Standing waves — formation, nodes, antinodes: Visual walkthrough

1.6.18 · D2 · Physics › Oscillations & Waves › Standing waves — formation, nodes, antinodes

Kuch bhi shuru karne se pehle, yeh tay kar lete hain ki pictures mein kya show ho raha hai.

Step 1 — Ek travelling wave se milo aur picture se har symbol padho

KYA. Hum ek aisi single wave draw karte hain jo right ki taraf slide karti hai. Iska formula hai $y_{1} = A sin (k x - ω t)$

YEH formula kyun, koi aur kyun nahi? Sine sabse simple smooth "upar-neeche-aur-repeat" shape hai, aur andar ka har symbol apni jagah justify karta hai:

$A$ — amplitude: rope ki sabse badi height. Picture mein yeh flat middle line se crest tak ki distance hai.
$x$ — rope ke saath aap kahan hain (horizontal axis).
$k$ — wave number: aap ek metre mein sine ke kitne radians cross karte ho. Bada $k$ = tightly packed wiggles. Yeh wavelength se $k = 2 π / λ$ se juda hai — dekho Wave number k and wavelength.
$λ$ (wavelength) — ek poori wiggle ki length, crest-to-crest.
$ω$ — angular frequency: poori shape time $t$ mein kitni tezi se sideways march karti hai.
$k x - ω t$ mein minus sign hi ise right ki taraf move karta hai: jaise-jaise $t$ badhta hai, $k x - ω t$ same rakhne ke liye aapko bada $x$ chahiye, to har feature bade $x$ ki taraf shift ho jaata hai.

PICTURE. Blue curve abhi ki rope hai; faded blue curve thodi der baad ki rope hai — right shift ho gayi. Green arrow direction of travel batata hai.

Figure — Standing waves — formation, nodes, antinodes

Step 2 — Doosri mirror twin wave add karo jo doosri taraf aa rahi hai

KYA. Ab ek doosri wave, $A$ , $k$ , $ω$ mein identical, lekin left ki taraf travel kar rahi hai: $y_{2} = A sin (k x + ω t)$

Plus sign kyun? $ω t$ ke aage sign flip karo aur Step 1 ki logic ulti ho jaati hai: jaise-jaise $t$ badhta hai, $k x + ω t$ fixed rakhne ke liye aapko chhota $x$ chahiye, to features left ki taraf slide karti hain. Physically yeh left-mover kahan se aata hai? Usually right-mover ek wall se takraati hai aur seedha wapas bounce ho jaati hai — woh returning wave hi $y_{2}$ hai. Dekho Reflection of waves at boundaries.

PICTURE. Blue = right-mover, coral = left-mover, usi instant par. Same height ( $A$ ), same wiggle-length ( $λ$ ), opposite arrows.

Step 3 — Superpose karo: rope sirf sum feel karti hai

KYA. Rope ke paas ek waqt mein do heights rakhne ka koi tarika nahi. Har $x$ par woh do waves ka sum leta hai: $y = y_{1} + y_{2} = A sin (k x - ω t) + A sin (k x + ω t)$

Bas add kyun? Yeh Superposition principle hai: jab do waves overlap karti hain, actual displacement plain arithmetic sum hota hai, height plus height. Kuch bhi fancy nahi.

PICTURE. Do thin curves (blue, coral) aur unka sum slate mein. Kuch jagahon par note karo ki do curves equal-and-opposite hain, to sum zero ho jaata hai — nodes ka pehla hint.

Step 4 — Trig trick jo space aur time ko alag karta hai

KYA. Hum sum ko rewrite karte hain is identity se: $sin P + sin Q = 2 sin \frac{P + Q}{2} cos \frac{P - Q}{2}$ jahan $P = k x - ω t$ aur $Q = k x + ω t$ .

Yeh identity kyun, brute force kyun nahi? Abhi $x$ aur $t$ ek hi bracket ke andar tangled baithe hain. Jab tak yeh tangled hain, shape time ke saath shift ho sakti hai — yeh travel karti hai. Identity woh exact tool hai jo sines ke sum ko factor karta hai ek product mein. Agar hum $y = (kuch sirf x mein) \times (kuch sirf t mein)$ pa sakein, to $x$ -shape frozen hai aur sirf uska size pulse karta hai. Yahi separation poora game hai.

