ω, T, f relationships: Worked examples — Hinglish deep dive

1.6.3 · D3 · Physics › Oscillations & Waves › ω, T, f relationships

Yeh ω–T–f subtopic ka drill page hai. Hum har tarah ke sawal ke through march karte hain jo yeh teen rhythm quantities throw kar sakti hain — normal cases, weird limits, degenerate zeros, word problems, ek exam trap, aur yahan tak ki ek negative $ω$ — aur hum har ek ko ground up se solve karte hain.

Shuru karne se pehle, ek second ka reminder — sirf teen tools jinhe hum yahan kabhi use karte hain (parent note mein banaye gaye hain):

Neeche ke examples mein kuch symbols aate hain, toh chaliye unhe use karne se pehle har ek build karte hain:

Definition

$x (t) = A cos (ω t + ϕ)$ ke andar ke symbols

$t$ time hai, seconds mein measure hota hai — yeh tumhare stopwatch ki reading hai. Is page par sab kuch is ek ticking variable ka function hai.
$x (t)$ time $t$ par position hai — oscillating cheez kahan baithe hai (jaise ki mass apne rest point se kitni door hai), kisi length unit jaise cm ya m mein measure hoti hai.
$A$ amplitude hai: motion jo sabse badi doori rest point se kabhi reach karta hai. Kyunki $cos$ $- 1$ aur $+ 1$ ke beech swing karta hai, position $x$ bhi $- A$ aur $+ A$ ke beech swing karti hai. Yeh ek size hai, rhythm nahi — yeh kabhi $ω$ , $f$ , ya $T$ ko affect nahi karta.
Phase woh poori quantity hai jo cosine ke andar hai, $ω t + ϕ$ . Merry-go-round point imagine karo: phase woh angle (radians mein) hai jo usne ab tak turn kiya hai. Jaise $t$ badhta hai, $ω t$ badhta hai — yahi rhythm aage tick karta hai.
$ϕ$ (Greek letter "phi") starting phase hai: woh angle jahan tak point $t = 0$ par pahle se turn kar chuka tha, stopwatch shuru hone se pehle. Yeh poori wave ko time mein left ya right shift karta hai; yeh $ω$ , $f$ , ya $T$ ko nahi badalta.

Phase ke baare mein aur Phase and Phase Difference mein; amplitude aur is poori form ke baare mein aur Simple Harmonic Motion mein.

Scenario matrix

Is topic ka har sawal in cells mein se kisi ek mein land karta hai. Neeche har worked example apni cell ke saath tagged hai taaki tum dekh sako ki coverage complete hai.

Cell	Ise distinct kya banata hai	Example
A. Forward chain	$T$ given → $f$ aur $ω$ nikalo	Ex 1
B. Backward chain	$ω$ given → $f$ aur $T$ nikalo	Ex 2
C. Read-off from equation	$x (t) = A cos (ω t + ϕ)$ given → $ω, f, T$ extract karo	Ex 3
D. Bahut bada $f$ (limit)	High frequency → chhota period, bada $ω$	Ex 4a
E. Bahut chhota $f$ (limit)	Low frequency → bada period, chhota $ω$	Ex 4b
F. Degenerate: $f = 0$	Bilkul oscillation nahi — $T, ω$ ka kya hoga?	Ex 5
G. Degenerate: $ω = 0$ inside cosine	Woh "motion" jo kabhi hilti nahi	Ex 5
H. Unit-trap	Frequency rpm / minutes mein di gayi hai, Hz mein nahi	Ex 6
I. Scaling / proportion	"Agar $f$ triple ho, toh $T, ω$ ka kya hoga?"	Ex 7
J. Real-world word problem	Merry-go-round / wheel, geometry involved	Ex 8 (figure)
K. Exam twist	Wrong-pairing trap ( $T = 2 π f$ ?) units se resolve hota hai	Ex 9
L. Sign / direction	Negative $ω$ — $ω < 0$ ka matlab kya hai?	Ex 10

Worked Examples

Ex 1 — Forward chain (Cell A)

Worked example Period diya hai, sab kuch nikalo

Ek mass on a spring ek poora up-and-down cycle $T = 0.25 s$ mein complete karta hai. $f$ aur $ω$ nikalo.

