Resonance — physical consequences, design implications: Question bank

1.6.12 · D5 · Physics › Oscillations & Waves › Resonance — physical consequences, design implications

Shuru karne se pehle, chaar anchors jo baar baar kaam aayenge. Inhe skip mat karo — neeche ke har trap mein inme se ek use hota hai.

Definition Chaar symbols jo is page par use hote hain

==Spring constant $k$ == — restoring pull ki stiffness: force $- k x$ is proportion mein badha jata hai jitna $x$ displace karo mass ko. Bada $k$ = stiff (zyada zyada kheechta hai), chhota $k$ = floppy. Iske units hain newtons per metre.
Natural frequency $ω_{0} = k / m$ — woh rhythm jis par system bina kisi push aur bina friction ke jhulta hai. Stiffer spring (bada $k$ ) ya halka mass (chhota $m$ ) matlab zyada tezi se jhulna. Banaya gaya hai Simple Harmonic Motion mein.
Damping $b$ — friction term $- b \overset{x}{˙}$ jo energy remove karta hai; ye akela cheez hai jo resonance ko infinity tak blow up hone se rokta hai.
Drive $F_{0} cos ω t$ — frequency $ω$ par baahri push. Resonance ka matlab hai $ω$ ko $ω_{0}$ se match karna.
==Phase lag $ϕ$ == — steady solution $x (t) = A cos (ω t - ϕ)$ mein, number $ϕ$ batata hai ki mass driver se kitna peeche hai, ek poore cycle ke angle mein measure kiya gaya ( $360°$ = ek poora cycle). $ϕ = 0$ matlab mass exactly push ke saath chalta hai; $ϕ = 90°$ matlab ek quarter cycle peeche hai; $ϕ = 180°$ matlab push ke bilkul ulta chalta hai.

Ab chalte hain dekhte hain resonance curve ko jo ye symbols describe karte hain.

Figure — Resonance — physical consequences, design implications

Blue curve hai amplitude $A (ω)$ . Notice karo ki iska peak $ω_{0}$ (dashed line) se thoda baayi taraf hai — woh offset yahi "displacement resonance" traps ke baare mein hai. Pale-yellow band, figure par directly $Δ ω$ label kiya gaya, do half-power points span karta hai: jahan $A$ apne peak se $1/ 2$ tak gir jaata hai, toh delivered power (∝ $A^{2}$ ) exactly half ho jaata hai.

Formula Do special formulas kahaan se aati hain (light-damping sketch)

Amplitude hai $A (ω) = \frac{F _{0} / m}{( ω _{0}^{2} - ω ^{2} ) ^{2} + ( bω / m ) ^{2}}$ . Peak location. $A$ wahan sabse bada hota hai jahan denominator sabse chhota hota hai. Andar ki cheez ko minimize karte hue, $(ω_{0}^{2} - ω^{2})^{2} + (bω / m)^{2}$ , iska derivative zero set karke milta hai $ω_{r es} = ω_{0}^{2} - \frac{b ^{2}}{2 m ^{2}} .$ $ω_{0}$ se neeche kyun? $(bω / m)^{2}$ term $ω$ ke saath badhta hai, isliye sweet spot ko $ω_{0}$ par pure spring–inertia balance se thoda baayi taraf kheeachta hai. Peak height. Light damping ke liye $ω_{r es} \approx ω_{0}$ , toh $ω = ω_{0}$ plug karo $A$ mein: pehla bracket vanish ho jaata hai aur sirf $b ω_{0} / m$ bachta hai, deta hai $A_{ma x} \approx \frac{F _{0} / m}{b ω _{0} / m} = \frac{F _{0}}{b ω _{0}} .$ Caveat: ye clean form sirf light-damping limit mein hoti hai; heavy damping ke liye poora $ω_{r es}$ use karna hoga.