Dono halves compute karo:

$dono brackets ka average \frac{P + Q}{2} = \frac{( k x - ω t ) + ( k x + ω t )}{2} = k x (ω t cancel ho jaata hai)$

$unka half difference \frac{P - Q}{2} = \frac{( k x - ω t ) - ( k x + ω t )}{2} = - ω t (k x cancel ho jaata hai)$

aur $cos (- ω t) = cos (ω t)$ kyunki cosine symmetric hai ( $t = 0$ par ek mirror). To

$y (x, t) = shape: sirf x par depend karta hai 2 A sin (k x) breathing: sirf t cos (ω t)$

$2 A sin (k x)$ — rope ke saath ek fixed picture draw ki gayi hai; ise $R (x)$ kaho, position $x$ par amplitude.
$cos (ω t)$ — ek single number, ek given instant par har jagah same, jo poori picture ko $+ 1$ aur $- 1$ ke beech upar-neeche scale karta hai.

PICTURE. Left panel: fixed shape $2 A sin (k x)$ . Right panel: breathing multiplier $cos (ω t)$ $\pm 1$ ke beech swing kar raha hai. Real rope = left picture times right number.

Step 5 — Ise breathe karte dekho ("standing" visible ho jaata hai)

KYA. Kai instants freeze karo aur overlay karo. Envelope $R (x) = 2 A sin (k x)$ wahi rehti hai; sirf scale badlta hai jab $cos (ω t)$ $1, 0, - 1, 0, \dots$ se guzarta hai.

Kyun frozen lagta hai. Kyunki $x$ sirf $sin (k x)$ mein rehta hai, crests aur zeros ki locations kabhi move nahi karti. Ek travelling wave apne crests slide karta hai; standing wave crests ko wahan rakhta hai jahan hain aur sirf unhe taller ya shorter karta hai. Isliye ise standing kehte hain.

PICTURE. Dashed grey lines fixed envelope $\pm 2 A sin (k x)$ hain. Coloured curves alag-alag times par rope hain, sab ek hi zero-crossings par pin kiye hue. Crossing points kabhi nahi hilte.

Step 6 — Shape se seedhe nodes padho

KYA. Node ek aisa point hai jo hamesha zero rehta hai, chahe $cos (ω t)$ kuch bhi kare. Uske liye shape khud vanish hona chahiye: $R (x) = 2 A sin (k x) = 0 \Rightarrow sin (k x) = 0$

$sin (k x) = 0$ kyun, cosine kyun nahi? $cos (ω t)$ time ke saath badlta hai — yeh sirf ek instant ke liye zero hota hai, phir wapas swing karta hai. Ek node hamesha ke liye still hona chahiye, isliye position-part zero hona chahiye. Sine har poore multiple of $π$ par zero hit karta hai: $k x = nπ \Rightarrow x = \frac{nπ}{k} = \frac{nλ}{2}, n = 0, 1, 2, \dots$ $k = 2 π / λ$ use karte hue. To nodes har $λ /2$ par hain.

$n$ — kaun sa zero-crossing hai yeh count karta hai (origin par 0th, phir 1st, 2nd, …).
Ek node se dusre tak ka step $λ /2$ hai — kyunki sine ek poore wavelength mein do baar zero cross karta hai.

PICTURE. Node dots ke saath envelope; spacing bracket $λ /2$ read karta hai.

Step 7 — Antinodes padho, aur $sin$ ke har quadrant ko check karo

KYA. Antinode woh jagah hai jahan shape sabse badi hoti hai, to rope sabse zyada swing karti hai ( $\pm 2 A$ ): $∣ R (x) ∣ = 2 A \Rightarrow ∣ sin (k x) ∣ = 1$

$∣ sin (k x) ∣ = 1$ kyun? Amplitude $2 A$ sirf tabhi reach hota hai jab $sin (k x)$ apni extremes $+ 1$ ya $- 1$ hit kare — aur hum dono signs include karne chahiye, kyunki ek trough ( $- 2 A$ ) utni hi violently swing karti hai jitni ek crest. Sine odd multiples of $π /2$ par $\pm 1$ reach karta hai: $k x = (n + \frac{1}{2}) π \Rightarrow x = (n + \frac{1}{2}) \frac{λ}{2} = (2 n + 1) \frac{λ}{4}$

$sin (k x)$ ke saare cases cover karte hue jab $x$ rope ke saath chalta hai:

$sin (k x) = 0$ at $k x = 0, π, 2 π, \dots$ → nodes.
$sin (k x) = + 1$ at $k x = π /2, 5 π /2, \dots$ → antinode, rope ek crest par.
$sin (k x) = - 1$ at $k x = 3 π /2, 7 π /2, \dots$ → antinode, rope ek trough par (phir bhi antinode hai!).
Beech mein har jagah: partial amplitude $0 < ∣ R ∣ < 2 A$ .

To node → nearest antinode quarter wavelength hai, $λ /4$ (node spacing ka aadha).