Forecast: Aage padhne se pehle — guess karo: kya $f$ 1 se bada hoga ya chhota? Kya $ω$ $f$ se bada hoga ya chhota?

Step 1 — $f = 1/ T = 1/0.25 = 4 Hz$ . Yeh step kyun? $f$ count karta hai ki ek second mein kitne cycles fit hote hain. Agar ek cycle $0.25 s$ leta hai, toh ek second mein chaar fit hote hain. Woh reciprocal hi frequency ki definition hai.

Step 2 — $ω = 2 π f = 2 π (4) = 8 π \approx 25.13 rad/s$ . Yeh step kyun? $ω$ wohi rhythm measure karta hai lekin radians mein. Har cycle $2 π$ radians sweep karta hai, toh 4 cycles/s ka matlab hai $4 \times 2 π$ rad/s. Isliye $ω$ yahan hamesha bada number hota hai — woh $2 π$ ka crown carry karta hai.

Forecast resolved: $f = 4$ bada hai 1 se (short period mein bahut saare cycles pack hote hain), aur $ω = 8 π \approx 25.1$ bada hai $f$ se — $2 π$ hamesha ise inflate karta hai. Dono guesses "bigger" wale the.

Verify: Loop back karo: $T = 2 π / ω = 2 π / (8 π) = 0.25 s$ ✔. Units: $[Hz] = s^{- 1}$ ✔, $[ω] = rad/s$ ✔.

Ex 2 — Backward chain (Cell B)

Worked example Angular frequency diya hai, usse wapas trace karo

Ek oscillator ka $ω = 15 rad/s$ hai. $T$ aur $f$ nikalo.

Forecast: Kya $T$ 1 second se zyada hoga ya kam? Kya $f$ $ω$ se zyada hoga ya kam?

Step 1 — $T = 2 π / ω = 2 π /15 \approx 0.4189 s$ . Yeh step kyun? $T$ woh time hai jisme angle ek poora turn, $2 π$ , sweep kare. Total angle ( $2 π$ ) ko angle sweep karne ki rate ( $ω$ ) se divide karne par time milta hai. Yeh bas distance ÷ speed hai, lekin angle ke liye.

Step 2 — $f = ω / (2 π) = 15/ (2 π) \approx 2.387 Hz$ . Yeh step kyun? Hum $2 π$ ko strip off karte hain. Ek cycle hai hi $2 π$ radians ka turning, toh ek radians-per-second rate ko pure number $2 π$ se divide karne par pata chalta hai ki per second kitne poore $2 π$ -turns (yaani cycles) hote hain — yahi Hz hai. $f$ $ω$ se exactly isi liye chhota hai jis liye $ω$ Ex 1 mein bada tha.

Forecast resolved: $T \approx 0.419$ 1 second se kam hai, aur $f \approx 2.39$ $ω = 15$ se kam hai — $2 π$ se divide karne par woh shrink ho gaya. Dono guesses "less" wale the.

Verify: $f = 1/ T = 1/0.4189 \approx 2.387 Hz$ ✔ — dono independent routes agree karte hain.

Ex 3 — Equation se Read-off (Cell C)

Worked example

$x (t)$ se seedha rhythm nikalo Ek object aise move karta hai: $x (t) = 5 cos (6 π t + \frac{π}{3}) cm$ . $ω$ , $f$ , $T$ nikalo.

Forecast: Us equation mein $ω$ kaun sa number hai — $6 π$ hai, $5$ hai, ya $π /3$ hai?

Step 1 — $ω = 6 π rad/s$ . Yeh step kyun? Standard SHM form $A cos (ω t + ϕ)$ se compare karo, jiske symbols humne upar define kiye. $t$ ko multiply karne wali cheez hi definition se $ω$ hai. $5$ amplitude $A$ hai (cm mein) — sabse badi distance jo reach hoti hai; $π /3$ starting phase $ϕ$ hai — $t = 0$ par angle kitna already turn ho chuka tha — aur inme se koi bhi rhythm ko affect nahi karta.

Step 2 — $f = ω / (2 π) = 6 π / (2 π) = 3 Hz$ . Yeh step kyun? Radian-rate ko cycle-rate mein convert karo; $π$ cleanly cancel ho jata hai, jo is baat ki tell-tale sign hai ki sahi kiya.