Formula Half-power condition ⇒ bandwidth

$Δ ω = ω_{0} / Q$ "Half-power" algebraically kya matlab hai. Delivered power $A^{2}$ ke scale hoti hai, toh half-power point koi bhi $ω$ hai jahan $A (ω) = \frac{A _{ma x}}{2} ⟺ A (ω)^{2} = \frac{1}{2} A_{ma x}^{2} .$ Use karke equation banao. $A_{ma x}^{2} = (F_{0} / m)^{2} / (b ω_{0} / m)^{2}$ ke saath, $A (ω)$ ko square karke aur half ke barabar demand karte hue, $(F_{0} / m)^{2}$ cancel hone ke baad milta hai, $(ω_{0}^{2} - ω^{2})^{2} + (\frac{bω}{m})^{2} = 2 (\frac{b ω _{0}}{m})^{2} .$ Peak ke paas solve karo. Light damping ke liye do solutions $ω_{0}$ ke kaafi paas rehte hain; likho $ω = ω_{0} \pm \frac{δ}{2}$ aur sirf leading terms rakho. Condition reduce hoti hai $(ω_{0}^{2} - ω^{2})^{2} \approx (b ω_{0} / m)^{2}$ mein, yaani $∣ ω_{0}^{2} - ω^{2} ∣ \approx b ω_{0} / m$ . Kyunki $ω_{0}^{2} - ω^{2} \approx 2 ω_{0} (ω_{0} - ω)$ , har half-power point $ω_{0}$ se $∣ ω_{0} - ω ∣ \approx b / (2 m)$ ki doori par hota hai. Unke beech ki full width isliye hai $Δ ω \approx \frac{b}{m} = \frac{ω _{0}}{Q}, using Q = \frac{m ω _{0}}{b} .$ Yahi se clean rule $Δ ω = ω_{0} / Q$ aata hai — aur figure par yellow band yahi hai.

Aakhir mein, phase. Dekho kaise $ϕ$ $0$ se $180°$ tak swing karta hai jab drive resonance se guzarti hai, exactly $90°$ par $ω_{0}$ cross karte hue:

Formula Phase-lag formula — aur kyun

$ω_{0}$ par $90°$ hota hai $cos$ aur $sin$ parts ko balance karte hue jab $x = A cos (ω t - ϕ)$ equation of motion mein substitute karte ho (kaam kiya gaya hai Forced Oscillations mein) milta hai $ϕ (ω) = arctan (\frac{bω / m}{ω _{0}^{2} - ω ^{2}}) .$ Teen regimes fraction se padho:

$ω_{0}$ se neeche ( $ω < ω_{0}$ ): denominator $ω_{0}^{2} - ω^{2}$ positive aur bada hai, ratio chhota hai ⇒ $ϕ \to 0$ (mass almost push ke saath chalta hai).
$ω = ω_{0}$ par: denominator exactly zero hai, toh argument $\to + \infty$ aur $arctan (+ \infty) = 90°$ . Ye crossover hai — koi algebra nahi chahiye "arctan of infinity is a right angle" se aage.
$ω_{0}$ se upar ( $ω > ω_{0}$ ): denominator negative flip ho jaata hai, toh $ϕ$ second quadrant mein $180°$ ki taraf jaata hai (mass push ka virodh karta hai, inertia dominate karta hai).

$ϕ = 90°$ par (crossing point) force displacement se ek quarter-cycle aage hai, jo ise velocity ke saath perfectly line up karta hai — yahi wajah hai ki wahan energy sabse zyada efficiently andar aati hai.

True ya false — justify karo

True ya false: resonance par driving force displacement ke saath in phase hoti hai.

False. Resonance par force displacement se 90° aage hoti hai (

ϕ = 90°

), jo ise velocity ke saath in phase rakhta hai — yahi exact wajah hai ki har push energy add karta hai aur swing badhti rehti hai.

True ya false: zero damping ke saath

ω_{0}

par steady-state amplitude infinite hogi.

True (mathematically). Denominator ban jaata hai

0 + 0 = 0

, toh

A \to \infty

. Physically impossible — aur yahi poora point hai ki real systems mein zaroor damping hona chahiye ise cap karne ke liye.

True ya false: displacement amplitude exactly

ω_{0}

par peak karta hai.

False. Displacement peak thoda neeche hota hai,

ω_{r es} = ω_{0}^{2} - b^{2} /2 m^{2}

par. Sirf velocity/power resonance exactly

ω_{0}

par hoti hai.

True ya false: ek bhaari oscillator resonance se safe hai kyunki woh sluggish hai.

False. Extra mass sirf

ω_{0} = k / m

lower karta hai; woh resonance relocate karta hai, remove nahi. Tum accidentally

ω_{0}

ko kisi aise driver par move kar sakte ho jisse pehle tum safe the.

True ya false: microwave oven water molecules ko resonate karke kaam karta hai.

