Forced oscillations — driving frequency: Question bank

1.6.11 · D5 · Physics › Oscillations & Waves › Forced oscillations — driving frequency

Traps se pehle, hum har formula pin down karte hain jis par questions rely karte hain, taaki koi bhi cheez earn kiye bina use na ho.

Notation reminder (taaki yahan koi bhi cheez earn kiye bina use na ho):

$ω_{d}$ = driving frequency, jo external push set karta hai.
$ω_{0} = k / m$ = natural frequency, system ki apni preferred rhythm.
$γ = b / (2 m)$ = damping rate; $b$ drag constant hai.
$A (ω_{d})$ = displacement amplitude; $A_{v} (ω_{d}) = ω_{d} A$ = velocity amplitude; $ϕ$ = force ke peeche displacement ka phase lag.
$P = F \overset{x}{˙}$ = instantaneous power jo driver mass ko feed karta hai: force times velocity, yaani push kitni tezi se usmein kaam karta hai. Yeh tab bada hota hai jab force aur velocity ek hi direction mein point karte hain.

Figure 1 — amplitude curve jo "true or false" items describe karti hai

Figure — Forced oscillations — driving frequency

Figure 2 — phase curve jo TF4, SE3, SE5 aur har "phase" question ke peeche hai

Figure 3 — right triangle jo $tan ϕ$ produce karta hai

$ω_{r es} = ω_{0}^{2} - 2 γ^{2}$ actually kahan se aata hai

Kai questions isse quote karte hain. Yeh ek-paragraph derivation hai taaki koi bhi ise blindly accept na kare. $A = \frac{F _{0} / m}{D}$ ek positive constant $F_{0} / m$ divided by $D$ hai. Kyunki woh numerator $ω_{d}$ contain hi nahi karta, $ω_{d}$ change karna use move nahi kar sakta — yeh sirf ek overall scale factor hai. Toh woh $ω_{d}$ jo $A$ ko sabse bada banata hai wahi $ω_{d}$ hai jo denominator ko sabse chhota banata hai; numerator extremum condition se bilkul bahar nikal jaata hai. Isliye hum minimise karte hain $D (ω_{d}) = (ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2} .$ $u = ω_{d}^{2}$ treat karo toh $D = (ω_{0}^{2} - u)^{2} + 4 γ^{2} u$ . $\frac{d D}{d u} = 0$ set karo: $\frac{d D}{d u} = - 2 (ω_{0}^{2} - u) + 4 γ^{2} = 0 \Rightarrow u = ω_{0}^{2} - 2 γ^{2} .$ Isliye $ω_{r es}^{2} = ω_{0}^{2} - 2 γ^{2}$ , yaani $ω_{r es} = ω_{0}^{2} - 2 γ^{2}$ — amplitude peak $ω_{0}$ se neeche hoti hai.

True or false — justify karo

TF1. Steady state mein ek driven oscillator apni natural frequency

ω_{0}

par vibrate karta hai.

False. Transient ke die hone ke baad,

ω_{d}

frequency ki ek periodic force sirf same frequency par response sustain kar sakti hai, toh motion

ω_{d}

par hoti hai;

ω_{0}

sirf control karta hai ki woh response kitna bada hoga.

TF2. Amplitude peak exactly

ω_{d} = ω_{0}

par hoti hai.

False. Denominator

D

minimise karne par peak

ω_{r es} = ω_{0}^{2} - 2 γ^{2}

par milti hai, damping se

ω_{0}

se neeche push hoti hai; sirf undamped limit

γ \to 0

mein dono coincide karte hain.

TF3. Driving force

F_{0}

double karne se resonance higher frequency par shift ho jaati hai.

False.

F_{0}

sirf

A

ke numerator mein appear karta hai, toh yeh amplitude ko linearly scale karta hai aur kabhi

ω_{r es} = ω_{0}^{2} - 2 γ^{2}

mein enter nahi karta; peak ki location sirf

ω_{0}

aur

γ

se fix hoti hai.

TF4. Resonance par driving force displacement ke saath in phase hoti hai.