Step 3 — $T = 1/ f = 1/3 \approx 0.3333 s$ . Yeh step kyun? Seedha reciprocal, possible sabse sasta last step.

Forecast resolved: $ω$ woh $6 π$ hai ( $t$ ka coefficient) — nahi amplitude $5$ aur nahi phase $π /3$ . Agar tumne $6 π$ guess kiya, tum sahi the.

Verify: Phase ek period mein $2 π$ advance karni chahiye: $ω T = 6 π \times \frac{1}{3} = 2 π$ ✔.

Ex 4 — Limiting values: bahut fast aur bahut slow (Cells D & E)

Worked example Frequency ko extremes tak push karo

(a) [Cell D — bahut fast.] Ek guitar string $f = 440 Hz$ par vibrate karti hai (musical note A). $T$ aur $ω$ nikalo. (b) [Cell E — bahut slow.] Ek slow ocean swell ka $f = 0.05 Hz$ hai. $T$ aur $ω$ nikalo.

Forecast: Fast string ke liye, kya $T$ bada hoga ya tiny? Slow swell ke liye, kya $T$ bada hoga ya tiny?

Step 1a (Cell D) — $T = 1/ f = 1/440 \approx 0.002273 s$ . Yeh step kyun? High frequency = zyada cycles per second = har cycle squeeze hokar chhoti ho jaati hai. Jaise $f \to$ large, $T = 1/ f \to 0$ .

Step 2a (Cell D) — $ω = 2 π f = 2 π (440) = 880 π \approx 2764.6 rad/s$ . Yeh step kyun? Fast rhythm ⇒ angle enormously fast sweep karta hai. $ω$ without bound badhta hai jaise $f$ badhta hai.

Step 1b (Cell E) — $T = 1/ f = 1/0.05 = 20 s$ . Yeh step kyun? Low frequency = kam cycles per second = har cycle stretch hokar lambi ho jaati hai. Jaise $f \to 0^{+}$ , $T = 1/ f \to \infty$ .

Step 2b (Cell E) — $ω = 2 π f = 2 π (0.05) = 0.1 π \approx 0.3142 rad/s$ . Yeh step kyun? Slow rhythm ⇒ angle barely aage creep karta hai.

Forecast resolved: Fast string → $T \approx 0.0023 s$ tiny hai; slow swell → $T = 20 s$ bada hai. High $f$ $T$ ko shrink karta hai, low $f$ use stretch karta hai.

Verify: (a) $ω T = 880 π \times 0.002273 \approx 2 π$ ✔. (b) $ω T = 0.1 π \times 20 = 2 π$ ✔. Product $ω T = 2 π$ hamesha sach hai — yahi invariant har case ko guard karta hai.

Ex 5 — Degenerate inputs: $f = 0$ aur $ω = 0$ (Cells F & G)

Worked example Agar oscillation hi nahi hai toh?

(a) Ek "system" ka $f = 0$ hai. $T$ aur $ω$ kya hain? (b) $x (t) = A cos (0 \cdot t + ϕ)$ kya describe karta hai? (Yaad karo $A$ = amplitude, $t$ = time, $ϕ$ = starting phase.)

Forecast: Kya non-oscillating cheez ka period ek real number ho sakta hai? Zero- $ω$ cosine kaunsa curve trace karta hai?

Step 1a — $ω = 2 π f = 2 π (0) = 0 rad/s$ . Yeh step kyun? Cycles per second nahi ⇒ radians swept per second nahi. Angle kabhi hilta hi nahi.

Step 2a — $T = 1/ f = 1/0 \to \infty$ (undefined / infinite). Yeh step kyun? Agar tum kabhi ek cycle complete nahi karte, toh "ek cycle ka time" infinite hai. Toh $T$ yahan finite number nahi hai — reciprocal exactly tab break ho jaata hai jab repeat karne ke liye kuch hota hi nahi. Yeh boundary case hai jise flag karna zaroori hai, koi number force nahi karna.