PICTURE. Node dots aur antinode stars ke saath poora envelope; brackets $λ /2$ (node-node) aur $λ /4$ (node-antinode) dikhate hain; ek $- 1$ antinode bhi starred hai yeh prove karne ke liye ki troughs count hote hain.

Step 8 — Degenerate case: agar waves har jagah cancel ho jaayein to?

KYA. Honest edge question poochho: kya poori rope dead flat ho sakti hai? Yeh hota agar hum add karne ki jagah do waves lete jo perfect opposites hain, $sin (k x) \equiv 0$ -style cancellation dete, ya agar hum $cos (ω t) = 0$ ke instant par dekhein.

Kyun matter karta hai. Har standing wave ke andar do limiting moments rehte hain:

Jab $cos (ω t) = 0$ (ek quarter-period andar), har point momentarily $y = 0$ par hota hai — poori rope ek instant ke liye flat hai. Lekin yeh dead nahi hai: yahan speed sabse fast hai (antinodes par maximum speed), saari energy kinetic hai. Yeh node jaisa nahi hai — node hamesha flat rehta hai, yeh ek instant ke liye flat hai.
Jab $cos (ω t) = \pm 1$ , rope poori tarah apni envelope shape mein stretched hai, momentarily still — saari energy potential hai.

To "flat" ka matlab do bilkul alag cheezein ho sakti hain, aur farq yeh hai ki hamesha vs ek instant ke liye.

PICTURE. Top: $cos (ω t) = 1$ par rope (max stretch, still, arrows zero velocity show karte hain, saari PE). Bottom: $cos (ω t) = 0$ par rope (flat, lekin antinodes par velocity arrows bade hain, saari KE). Dono pictures par nodes marked hain — dono pictures mein flat, yahi unhe nodes banata hai.

Ek-picture summary

Upar sab kuch compress kiya: do travelling arrows (top) superposition ko feed karte hain, trig identity ise shape × breathing mein factor karti hai, aur shape aapko nodes ( $λ /2$ apart) aur antinodes ( $λ /4$ har node se) deta hai.

Recall Feynman retelling — poora walkthrough plain words mein

Socho ek wiggle ek rope par right ki taraf daud rahi hai, aur uski exact twin left ki taraf. Jahan bhi tum khade ho, rope sirf ek height par ho sakti hai, to woh dono heights leta hai aur unhe add karta hai. Add ko carefully karo aur trig magic ka ek chota sa tukda hota hai: answer saaf-saaf split ho jaata hai "rope ke saath draw ki gayi ek fixed shape" times "ek single number jo time mein pulse karta hai." Kyunki shape mein koi time nahi hai, uske crests aur still-spots kabhi sideways nahi hilte — poori cheez bas taller aur shorter hoti hai jagah par, jaise saans lena. Still-spots, jahan shape khud zero hai, nodes hain; yeh aadhe wavelength apart hote hain kyunki sine ek wavelength mein do baar zero se guzarta hai. Kisi bhi do nodes ke beech mein shape poori stretch par hoti hai — yeh antinodes hain, sabse zyada swing kar rahe hain, nearest node se quarter wavelength door. Aur "flat rope" mein ek trick chhipi hai: node hamesha flat hota hai, lekin ek aisa bhi instant hota hai jab poori rope flat hoti hai — dead nahi, balki top speed par middle se fly kar rahi hoti hai. Yahi hai standing wave, shuru se ant tak.

Recall Derivation ko memory se rebuild karo

Do waves se shuru karo? ::: $A sin (k x - ω t)$ (right) aur $A sin (k x + ω t)$ (left). Kaunsa law unhe add karne deta hai? ::: Superposition — displacements bas sum ho jaate hain. Kaunsi identity $x$ ko $t$ se alag karti hai? ::: $sin P + sin Q = 2 sin \frac{P + Q}{2} cos \frac{P - Q}{2}$ . Result? ::: $y = 2 A sin (k x) cos (ω t)$ . Node condition aur spacing? ::: $sin (k x) = 0$ , nodes har $λ /2$ par. Antinode condition aur node-to-antinode gap? ::: $∣ sin (k x) ∣ = 1$ , gap $λ /4$ .

Connections

Superposition principle — Step 3 mein use kiya "bas add karo" wala law.
Travelling waves — Steps 1–2 ke do ingredients.
Reflection of waves at boundaries — left-mover kahan se aata hai.
Wave number k and wavelength — $k = 2 π / λ$ link jo Steps 6–7 mein use hua.
Resonance and normal modes — kaun si wavelengths actually bounded string par survive karti hain.
Waves on a string / Sound in pipes — real systems jinmein yeh pattern rehta hai.