False. Resonance par

ϕ = π /2

(90°): force displacement se 90° lead karti hai, jo ise velocity ke saath exactly in phase rakhta hai — toh power

P = F \overset{x}{˙}

(force times velocity) maximal hoti hai kyunki dono har instant ek hi taraf point karte hain.

TF5. Damping badhane se har driving frequency par amplitude hamesha kam hoti hai.

Mostly true, par resonance ke paas sabse zyada. Zyada

γ

denominator mein

2 γ ω_{d}

term badhata hai,

A

kam karta hai; yeh effect resonance se door bahut kam hoti hai (jahan mismatch term dominate karta hai) aur peak par bahut zyada hoti hai.

TF6. Resonance se neeche mass driving force ke saath in phase move karta hai.

Approximately true sirf $ω_{d} \to 0$ ke liye. Slow limit mein

tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}) \to 0

toh

ϕ \to 0

(spring-dominated); jaise

ω_{d}

ω_{0}

ki taraf climb karta hai,

ϕ

continuously

π /2

(90°) ki taraf badhta hai, toh "in phase" ek limiting statement hai, below-resonance rule nahi.

TF7. Ek forced oscillator same energy input ke saath ek free oscillator se zyada amplitude le sakta hai.

True at resonance. Kyunki energy har cycle velocity ke saath step mein inject hoti hai (power

P = F \overset{x}{˙}

positive rehti hai), amplitude tab tak build hoti hai jab tak per-cycle input per-cycle damping loss ke barabar na ho jaaye — ek balance jo single free release kabhi nahi pahunchta.

TF8. Zero damping ke saath

ω_{d} = ω_{0}

par steady-state amplitude finite hoti hai.

False.

γ = 0

set karne par denominator

(ω_{0}^{2} - ω_{d}^{2})^{2}

ho jaata hai, jo

ω_{d} = ω_{0}

par vanish karta hai, toh

A \to \infty

— woh idealised runaway jo real damping tame karta hai.

Error dhundho

SE1. "Transient die ho jaata hai, isliye natural frequency system mein exist karna band kar deti hai."

Natural frequency abhi bhi transient ki oscillation rate govern karti hai jab tak woh rehta hai aur abhi bhi set karti hai ki amplitude peak kahan hogi; sirf motion mein transient ka contribution

e^{- γ t}

ki tarah decay karta hai,

ω_{0}

khud nahi.

SE2. "Kyunki

A = F_{0} / k

jab

ω_{d} \to 0

, toh amplitude sirf spring par depend karti hai."

Yeh sirf static limit hai.

A = F_{0} / (k - m ω_{d}^{2})^{2} + (b ω_{d})^{2}

mein general

ω_{d}

ke liye inertia (

m ω_{d}^{2}

) aur damping (

b ω_{d}

) dono enter karte hain; spring-only formula woh special case hai jahan drive itni slow hai ki woh terms matter nahi karti.

SE3. "Resonance ka matlab hai force aur displacement saath mein build up karte hain, toh woh in phase honge."

Woh magnitude mein saath build up karte hain, lekin phase mein nahi — resonance par

tan ϕ \to \infty

toh force displacement se

π /2

(90°) ahead hai. "Saath grow karna" aur "phase-aligned hona" ko confuse karna yahi trap hai.

SE4. "

ω_{r es} < ω_{0}

, toh velocity amplitude bhi

ω_{0}

se neeche peak karti hai."

Displacement amplitude

A

ω_{r es} = ω_{0}^{2} - 2 γ^{2}

par peak karti hai, lekin velocity amplitude

A_{v} = ω_{d} A

mein extra factor

ω_{d}

hota hai jo uska maximum exactly

ω_{d} = ω_{0}

tak push kar deta hai; alag quantities alag frequencies par peak karti hain.

SE5. "Kyunki driver system ko enslave karta hai, phase lag

ϕ

hamesha zero hota hai."

Enslavement frequency fix karta hai, phase nahi. Kyunki

ϕ = arctan (2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}))

, damping

ϕ

0

(slow) se

π /2

(resonance) se

π

(fast) tak run karata hai.