Step 1b — $x (t) = A cos (0 + ϕ) = A cos ϕ = constant$ . Yeh step kyun? $ω = 0$ hone par, argument kabhi nahi badalta jaise $t$ tick karta hai, toh cosine frozen hai. "Motion" ek stationary point hai jo height $A cos ϕ$ par baitha hai (amplitude $A$ ka ek fixed fraction) — $x$ – $t$ graph par ek flat horizontal line. Yeh SHM ka degenerate limit hai: infinite period wala oscillation simply nahi hilta.

Forecast resolved: Nahi — period finite real number nahi hai; woh infinite hai. Aur zero- $ω$ cosine ek flat horizontal line trace karta hai (ek constant), wave nahi.

Verify: Invariant ko dheerey check karo: $ω T = 2 π$ ke liye $0 \cdot \infty$ chahiye, jo ek indeterminate form hai — consistent hai $T$ ke undefined hone ke saath, kisi finite value ke saath nahi. Toh sirf ek self-consistent reading hai: $ω = 0, f = 0, T = \infty$ ✔.

Ex 6 — Unit trap: rpm nahi Hz (Cell H)

Worked example Minutes mein chhupa hua frequency

Ek record player $33 \frac{1}{3}$ revolutions per minute (rpm) par spin karta hai. $f$ (Hz), $T$ , aur $ω$ nikalo.

Forecast: Kya $f$ roughly $33$ hoga, ya roughly $0.5$ ? "Per minute" dhyan se dekho.

Step 1 — Hz mein convert karo: $f = \frac{33.333 rev}{60 s} = 0.5556 Hz$ . Yeh step kyun? Hz ka matlab cycles per second hai, lekin humein cycles per minute diya gaya tha. $60 s/min$ se divide karo taaki "minutes" cancel ho jayein aur seconds aayein. Isko skip karna classic unit blunder hai.

Step 2 — $T = 1/ f = 1/0.5556 = 1.8 s$ . Yeh step kyun? Reciprocal. Sanity check: ek full turn ko 2 seconds se thoda kam lena chahiye — slow turntable ke feel se match karta hai.

Step 3 — $ω = 2 π f = 2 π (0.5556) \approx 3.491 rad/s$ . Yeh step kyun? Standard conversion jab $f$ finally Hz mein aa jaata hai.

Forecast resolved: $f \approx 0.56 Hz$ — roughly $0.5$ wala guess jeeta. " $33$ " cycles per minute tha, toh Hz banne ke liye ise 60 se shrink karna pada.

Verify: $T = 60/33.333 = 1.8 s$ directly (60 s ÷ revs-per-minute) ✔, aur $ω T = 3.491 \times 1.8 \approx 2 π$ ✔.

Ex 7 — Scaling aur proportion (Cell I)

Worked example Frequency triple karo

Ek oscillator ki frequency triple ho jaati hai ( $f$ se $3 f$ ). $ω$ aur $T$ ka kya hoga? Multiplying factors do.

Forecast: Do numbers guess karo: " $ω$ ___× ho jaayega aur $T$ ___× ho jaayega."

Step 1 — $ω = 2 π f$ $f$ mein linear hai. Yeh step kyun? $ω$ aur $f$ ek plain multiplication ( $2 π$ , ek constant) se bande hain. Input ko 3 se multiply karne par output bhi 3 se multiply hota hai. Toh $ω \to 3 ω$ .

Step 2 — $T = 1/ f$ $f$ mein inverse hai. Yeh step kyun? Reciprocals ek multiplication ko division mein flip kar dete hain. $f$ ko triple karne par $T$ 3 se divide ho jaata hai, toh $T \to T /3$ .

Forecast resolved: $ω$ 3× ho jaata hai (linear) aur $T$ ⅓× ho jaata hai (inverse). Agar tumne 3 aur one-third guess kiya, bilkul sahi.

Verify (numbers): Start karo $f = 2 Hz \Rightarrow ω = 4 π, T = 0.5$ . Ab $f = 6 Hz \Rightarrow ω = 12 π (= 3 \times 4 π)$ ✔ aur $T = 1/6 \approx 0.1667 (= 0.5/3)$ ✔.

Ex 8 — Real-world word problem with geometry (Cell J)

Worked example Merry-go-round rider

Ek bacha $r = 2 m$ radius par ek merry-go-round par baitha hai jo har $T = 4 s$ mein ek full turn karta hai. $f$ , $ω$ , aur baache ki linear speed $v$ circle ke around nikalo.

