Wave parameters — amplitude, wavelength, frequency, period, wave speed: Worked examples

1.6.14 · D3 · Physics › Oscillations & Waves › Wave parameters — amplitude, wavelength, frequency, period,

Yahan use kiye gaye har quantity parent note mein already explain ho chuke hain. Ek line ka reminder, taaki koi symbol unexplained na lage:

Scenario matrix

Har wave-parameter problem inhi cells mein se ek hoti hai. Neeche ke examples tagged hain un cell(s) ke saath jo woh cover karte hain.

#	Cell (case class)	Isme kya tricky hai	Example
C1	Direct — given $f, λ$ → find $v$	bas plug in karo	Ex 1
C2	Inverse — given $v, f$ → find $λ$	rearrange karo	Ex 2
C3	Same medium, change $f$	$v$ is FIXED, $λ$ moves	Ex 3
C4	Amplitude from a swing	peak-to-peak $= 2 A$ , aadha karna zaroori	Ex 4
C5	Period ↔ frequency reciprocal	$f = 1/ T$ , chhote/bade numbers dhyan se	Ex 4, Ex 5
C6	Unit-prefix trap (kHz, MHz, cm, nm)	plug in karne se pehle convert karo	Ex 6, Ex 7
C7	Two-route cross-check ( $v T$ vs $v / f$ )	do tarike se answer confirm karo	Ex 5
C8	Real-world word problem	kahani mein se numbers nikalho	Ex 8
C9	Degenerate / limiting inputs ( $f \to 0$ , $λ \to \infty$ , $f \to \infty$ )	kya toot ta hai, kya finite rehta hai	Ex 9
C10	Exam twist — reading period off a graph, changing medium	$v$ does change when medium changes	Ex 10

Intuition "Negative" ya "quadrant" cells kyun nahi hain

Ek angle ke unlike (jo charon quadrants mein point kar sakta hai aur sign carry karta hai), paanch wave parameters sab non-negative magnitudes hain — tumhare paas $- 2 m$ wavelength ya $- 5 Hz$ frequency nahi ho sakti. Toh is matrix mein koi signed ya quadrant case nahi hai. Do edge-of-the-sign-line situations phir bhi dhyan deserve karti hain, aur hum unhe explicitly cover karte hain:

Zero values — jaise $f \to 0$ (koi wave hi nahi): cell C9 / Example 9 mein handle kiya.
Negative displacement: ek particle ka displacement $y$ negative ho sakta hai (beech wali line ke neeche) — yeh ek phase inversion hai, naam dena zaroori hai. Lekin woh $y$ par sign hai, na $A, λ, T, f, v$ par. Amplitude $A$ = maximum $∣ y ∣$ , hamesha $\geq 0$ . Hum yeh Example 9 ke baad flag karte hain taaki koi reader na sooche "minus signs kahan gaye."

Example 1 — Direct (cell C1)

Worked example Given

$f$ aur $λ$ , find $v$ Ek guitar string $f = 200 Hz$ par vibrate karti hai aur wavelength $λ = 1.6 m$ hai. Wave speed $v$ nikalo.

Forecast: har second ke $200$ bobs mein se har ek $1.6 m$ lamba crest launch karta hai, toh guess: kuch sau m/s.

Master equation likho. $v = f λ$ . Yeh step kyun? Hume $v$ chahiye; humare paas woh do cheezein hain ( $f, λ$ ) jo milake $v$ deti hain. Yeh wahi ek equation hai jo inhe teeno ko ek saath jodte hain.
Numbers substitute karo. $v = 200 \times 1.6$ . Yeh step kyun? Symbols ki jagah given values rakhte hain taaki unknown $v$ ek simple arithmetic product ban jaaye jise hum evaluate kar sakein.
Compute karo. $v = 320 m/s$ . Yeh step kyun? Multiply karne se expression single final number ban jaata hai jo question maang raha hai.

Verify: units hain $Hz \times m = s^{- 1} \cdot m = m/s$ ✓. Ek second mein, $200$ crests mein se har ek $1.6 m$ move karta hai: $200 \times 1.6 = 320 m$ per second ✓.