SE6. "

ω_{d} ≫ ω_{0}

par amplitude slow limit ki tarah

F_{0} / k

approach karti hai."

Nahi — fast limit mein

m ω_{d}^{2}

term denominator dominate karta hai aur

A \to F_{0} / (m ω_{d}^{2}) \to 0

hota hai. Sirf slow limit

F_{0} / k

deta hai; dono extremes opposite hain (large vs vanishing amplitude).

SE7. "Sharper resonance peaks ka matlab hai system less total energy absorb karta hai."

Sharper peak ka matlab higher quality factor hai aur resonance par taller response — yeh energy ko ek narrow band mein bahut efficiently absorb karta hai, kam nahi. Resonance and Quality Factor dekho.

Why questions

WHY1. Steady-state response ka frequency drive ke saath same kyun hona chahiye?

Equation of motion linear hai;

ω_{d}

par sinusoidal input sirf

ω_{d}

par output produce kar sakta hai (linear systems naye frequencies create nahi karte), toh koi bhi lasting response

ω_{d}

par lock hota hai.

WHY2. Damping amplitude peak ko

ω_{0}

se neeche kyun shift karta hai, upar nahi?

u = ω_{d}^{2}

ke saath

d D / d u = 0

set karne par

u = ω_{0}^{2} - 2 γ^{2}

milta hai; damping term

4 γ^{2} u

frequency ke saath badhta hai aur higher

ω_{d}

ko penalise karta hai, toh denominator ka minimum

ω_{0}

se neeche land karta hai.

WHY3. Power transfer exactly

ω_{0}

par maximal kyun hai

ω_{r es}

par nahi?

Power

P = F \overset{x}{˙}

tab maximal hoti hai jab force aur velocity in phase hote hain, jo

ϕ = π /2

(90°) hai; woh phase condition tab hold hoti hai jab

ω_{0}^{2} - ω_{d}^{2} = 0

, yaani exactly

ω_{d} = ω_{0}

par, displacement peak ki location se regardless. (Upar boxed intuition mein velocity-maximisation same

ω_{0}

confirm karta hai.)

WHY4. Phase lag

ϕ

exist kyun karta hai?

Damping ka matlab hai system instantly respond nahi kar sakta;

tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2})

formula dikhata hai ki lag damping term ke proportional hai, toh

γ > 0

ke saath displacement hamesha force se badhte angle par trail karta hai.

WHY5. Amplitude resonance ke dono sides par kyun fall off hoti hai?

D = (ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2}

dekho. Jaise

ω_{d}

ω_{0}

se upar ya neeche move karta hai, difference

ω_{0}^{2} - ω_{d}^{2}

zero se door ho jaata hai; kyunki yeh squared hai, uska sign irrelevant hai aur

D

dono directions mein grow karta hai, toh

A = (F_{0} / m) / D

dono taraf shrink karta hai.

WHY6. Steady-state amplitude compute karte waqt transient ko kyun ignore kar sakte hain?

Transient

e^{- γ t}

ki tarah decay karta hai aur kuch time constants ke baad chala jaata hai; amplitude formula surviving driven part describe karta hai, jo koi bhi long-time measurement dekhta hai.

WHY7. Agar system

ω_{d}

par oscillate karta hai toh

ω_{0}

jaanna abhi bhi kyun zaroori hai?

Kyunki

ω_{0}

response ki scale set karta hai:

D

ke zariye yeh fix karta hai ki amplitude peak kahan hogi, peak kitni sharp hai, aur

tan ϕ

ke zariye phase lag ka sign — poora

A (ω_{d})

curve ka shape isi se anchor hota hai. Simple Harmonic Motion dekho.

Edge cases

EC1.

ω_{d} \to 0

par amplitude kya hoti hai, aur physically kyun?

A = F_{0} / (k - m ω_{d}^{2})^{2} + (b ω_{d})^{2}

mein denominator

\to k

ho jaata hai, toh

A \to F_{0} / k

: infinitely slow drive ek constant force hai aur mass sirf Hooke's law se displaced ho ke baith jaata hai.

EC2.

ω_{d} \to \infty

par phase lag

ϕ

kya hota hai, aur woh kaisa dikhta hai?

