }_0^2-{\omega }^2)$ term vanish ho jaata hai): $A=\frac{3.0\times 10^5}{4.0\times 10^5\times 1.2}=\frac{3.0\times 10^5}{4.8\times 10^5}=0.625\text{m}.$ $0.625\text{m}$ ka sway — uncomfortable aur structurally serious.

(c) $b$ ko double karke $8.0\times 10^5\text{kg/s}$ karna: kyunki resonance par $A\propto 1/b$ hai, amplitude half hokar $0.3125\text{m}$ ho jaati hai. Damper sway ko 2 ke factor se kum karta hai aur, $Q$ ko lower karke, peak ko bhi broader banata hai toh building ${\omega }_0$ ko exactly hit karne ke liye kam sensitive hoti hai. Yahi Taipei 101 ke giant pendulum ke peeche ki physics hai.

Agar heating high-$Q$ (sharp) resonance par depend karti, toh oven ki frequency ko water transition se ek tiny fraction of a percent tak match karna padta — hold karna impossible, aur koi bhi drift heating band kar deta. Iska matlab yeh hota ki oven dielectric (dipole-relaxation) heating exploit karta hai: water ke permanent dipoles oscillating field follow karne ke liye twist karne ki koshish karte hain lekin peeche reh jaate hain, aur woh lag energy ko heat ke roop mein dump karta hai. Yeh ek broad, low-$Q$ process hai — yeh frequencies ki wide band par kaam karta hai (roughly $1$–$100\text{GHz}$), isliye $2.45\text{GHz}$ regulatory aur penetration reasons ke liye choose kiya gaya hai, resonant peak hit karne ke liye nahi. Robustness exactly low $Q$ se aati hai: koi sharp tuning zaroori nahi.

Resonance par peak amplitude formula? ::: $A_{max}=F_0/(b{\omega }_0)$ $Q$ in terms of ${\omega }_0,m,b$? ::: $Q={\omega }_{0}m/b$ $Q$ se bandwidth? ::: $\Delta \omega ={\omega }_0/Q$ Displacement resonance frequency? ::: ${\omega }_{res}=\sqrt{{\omega }_0^2-b^2/2m^2}$, ${\omega }_0$ se thoda neeche Damping double karne se resonant amplitude ka kya hota hai? ::: half ho jaata hai (kyunki $A\propto 1/b$) LC natural frequency? ::: ${\omega }_0=1/\sqrt{LC}$