Standing waves — formation, nodes, antinodes: Worked examples

1.6.18 · D3 · Physics › Oscillations & Waves › Standing waves — formation, nodes, antinodes

Shuru karne se pehle, teen reminders jo poori journey mein kaam aayenge:

$k$ wave number hai: yeh metre per metre mein wave ke radians count karta hai. Yeh wavelength se $k = \frac{2 π}{λ}$ ke zariye juda hai, isliye $λ = \frac{2 π}{k}$ . (Dekho Wave number k and wavelength.)
$ω$ angular frequency hai: breathing ke radians per second. Yeh ordinary frequency $f$ se $ω = 2 π f$ ke zariye juda hai, aur period $T$ se $ω = \frac{2 π}{T}$ ke zariye.
$v$ wave speed hai: ek travelling wave ka crest kitne metres per second aage badhta hai. Yeh medium se fix hoti hai (string ka tension aur mass, sound ke liye air). Ise compute karne ke do tarike hain: $v = \frac{ω}{k}$ (wave equation se) ya $v = f λ$ (frequency times wavelength). Dono agree karne chahiye.

$k$ aur $ω$ dono ko apni given equation ko template $y = 2 A sin (k x) cos (ω t)$ se compare karke nikala jaata hai: sine ke andar $x$ ke saath jo multiply ho raha hai woh $k$ hai; cosine ke andar $t$ ke saath jo multiply ho raha hai woh $ω$ hai.

Scenario matrix

Har standing-wave problem in case classes mein se ek (ya blend) hoti hai. Har row ek alag cheez hai jo galat ho sakti hai ya surprise kar sakti hai; neeche ke examples har cell ko cover karte hain.

Cell	Case class	Isme kya tricky hai	Example
C1	Diye gaye standing wave se $k, ω, A$ padhna	template se matching; $λ, T, f$	Ex 1
C2	Do travelling waves se standing wave banana	sum-to-product, sahi final $A$	Ex 2
C3	Node/antinode positions (indexing $n$ )	off-by-one; "origin se 3rd"	Ex 3
C4	Zero / degenerate node at $x = 0$	$n = 0$ node origin par baitta hai; har setup mein nahi hota	Ex 4
C5	Cosine-in- $x$ standing wave (free end)	node/antinode conditions swap	Ex 5
C6	Limiting / snapshot in time	jab $t$ par $cos ω t = 0$ hota hai to poori string flat ho jaati hai	Ex 6
C7	Boundary-fit / normal mode (real-world)	sirf kuch khaas $λ$ hi length $L$ mein fit hoti hai	Ex 7
C8	Exam twist: ek point ki velocity, node par energy	node mein max strain hoti hai, antinode mein max speed	Ex 8

Example 1 — Cell C1: equation se sab kuch padhna

Step 1. Template $y = 2 A sin (k x) cos (ω t)$ se compare karo. Toh $2 A = 0.02 \Rightarrow A = 0.01 m$ ; $k = 5 π rad/m$ ; $ω = 200 π rad/s$ . Yeh step kyun? Kyunki template ne shape aur breathing ko pehle se alag kar diya hai, har coefficient exactly ek jagah baitta hai — matching sirf pattern recognition hai, koi algebra nahi.

Step 2. $k = \frac{2 π}{λ} \Rightarrow λ = \frac{2 π}{k} = \frac{2 π}{5 π} = 0.4 m$ se $λ$ nikalo. Yeh step kyun? "Ek poora spatial repeat kitna lamba hai?" yeh sawaal exactly wahi hai jo $λ$ answer karta hai, aur $k$ woh ek cheez hai jo yeh info store karta hai.

Step 3. $ω = \frac{2 π}{T}$ se $T$ aur $f$ nikalo: $T = \frac{2 π}{200 π} = 0.01 s$ , aur $f = \frac{1}{T} = 100 Hz$ . Yeh step kyun? $ω$ timing information ka ek maatra carrier hai; $T$ "ek breath kitni lambi hai" ka jawab deta hai.

Verify: $A = 0.01$ , $0.02$ ka aadha hai ✓ (forecast sahi nikla). $λ = 0.4 m < 0.5 m$ ✓. Units check: ratio $\frac{[ ω ]}{[ k ]} = \frac{rad/s}{rad/m} = m/s$ ek speed hai — aur actually woh ratio wave speed $v$ hai (upar define ki gayi), $v = \frac{ω}{k} = \frac{200 π}{5 π} = 40 m/s$ . Doosre tarike se cross-check karo: $v = f λ = 100 \times 0.4 = 40 m/s$ ✓ — dono routes agree karte hain, toh hamare $k, ω$ consistent hain.