ϕ \to π

(180°): displacement exactly force ke opposite hai. Arctan trap se bachna — raw ratio

tan ϕ = 2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2})

negative side se

0

approach karta hai (denominator ek large negative number hai), aur naive

arctan

use karne par

ϕ \to 0

padha jaayega; correct branch (denominator negative ⇒

π

add karo, EC7 dekho)

ϕ \to π

deta hai, jo opposite motion ki physics se match karta hai.

EC3. Jab

2 γ^{2} \geq ω_{0}^{2}

ho toh amplitude curve ka kya hota hai?

ω_{r es}^{2} = ω_{0}^{2} - 2 γ^{2} \leq 0

non-positive ho jaata hai, toh koi real peak frequency nahi hoti.

D = (ω_{0}^{2} - ω_{d}^{2})^{2} + (2 γ ω_{d})^{2}

trace karo: uski derivative

d D / d u = - 2 (ω_{0}^{2} - u) + 4 γ^{2}

already

u = 0

par positive hai jab

2 γ^{2} \geq ω_{0}^{2}

, toh

D

sirf badhta hai aur

A

monotonically apni static value

F_{0} / k

se fall karta hai. Yeh heavily-overdamped regime hai, Damped Oscillations se linked.

EC4. Undamped limit

γ \to 0

mein, peak kahan hai aur kitna tall hai?

ω_{r es} = ω_{0}^{2} - 0 = ω_{0}

aur denominator wahan zero hit karta hai, toh

A \to \infty

— true resonance ka mathematical signature jisme kuch limit karne wala nahi.

EC5. Agar

ω_{d} = ω_{0}

hai lekin damping heavy hai, kya system "at resonance" hai?

Aap maximum power transfer ki frequency par hain (velocity/

A_{v}

peak, hamesha

ω_{0}

par), lekin amplitude peak par nahi (jo

ω_{r es} < ω_{0}

par hai); "resonance" ke liye aapko specify karna hoga ki aap kaunsi specific quantity ki baat kar rahe hain.

EC6. Exactly

ω_{d} = ω_{0}

par,

tan ϕ

kya hai, aur hum

ϕ

wahan kaise padhte hain?

Mismatch

ω_{0}^{2} - ω_{d}^{2} = 0

ho jaata hai, toh

tan ϕ = 2 γ ω_{d} /0 \to + \infty

, jis se

ϕ = π /2

(90°) exactly force hota hai — woh crossover jahan response spring-led se inertia-led ho jaata hai.

EC7. Naive

ϕ = arctan (2 γ ω_{d} / (ω_{0}^{2} - ω_{d}^{2}))

values

(- π /2, π /2)

mein return karta hai. Hum true lag jo

0 \to π

run karta hai kaise paate hain?

ω_{0}

se neeche denominator

ω_{0}^{2} - ω_{d}^{2} > 0

hota hai toh

arctan

correct acute lag deta hai;

ω_{0}

se upar yeh negative ho jaata hai aur

arctan

negative angle par jump karta hai, toh hum $π$ add karte hain taaki

ϕ

π /2

(90°) se

π

(180°) tak smoothly badhta rahe. Yeh standard arctan branch-fix hai — phase physically kabhi backward nahi jaata. Complex Exponential Method dekho.

EC8. Agar driving force switch off kar di jaaye (

F_{0} = 0

) toh steady-state amplitude kya hoti hai?

A = 0

F_{0} = 0

ke saath numerator vanish ho jaata hai, toh koi steady state nahi hoti — sirf free transient jo decay ho jaata hai, Energy in Oscillations ke damping se energy lose karne ke consistent.

Recall Ek-line self-test

Upar ke har answer ko cover karo, top se bottom sweep karo, aur koi bhi flag karo jahan aapne sirf "true/false" kaha bina because ke — aur check karo ki aapki reasoning actual formula ( $D$ , $tan ϕ$ , ya $A_{v} = ω_{d} A$ ) cite karta hai, sirf words nahi. Reasoning hi exam answer hai; sirf verdict se kuch nahi milta.