Forecast: Baache ki speed m/s mein badi hogi ya angular frequency rad/s mein?

Step 1 — $f = 1/ T = 1/4 = 0.25 Hz$ . Yeh step kyun? Har 4 s mein ek turn ⇒ har second mein quarter turn.

Step 2 — $ω = 2 π / T = 2 π /4 = π /2 \approx 1.571 rad/s$ . Yeh step kyun? Uniform Circular Motion $ω$ ka birthplace hai: ek lap $2 π$ radians sweep karta hai, toh sweep-rate $2 π$ divided by lap time hai. Figure mein blue arc dekhо jis par "angle sweeps at rate omega" likha hai — woh angle $ω$ rate par badh raha hai.

Step 3 — Linear speed $v = ω r = (π /2) (2) = π \approx 3.142 m/s$ . Yeh step kyun? Yahan geometry enter karta hai. Radius $r$ par koi point angle $Δ θ$ se guzar kar arc-length $r Δ θ$ travel karta hai (radian ki definition!). Time se divide karne par $Δ θ / time = ω$ speed $= r ω$ ban jaata hai. Figure mein yellow arrow jis par "v = omega r" likha hai woh yahi tangential velocity hai. Note karo yeh $r$ par depend karta hai, jabki $ω$ nahi karta — bahar ka bacha faster move karta hai lekin wohi angle-rate sweep karta hai.

Forecast resolved: Yahan $v = π \approx 3.14 m/s$ bada hai $ω = π /2 \approx 1.57 rad/s$ se — kyunki radius $r = 2$ (1 se bada) $ω$ ko magnify karke $v$ banata hai. ( $r < 1$ par yeh flip ho jaata, toh answer genuinely $r$ par depend karta hai.)

Verify: Poora circumference $= 2 π r = 4 π m$ , $T = 4 s$ mein cover hota hai, toh $v = 4 π /4 = π m/s$ ✔ — $ω r$ se match karta hai.

Ex 9 — Exam twist: wrong-pairing trap (Cell K)

Worked example Kaun sa formula legal hai?

Ek student se pucha jaata hai: "Given $f = 5 Hz$ , ek classmate likhta hai $T = 2 π f$ aur $ω = 2 π / f$ . Kya yeh sahi hain? Sahi $T$ aur $ω$ do."

Forecast: Apne gut par trust karo — kya $2 π$ $T$ ke saath jaata hai, ya sirf $ω$ ke saath?

Step 1 — Units use karke wrong pairing kill karo. Yeh step kyun? $T$ seconds mein aana chahiye. Factor $2 π$ sirf ek pure number hai (radians dimensionless hote hain), toh $2 π f$ ke units wahi hain jo $f$ ke hain, yaani per-second — yeh ek frequency hai, time nahi. Ek time ka formula jo per-second quantity produce kare, impossible hai, toh $T = 2 π f$ arrive-on-arrival dead hai. $2 π$ sirf $ω$ ke saath isliye jaata hai kyunki $ω$ ko define kiya gaya hai radians per second count karne ke liye, na isliye ki $2 π$ koi unit carry karta ho.

Step 2 — Correct period: $T = 1/ f = 1/5 = 0.2 s$ . Yeh step kyun? $T$ aur $f$ pure reciprocals hain; inhe $2 π$ kabhi touch nahi karta. ("f Tlips T.")

Step 3 — Correct angular frequency: $ω = 2 π f = 2 π (5) = 10 π \approx 31.42 rad/s$ . Yeh step kyun? $ω = 2 π / f$ deta $\approx 1.257$ , jo 5 Hz motion ke liye bahut chhota hai — ek aur sanity red flag ( $1/ f$ bas seconds hai, period ki size, fast radian-rate nahi). Sahi form $2 π$ se multiply karta hai.

Forecast resolved: $2 π$ sirf $ω$ ke saath jaata hai. Classmate ke dono formulas galat the; $T = 1/ f$ aur $ω = 2 π f$ legal pairings hain.

Verify: $ω T = 10 π \times 0.2 = 2 π$ ✔. Student ka $T = 2 π f = 10 π \approx 31.4$ "s" 5 cycles-per-second ke liye absurdly lamba hai — clearly galat, confirmed.