Example 2 — Cell C2: do travelling waves ko superpose karna

Step 1. Unhe add karo aur $sin P + sin Q = 2 sin \frac{P + Q}{2} cos \frac{P - Q}{2}$ apply karo (yeh Superposition principle plus ek trig identity hai): $P = 2 x - 6 t, Q = 2 x + 6 t$ ke saath, hame $\frac{P + Q}{2} = 2 x$ aur $\frac{P - Q}{2} = - 6 t$ milta hai. $y = 2 (0.3) sin (2 x) cos (- 6 t) = 0.6 sin (2 x) cos (6 t)$ Yeh step kyun? $cos (- θ) = cos θ$ use karke sign theek ho jaata hai; sum-to-product woh ek hi identity hai jo " $x$ aur $t$ ek bracket mein tangled" ko " $x$ akela times $t$ akela" mein badal sakti hai.

Step 2. Max amplitude front number hai: $2 A = 0.6 m$ . Yeh step kyun? $∣ sin (2 x) ∣$ 1 tak pahunch sakta hai, toh ek point ko kabhi bhi milne wala sabse bada swing $0.6$ hai.

Step 3. $k = 2 \Rightarrow λ = \frac{2 π}{2} = π m$ .

Verify: Yeh Travelling waves se aaye the, dono $0.3$ amplitude ke; antinodes par do ka stack hokar $0.6$ hona exactly $2 A$ hai ✓. $t$ -part pure $cos (6 t)$ hai jisme andar koi $x$ nahi — space aur time separate hain ✓.

Example 3 — Cell C3: node & antinode positions (indexing carefully)

Figure — Standing waves — formation, nodes, antinodes

Step 1. Nodes: $sin (2 x) = 0 \Rightarrow 2 x = nπ \Rightarrow x = \frac{nπ}{2}, n = 0, 1, 2, \dots$ Yeh step kyun? Node matlab "kabhi nahi hilta," yaani shape factor zero hai, aur $sin$ exactly $π$ ke integer multiples par zero hota hai.

Step 2. Unhe index karo: $n = 0 \to x = 0$ (origin khud, Figure 1 mein hollow dot), $n = 1 \to x = π /2$ , $n = 2 \to x = π$ . Origin ke baad 2nd node $n = 2$ hai: $x = π \approx 3.14 m$ . Yeh step kyun? "Origin ke baad" ka matlab hai $n = 0$ ko unme se count nahi karte — yeh wahi classic off-by-one trap hai jiske baare mein figure mein hollow dot warn karta hai.

Step 3. Antinodes: $∣ sin (2 x) ∣ = 1 \Rightarrow 2 x = \frac{π}{2} \Rightarrow x = \frac{π}{4} \approx 0.785 m$ (1st wala — Figure 1 mein orange square). Yeh step kyun? Max swing ke liye shape factor $\pm 1$ par chahiye; $sin$ pehli baar $π /2$ par $1$ hit karta hai.

Verify: Node(0) $\to$ antinode( $π /4$ ) $= π /4 = λ /4$ kyunki $λ = π$ hai ✓. Node $\to$ node $= π - π /2 = π /2 = λ /2$ ✓. Forecast aur Figure 1 mein draw ki gayi spacings se match karta hai.

Example 4 — Cell C4: origin par degenerate node

Step 1. (a) $R (0) = 2 A sin (k \cdot 0) = 2 A sin 0 = 0$ . Toh $x = 0$ sine form ke liye ek node hai. Yeh step kyun? Origin par evaluate ki gayi amplitude function decide karti hai — koi motion nahi matlab node.

Step 2. (b) Ek fixed end kabhi nahi hil sakta: physically uska displacement clamped to zero hai. Toh boundary demand karti hai $R (0) = 0$ . Sine form automatically ise satisfy karta hai — isliye ek fixed end par reflection (dekho Reflection of waves at boundaries) ek $sin (k x)$ standing wave produce karta hai, cosine wali nahi. Yeh step kyun? Hum maths ko ek real constraint se match kar rahe hain; boundary form choose karti hai, ulta nahi.