False. Ye dielectric (dipole-relaxation) heating use karta hai — dipoles lag karte hain jab field follow karne ki koshish karte hain, aur ek broad band mein heat dump karte hain. Agar ye sharp resonance hoti, toh frequency precise honi chahiye, jo deliberately nahi hai.

True ya false: high-

Q

system mein tall, narrow resonance peak hoti hai.

True. High

Q = m ω_{0} / b

matlab low damping, toh energy dheere dheere leak hoti hai, peak sharp hai aur bandwidth

Δ ω = ω_{0} / Q

narrow hai — selectivity ke liye great hai (radio tuning, quartz clocks).

True ya false: damping badhana hamesha peak amplitude ko lower karta hai.

True. Bada

b

denominator ki minimum value

bω / m

badhata hai, toh

A_{ma x} = F_{0} / (b ω_{0})

gir jaata hai. Ye exactly engineer ka lever hai.

True ya false: soldiers break step karte hain kyunki unka combined weight bridge ke liye bahut bhaari hota hai.

False. Static weight nahi balki ~2 Hz par coherent periodic force wajah hai. Break step timing randomise karta hai toh koi ek frequency drive ko dominate nahi karti.

True ya false: half-power points par amplitude peak ki aadhi ho jaati hai.

False. Amplitude peak ki

1/ 2 \approx 0.707

tak girta hai; power (∝ amplitude²) half hoti hai, kyunki

(1/ 2)^{2} = \frac{1}{2}

True ya false: tuned mass damper building ko rigid banane ke liye stiffness add karke kaam karta hai.

False. Ye ek doosra oscillator add karta hai jo building ki frequency par tuned hota hai; woh energy absorb aur dissipate karta hai, structure ko stiffen karne ki bajaye resonance peak ko split aur lower karta hai.

Error dhundho

"Resonance dangerous hai, toh acchi engineering matlab hai ise poori tarah avoid karna." — flaw dhundho.

Resonance aksar chahiye hoti hai: radios, MRI, aur quartz watches sharp resonance exploit karte hain. Goal hai ise deliberately place karna — high

Q

jahan selectivity help kare, damped aur detuned jahan amplitude structures ko threaten kare.

"Kyunki

A_{ma x} = F_{0} / (b ω_{0})

, drive force

F_{0}

double karna hamesha danger double karta hai." — flaw dhundho.

Amplitude do scale karta hai linearly

F_{0}

ke saath, lekin "danger" is par bhi depend karta hai ki system apni resonant frequency par hai bhi ya nahi.

ω_{0}

se bahut door ek badi force harmless ho sakti hai;

ω_{0}

par ek chhoti si force catastrophic ho sakti hai.

"Tacoma Narrows bridge isliye fail hua kyunki steady wind ne use gira diya." — flaw dhundho.

Steady wind periodic nahi hai aur resonance drive nahi kar sakti. Rhythmic driver tha vortex shedding (ek Kármán vortex street) jiska frequency ek torsional mode se match karta tha. Relevant force oscillate karta tha even though wind speed roughly constant tha.

"Power resonance aur displacement resonance dono

ω_{0}

par hote hain." — flaw dhundho.

Sirf power/velocity resonance exactly

ω_{0}

par hai. Displacement resonance

ω_{r es} = ω_{0}^{2} - b^{2} /2 m^{2}

par hai, thoda neeche — dono sirf

b \to 0

limit mein coincide karte hain.

"Low-

Q

system sirf ek poorly built high-

Q

system hai." — flaw dhundho.

Low

Q

aksar design goal hota hai: shock absorbers, door dampers, aur microwave oven sab broad, robust response chahte hain taaki behaviour chhoti frequency drift se barely change ho. "Good" poori tarah kaam par depend karta hai.

"Kyunki resonance par

ϕ = 90°

hai, force koi kaam nahi karta — ye displacement ke perpendicular hai." — flaw dhundho.

Kaam force ke velocity ke saath aligned hone par depend karta hai, displacement par nahi.

ϕ = 90°

par force exactly velocity ke saath in phase hai, toh ye har cycle mein maximum kaam karta hai — bilkul ulta conclusion.

Why questions

Resonance par ek tiny periodic push bhi huge amplitude kyun build kar leta hai?

Kyunki har push motion ke saath step mein aata hai (force velocity ke saath in phase hai), toh har push energy coherently add karta hai instead of kuch add karne aur kuch subtract karne ke. Growth sirf tab limit hoti hai jab energy har cycle mein fed hona energy lost to damping ke barabar ho jaata hai.