Example 2 — Inverse (cell C2)

Example 3 — Same medium, frequency change (cell C3)

Worked example "

$v$ is fixed" wala trap Example 1 ki usi string par ( $v = 320 m/s$ ), ab tum $f = 800 Hz$ par ek harmonic bajate ho. Naya $λ$ nikalo.

Forecast: humne $f$ chaar guna bada kiya. Medium nahi badla, toh $v$ nahi badal sakta → $λ$ ek-chauthai ho jaayega: $1.6$ se lagbhag $0.4 m$ .

State karo kya fixed hai. Same string ⇒ same tension aur density ⇒ $v = 320 m/s$ unchanged. Yeh step kyun? Kyunki speed medium se set hoti hai, na isko drive karne ki speed se. Dekho Speed of Waves on a String. Yahi is problem ka poora trick hai.
$λ$ ke liye rearrange karo. $λ = \frac{v}{f} = \frac{320}{800}$ . Yeh step kyun? Hume $λ$ chahiye aur ab humare paas $v$ (fixed) aur naya $f$ dono hain, toh $λ$ isolate karke values daalo.
Compute karo. $λ = 0.40 m$ . Yeh step kyun? Division karne se higher harmonic ka naya wavelength milta hai.

Verify: ratio check — $f$ ×4 hua, toh $λ$ ÷4 hona chahiye: $1.6/4 = 0.40$ ✓. Aur $f λ = 800 \times 0.40 = 320 = v$ ✓.

Figure dekho: dono curves ek hi box mein fit hain (same $v$ -per-second travel), lekin burnt-orange high-frequency wave zyada crests thoos deta hai, toh har crest chota hai. Same speed, different $λ$ .

Example 4 — Amplitude from a swing, plus reciprocal (cells C4, C5)

Worked example Peak-to-peak aur period

Samudra par ek cork kul $0.90 m$ vertical swing mein move karta hai, aur ek poora upar-neeche karne mein $T = 3.0 s$ leta hai. Amplitude $A$ aur frequency $f$ nikalo.

Forecast: amplitude aadhi hogi poori swing ki (classic galti hai halve na karna) → lagbhag $0.45 m$ . Aur slow $3 s$ bob matlab chhoti frequency, $1 Hz$ se kaafi kam.

Swing ko aadha karo. Poori swing top-to-bottom peak-to-peak $= 2 A$ hai, toh $A = \frac{0.90}{2} = 0.45 m$ . Yeh step kyun? Amplitude middle line se measure hoti hai, tip to tip se nahi — cork shant surface se $A$ upar uthta hai aur $A$ neeche jaata hai. Middle-line picture ke liye Simple Harmonic Motion dekho.
Period ko invert karo. $f = \frac{1}{T} = \frac{1}{3.0}$ . Yeh step kyun? Frequency matlab "bobs per second"; agar ek bob $T$ seconds leta hai toh ek second mein $1/ T$ bobs aate hain.
Compute karo. $f = 0.33 Hz$ (2 s.f. tak; exactly $\frac{1}{3} Hz$ ). Yeh step kyun? $1/3.0$ evaluate karne se numerical frequency milti hai; exact $\frac{1}{3}$ yaad rakho taaki neeche verify clean ho.

Verify: $A$ poori swing se chhoti hai ✓ (hona hi chahiye, kyunki $A = \frac{1}{2} \times$ swing). Aur $f \cdot T = \frac{1}{3} \times 3 = 1$ ✓ — exactly ek cycle per period, jaisa hona chahiye.

Figure mein plum peak-to-peak span ( $0.90 m$ ) aur teal amplitude ( $0.45 m$ ) dashed middle line se mark hai — ek literally doosre ka double hai.

Example 5 — Two-route cross-check (cells C5, C7)

Worked example

$T$ aur $v$ se $λ$ do tareekon se nikalo Ek water wave ka period $T = 0.50 s$ aur speed $v = 6.0 m/s$ hai. $λ$ dono $λ = v T$ aur $λ = v / f$ se nikalo, aur check karo ki agree karte hain.

Forecast: aadhe second mein crest $6 m$ ka aadha move karta hai → lagbhag $3 m$ .