Ex 10 — Sign aur direction: $ω < 0$ ka matlab kya hai? (Cell L)

Worked example Ek negative angular frequency

Ek rotating point $x (t) = A cos (ω t)$ se describe hota hai jahan $ω = - 4 π rad/s$ hai. $f$ aur $T$ kya hain, aur minus sign ka physically kya matlab hai?

Forecast: Kya negative $ω$ se period bhi negative ho jaata hai? Aur kya rhythm $+ 4 π$ oscillator se faster hoti hai ya slower?

Step 1 — Rhythm sirf size $∣ ω ∣$ par depend karti hai. Yeh step kyun? $cos$ ek even function hai: $cos (- θ) = cos (θ)$ . Toh $cos (- 4 π t) = cos (4 π t)$ — position-vs-time graph $+ 4 π$ case se identical hai. Sign kabhi nahi badalta ki wave kitni fast repeat hoti hai; sirf $∣ ω ∣$ karta hai.

Step 2 — $f = ∣ ω ∣/ (2 π) = 4 π / (2 π) = 2 Hz$ . Yeh step kyun? Frequency cycles per second count karti hai — ek count, jo negative nahi ho sakti. Hum magnitude feed karte hain. Isi tarah $T = 2 π /∣ ω ∣ = 2 π / (4 π) = 0.5 s$ ; time-per-cycle kabhi negative nahi hota.

Step 3 — Minus actually kya karta hai. Yeh step kyun? Uniform Circular Motion mein, $ω$ underlying spinning point ki angular velocity hai, aur uska sign direction hai: $ω > 0$ = counter-clockwise, $ω < 0$ = clockwise. Toh $ω = - 4 π$ wohi speed, opposite spin direction hai. Oscillation ka shadow waise hi dikhta hai; uske peeche ka wheel doosri taraf ghoomta hai. Direction sign mein rehta hai, magnitude $∣ ω ∣$ mein.

Forecast resolved: $T$ negative nahi hai — woh $+ 0.5 s$ hai, $∣ ω ∣$ use karke. Aur rhythm wohi speed hai jaise $+ 4 π$ ; minus sirf rotation ki direction flip karta hai, pace nahi.

Verify: $∣ ω ∣ T = 4 π \times 0.5 = 2 π$ ✔, aur $cos (- 4 π (0.5)) = cos (- 2 π) = 1 = cos (2 π)$ — graphs coincide karte hain ✔.

Recall Matrix ke across quick self-test

$T = 0.25$ s → $f$ ? ::: $4$ Hz $ω = 15$ rad/s → $f$ ? ::: $\approx 2.387$ Hz $x = 5 cos (6 π t + π /3)$ → $T$ ? ::: $1/3 \approx 0.333$ s $f = 0$ → $T$ ? ::: infinite / undefined (complete karne ke liye koi cycle nahi) $ω = 0$ cosine → kaun si motion? ::: ek frozen constant, koi oscillation nahi $33 \frac{1}{3}$ rpm → $T$ ? ::: $1.8$ s Triple $f$ → $T$ kya ho jaata hai? ::: one-third Merry-go-round $T = 4$ s, $r = 2$ m → $v$ ? ::: $π \approx 3.14$ m/s Kya $T = 2 π f$ kabhi valid hai? ::: kabhi nahi — units force karte hain $T = 1/ f$ $ω = - 4 π$ → $T$ aur sign ka matlab? ::: $T = 0.5$ s ( $∣ ω ∣$ use karta hai); minus ka matlab clockwise rotation hai, same speed

Connections

ω, T, f relationships — woh parent jise yeh page drill karta hai.
Uniform Circular Motion — Ex 8 ka arc aur $v = ω r$ , aur Ex 10 ka sign-as-direction yahan se aata hai.
Simple Harmonic Motion — Ex 3, Ex 5, Ex 10 $A cos (ω t + ϕ)$ se $ω$ (aur amplitude $A$ ) read karte hain.
Phase and Phase Difference — argument $ω t + ϕ$ aur starting phase $ϕ$ jo humne define kiya.
Wave Speed v = fλ — agla step: is time-rhythm ko spatial wavelength mein convert karo.
Springs and Pendulums — jahan numerical $T$ aur $ω$ physically actually aate hain.