Damping woh quantity kyun hai jo "save the day" karta hai, mass ya stiffness nahi?

Damping akela term hai jo energy remove karta hai. Mass aur stiffness sirf set karte hain ki

ω_{0}

kahaan baithta hai; kisi bhi

b

ke bina energy in aur out ka balance kabhi close nahi hota, toh amplitude bhaag jaata hai.

Higher

Q

ek radio ko better station selectivity kyun deta hai?

High

Q

matlab narrow bandwidth

Δ ω = ω_{0} / Q

, toh sirf

ω_{0}

ke bahut karib ki frequencies strongly amplify hoti hain aur neighbouring stations reject ho jaate hain. Dekho LC Circuits & AC Resonance.

Displacement peak

ω_{r es}

par

ω_{0}

se neeche kyun hoti hai, us par nahi?

Kyunki denominator mein damping term

(bω / m)^{2}

ω

ke saath badhta hai; ye higher frequencies ko penalise karta hai, smallest-denominator (largest-amplitude) point ko

ω_{0}

par spring–inertia balance se baayi taraf nudge karta hai.

Ek sharp driver ek matching normal mode ko strongly kyun excite karta hai?

Single-frequency drive apni energy almost poori tarah us mode mein daalta hai jiska

ω_{0}

match karta hai; via Fourier Analysis, ek mismatched ya broadband drive apni energy bahut saare modes mein thinly spread karta hai, kisi ko bhi strongly excite nahi karta.

Phase lag

ϕ

resonance se sweep karte waqt

90°

se kyun guzarta hai?

ϕ = arctan [(bω / m) / (ω_{0}^{2} - ω^{2})]

mein denominator

ω_{0}

se neeche positive hai (

ϕ

chhota), at

ω_{0}

zero hai (

arctan \infty = 90°

), aur upar negative (

ϕ \to 180°

90°

par crossover perfect energy transfer ki frequency mark karta hai.

Edge cases

ω = 0

par (constant push, koi oscillation nahi) amplitude kya hogi?

Formula deta hai

A = F_{0} / k

— sirf static spring stretch. Koi resonance nahi, koi damping effect nahi: ek steady force simply spring ko Hooke's law se displace karta hai.

ω \to \infty

par (natural se bahut zyada tezi se drive karte hue)

A

ka kya hota hai?

Amplitude

\to 0

. Mass ki inertia rapid reversals ke saath keep up nahi kar sakti, toh

m ω^{2}

term denominator mein dominate karta hai aur sab kuch swamp kar deta hai — oscillator barely move karta hai.

Jab damping

b^{2} = 2 m^{2} ω_{0}^{2}

ki taraf badhti hai toh

ω_{r es}

ka kya hota hai?

ω_{r es} = ω_{0}^{2} - b^{2} /2 m^{2} \to 0

: displacement-resonance peak zero frequency ki taraf saari tarah slide ho jaata hai. Us damping level se aage koi peak nahi hoti — response sirf monotonically apni static value se girta hai (neeche figure par overdamped curve dekho).

Light-damping limit

b \to 0

mein, peak kahaan jaata hai aur kitna tall?

ω_{r es} \to ω_{0}

aur

A_{ma x} = F_{0} / (b ω_{0}) \to \infty

. Ye idealised undamped picture hai jo resonance ko infinitely dangerous dikhata hai — ek limit, kabhi reality nahi.

Ek perfectly rigid, undriven, undamped mass on a spring ke liye, kya "resonance" define bhi hoti hai?

Koi drive nahi matlab koi resonance nahi — tumhare paas sirf

ω_{0}

par free Simple Harmonic Motion hai. Resonance inherently ek forced-oscillation phenomenon hai: ise

ω_{0}

ke saath match karne ke liye ek baahri periodic push chahiye.

Connections

Forced Oscillations — steady-state amplitude aur phase jo ye traps probe karte hain.
Damped Oscillations — jahan se $b$ , $Q$ , aur amplitude cap aate hain.
Simple Harmonic Motion — $ω_{0} = k / m$ ka source.
LC Circuits & AC Resonance — electrical form mein selectivity questions.
Standing Waves & Normal Modes — "why" section ke peeche mode-matching.
Fourier Analysis — kyun sharp drivers matching modes excite karte hain.
Resonance — physical consequences, design implications — parent topic.