Route A: ek period mein distance. Ek period mein crest exactly ek wavelength aage badhta hai, toh $λ = v T = 6.0 \times 0.50 = 3.0 m$ . Yeh step kyun? Yeh definition-level route hai: distance $=$ speed $\times$ time, aur ek wavelength ka time $T$ hota hai.
Route B: frequency ke zariye. Pehle $f = 1/ T = 1/0.50 = 2.0 Hz$ , phir $λ = v / f = 6.0/2.0 = 3.0 m$ . Yeh step kyun? Prove karne ke liye ki dono forms same identity hain, kyunki $1/ f = T$ .
Compare karo. Dono se $3.0 m$ milta hai. Yeh step kyun? Do independent routes ka agree karna hamara confidence check hai ki koi algebra slip nahi hua.

Verify: dono routes algebraically identical hain: $v T = v \cdot \frac{1}{f} = v / f$ ✓. Numeric match $3.0 = 3.0$ ✓.

Example 6 — Unit-prefix trap: radio (cell C6)

Worked example Megahertz aur speed of light

Ek FM radio station $f = 100 MHz$ par broadcast karta hai. Radio waves electromagnetic hain, toh yeh speed of light $v = 3.0 \times 1 0^{8} m/s$ par travel karti hain (dekho Electromagnetic Spectrum). $λ$ nikalo.

Forecast: enormous speed, high frequency — FM ke metre-scale wavelengths famous hain ("kuch metres"), toh guess around $3 m$ .

Pehle prefix convert karo. $100 MHz = 100 \times 1 0^{6} Hz = 1.0 \times 1 0^{8} Hz$ . Yeh step kyun? Equation sirf base SI units mein kaam karti hai; "M" matlab mega $= 1 0^{6}$ . Yeh bhoolna #1 galti hai.
$λ$ ke liye rearrange karo aur substitute karo. $λ = \frac{v}{f} = \frac{3.0 \times 1 0 ^{8}}{1.0 \times 1 0 ^{8}}$ . Yeh step kyun? $λ$ chahiye aur $v, f$ hamare paas hain, toh $λ$ isolate karo; numbers daalne se single division ban jaata hai.
Compute karo. $λ = 3.0 m$ . Yeh step kyun? Powers of ten cancel ho jaate hain, simple metre answer milta hai jo question chahta hai.

Verify: $f λ = (1.0 \times 1 0^{8}) (3.0) = 3.0 \times 1 0^{8} = v$ ✓. Forecast se match ✓.

Example 7 — Unit-prefix trap: light (cell C6)

Worked example Nanometres aur colour

Green light ka wavelength vacuum mein $λ = 500 nm$ hai, $v = 3.0 \times 1 0^{8} m/s$ ki speed se travel karta hai. Iska frequency $f$ nikalo.

Forecast: light waves inanely fast repeaters hain — expect a huge number, around $1 0^{14} Hz$ .

Nanometres convert karo. $500 nm = 500 \times 1 0^{- 9} m = 5.0 \times 1 0^{- 7} m$ . Yeh step kyun? "n" (nano) $= 1 0^{- 9}$ ; equation ko metres chahiye.
$f$ ke liye rearrange karo aur substitute karo. $f = \frac{v}{λ} = \frac{3.0 \times 1 0 ^{8}}{5.0 \times 1 0 ^{- 7}}$ . Yeh step kyun? Humare paas $v$ aur $λ$ hain aur $f$ chahiye, toh divide karo; values daalne se single arithmetic operation ban jaata hai.
Compute karo. $f = 6.0 \times 1 0^{14} Hz$ . Yeh step kyun? Mantissas divide karne aur exponents subtract karne se ( $8 - (- 7) = 15$ , $3/5 = 0.6$ se adjust) final frequency milti hai.

Verify: $f λ = (6.0 \times 1 0^{14}) (5.0 \times 1 0^{- 7}) = 3.0 \times 1 0^{8} = v$ ✓. Forecast ke order of magnitude par aaya ✓.

Example 8 — Real-world word problem (cell C8)

Worked example Pier par waves count karna

Ek pier par khade hokar, tum ek post ke paas se $30 s$ mein exactly $12$ wave crests guzarte dekho. Crests $8.0 m$ ki doori par hain. Waves kitni tezi se move kar rahi hain?

Forecast: roughly har kuch seconds mein ek crest, har ek $8 m$ lamba → kuch m/s.

Kahani se frequency nikalo. $30 s$ mein $12$ crests matlab $f = \frac{12}{30} = 0.40 Hz$ . Yeh step kyun? "Crests per second" hi frequency hai; kahani humein count aur time deti hai.
Wavelength identify karo. Crest spacing = ek poora space-repeat = $λ = 8.0 m$ . Yeh step kyun? Crest-to-crest distance $λ$ ki definition hai.
Master equation apply karo. $v = f λ = 0.40 \times 8.0 = 3.2 m/s$ . Yeh step kyun? Ab $f$ aur $λ$ dono pin ho gaye hain, toh $v = f λ$ woh speed deta hai jo question chahta hai.

Verify: period $T = 1/ f = 2.5 s$ per crest; $2.5 s$ mein crest $λ = 8.0 m$ move karta hai, giving $8.0/2.5 = 3.2 m/s$ ✓. Forecast se match ✓.

Example 9 — Degenerate aur limiting inputs (cell C9)

Worked example Extremes par kya hota hai

Ek fixed-medium wave ke liye $v = 20 m/s$ , har limit examine karo aur batao $λ$ ya $T$ ka kya hota hai. (a) $f \to 0$ (kabhi wiggle nahi karta). (b) $f \to \infty$ (infinitely fast wiggling). (c) $A \to 0$ .

Forecast: slow shaking → lambi waves; fast shaking → chhoti waves; zero amplitude → koi wave nahi, chahe maths for $λ$ "kaam" kare.

Limit (a): $f \to 0$ . $λ = v / f = 20/ f \to \infty$ ; equivalently $T = 1/ f \to \infty$ . Yeh step kyun? Agar source kabhi ek cycle complete nahi karta, "agla crest" infinitely door hai — pattern bina kisi seema ke stretch karta hai. Physically: koi wave nahi (flat medium).
Limit (b): $f \to \infty$ . $λ = 20/ f \to 0$ aur $T = 1/ f \to 0$ . Yeh step kyun? Frequency badhao aur, kyunki $v$ medium se pin hai, sirf free quantity $λ$ zero ki taraf squeeze hoti hai. Real media yeh eventually tod dete hain (particle spacing ek floor set karta hai), lekin ideal formula mein $λ \to 0$ .
Limit (c): $A \to 0$ . $A$ kisi bhi $v = f λ$ , $f = 1/ T$ mein appear nahi karta. Toh $v, f, λ, T$ sab unaffected rehte hain — lekin wave zero displacement carry karti hai, yaani physical disturbance ke roop mein woh gayab ho jaati hai. Yeh step kyun? Yeh dikhane ke liye ki amplitude ek independent knob hai: yeh wave ka size/energy set karta hai, kabhi speed ya spacing nahi.

Verify: $λ \cdot f = 20$ har finite $f$ ke liye hold karta hai (jaise $f = 0.001 \Rightarrow λ = 20000$ ; $f = 1 0^{6} \Rightarrow λ = 2 \times 1 0^{- 5}$ ) ✓. Amplitude dono relations se absent ✓.

Example 10 — Exam twist: graph se period padho, phir medium change karo (cell C10)

Worked example Graph-reading + ek real speed change

Ek particle ka displacement-vs-time graph dikhata hai ki woh har $0.20 s$ mein same state par wapas aata hai, aur wave (medium 1 mein) ki speed $v_{1} = 40 m/s$ hai. (a) $f$ aur $λ_{1}$ nikalo. (b) Wave ab medium 2 mein jaati hai jahan speed $v_{2} = 60 m/s$ ho jaati hai. Frequency same rehti hai (source ise set karta hai). $λ_{2}$ nikalo.

Forecast: same frequency par tezi medium → lamba wavelength, toh $λ_{2} > λ_{1}$ .

Part (a).

Graph se $T$ padho. Pattern har $0.20 s$ mein repeat hota hai, toh $T = 0.20 s$ . Yeh step kyun? Ek displacement–time graph har period mein repeat hota hai (horizontal axis time hai), kabhi wavelength mein nahi. Dono axes confuse karna classic exam trap hai.
$T$ se $f$ nikalo. $f = \frac{1}{T} = \frac{1}{0.20} = 5.0 Hz$ . Yeh step kyun? Frequency period ka reciprocal hai; read-off time invert karne se bobs per second milte hain.
Medium 1 mein wavelength. $λ_{1} = \frac{v _{1}}{f} = \frac{40}{5.0} = 8.0 m$ . Yeh step kyun? Ab humare paas $v_{1}$ aur $f$ dono hain, toh $λ = v / f$ space-repeat length deta hai; values daalo aur divide karo.

Part (b). 4. Decide karo kya rehta hai aur kya badlta hai. Source same rate par beat karta rehta hai, toh $f = 5.0 Hz$ unchanged hai; lekin humne medium badla, toh $v$ sach mein $v_{2} = 60 m/s$ ho jaata hai. Yeh step kyun? Example 3 se contrast karo: wahan humne medium rakha aur $v$ fixed tha jabki $f$ badla. Yahan hum medium swap karte hain, toh $v$ badlta hai jabki $f$ fixed rehta hai. Yeh "kya fixed hai?" call sahi karna is twist ka poora point hai. 5. Medium 2 mein wavelength. $λ_{2} = \frac{v _{2}}{f} = \frac{60}{5.0} = 12 m$ . Yeh step kyun? Same relation $λ = v / f$ , ab naye speed aur unchanged frequency ke saath; divide karne se naya wavelength milta hai.

Verify: $f$ dono media mein identical hai ✓. Ratio check: $\frac{λ _{2}}{λ _{1}} = \frac{12}{8} = 1.5 = \frac{v _{2}}{v _{1}} = \frac{60}{40}$ ✓ — wavelength fixed frequency par speed ke saath scale karta hai, forecast se match karta hai ki $λ_{2} > λ_{1}$ . Related case ke liye jahan $f$ (medium nahi) badlta hai, dekho Doppler Effect.

Active recall

Recall Which cell is which?

Match karo: "30 s mein 12 crests" ::: word problem, C8 (count se $f$ extract karo). Match karo: "500 nm green light, find f" ::: unit-prefix trap, C6/C7. Match karo: "same string, quadruple f" ::: same-medium, $v$ fixed, C3. Match karo: "wave tezi medium mein jaati hai" ::: exam twist, $v$ badlta hai, $f$ fixed, C10. Match karo: "0.90 m total swing" ::: amplitude-from-swing, aadha karo, C4.

Recall Self-test

Ek displacement–time graph har 0.25 s mein repeat hota hai — kya yeh $λ$ hai ya $T$ ? ::: $T$ (period), kyunki axis time hai. Fixed medium, $f$ teen guna — $λ$ ka kya hoga? ::: Yeh ek-tehaai ho jaayega ( $λ = v / f$ ). Kya $A$ $v = f λ$ mein appear karta hai? ::: Nahi — amplitude speed, frequency aur wavelength se independent hai. Wave slower medium mein same $f$ par jaati hai — kya $λ$ badhega ya ghataega? ::: Ghataega, kyunki $λ = v / f$ aur $v$ ghat gaya. Kya $A, λ, T, f, v$ mein se koi negative ho sakta hai? ::: Nahi — yeh non-negative magnitudes hain; sirf displacement $y$ negative ja sakta hai (phase inversion).

Connections

Wave parameters — amplitude, wavelength, frequency, period, wave speed — woh parent note jinhe yeh examples drill karte hain.
Simple Harmonic Motion — amplitude ke peeche middle-line picture (Ex 4).
Speed of Waves on a String — kyun $v$ medium-set hai (Ex 3).
Sound Waves — $340 m/s$ air ki value (Ex 2).
Electromagnetic Spectrum — Ex 6–7 mein speed of light.
The Wave Equation y(x,t) — in sab parameters ko $y = A sin (k x - ω t)$ mein package karta hai, jahan negative $y$ (phase inversion) rehti hai.
Doppler Effect — woh case jahan $f$ khud badlta hai, Ex 10 ko complement karta hai.