Step 3. Degenerate check: agar $A = 0$ (koi wave hi nahi) ho, toh $R (x) = 0$ har jagah — har point trivially ek "node" hai, lekin koi wave nahi hai. Yeh degenerate limit hai.

Verify: $sin 0 = 0$ exactly ✓. Aur ek fixed end jahan $R (0) \neq = 0$ ho woh "clamped" ko contradict karega, toh sine form ek hi consistent form hai ✓. $A = 0$ case flat, wave-free string deta hai ✓.

Example 5 — Cell C5: cosine form (free end) — conditions swap ho jaati hain

Step 1. Antinodes (max swing): $∣ cos (4 x) ∣ = 1 \Rightarrow 4 x = nπ \Rightarrow x = \frac{nπ}{4}, n = 0, 1, 2, \dots$ $n = 0$ par candidate $x = 0$ hai — free end — lekin domain $x > 0$ hai, toh hum ise discard karte hain aur $n = 1$ lete hain: $x > 0$ ke saath pehla antinode $x = \frac{π}{4} \approx 0.785 m$ hai. Yeh step kyun? $cos$ , $π$ ke integer multiples par $\pm 1$ hit karta hai — wahan cosine extreme hota hai, sine ki tarah nahi; aur restriction " $x > 0$ " ek genuine constraint hai jo $n = 0$ endpoint ko rule out karti hai.

Step 2. Nodes ( $R = 0$ ): $cos (4 x) = 0 \Rightarrow 4 x = \frac{π}{2} + nπ \Rightarrow x = \frac{π}{8} + \frac{nπ}{4}$ . $x > 0$ ke saath pehla node ( $n = 0$ ): $x = \frac{π}{8} \approx 0.393 m$ . Yeh step kyun? Cosine apne peak ke quarter-period baad zero cross karta hai, toh node, free-end antinode se $λ /4$ offset hai — geometry same hai, sirf kaunsa point $x = 0$ par hai yeh flip ho gaya hai.

Step 3. Wavelength: $k = 4 \Rightarrow λ = \frac{2 π}{4} = \frac{π}{2}$ .

Verify: Free-end antinode(0) $\to$ node( $π /8$ ) $= π /8 = λ /4$ ✓, aur node( $π /8$ ) $\to$ next antinode( $π /4$ ) $= π /8 = λ /4$ ✓ (spacing law sine case se identical hai — sirf labels swap ho gaye). Yeh exactly open end ke liye Sound in pipes wali situation hai.

Example 6 — Cell C6: jab string flat ho, woh limiting snapshot

Step 1. (a) $t = 0$ : $cos (0) = 1$ , toh $y (x, 0) = 0.02 sin (5 π x)$ — poora spatial pattern, antinodes max displacement par. Yeh step kyun? Breathing factor apne peak par hai, toh hame full size par shape dikhti hai.

Step 2. (b) $t = T /4$ : $ω t = 200 π \cdot 0.0025 = 0.5 π$ , aur $cos (0.5 π) = 0$ . Toh har $x$ ke liye $y (x, T /4) = 0$ — string bilkul flat hai. Yeh step kyun? Jab breathing factor zero se guzarta hai, toh displacement mein poora standing pattern momentarily gayab ho jaata hai.

Step 3. Energy kahan gayi? Is instant mein har point equilibrium displacement par hai lekin sabse tezi se chal raha hai — poori energy kinetic hai. (Antinodes yahan maximum speed se guzarte hain.) Yeh step kyun? Zero displacement $\neq =$ zero energy; parent note ka mistake-callout exactly isi cheez ke baare mein warn karta tha.

Verify: $cos (0.5 π) = 0$ exactly ✓. $cos (0) = 1$ ✓. Flat snapshot ek real, well-known instant hai — energy ke KE $\leftrightarrow$ PE sloshing ke saath consistent hai, string kabhi nahi chhodti.

Example 7 — Cell C7: boundary-fit / normal mode (word problem)

Intuition Figure 2 — kyun sirf kuch khaas waves fit hoti hain

Teen stacked strings, har ek dono ends par fixed (thick black end-bars). Top (orange), $n = 1$ : exactly ek hump length fill karta hai — string aadhi wavelength span karti hai, toh $λ_{1} = 2 L$ . Yeh fundamental hai. Middle (teal), $n = 2$ : do humps, beech mein ek internal node (hollow dot) appear hona forced hai; string ab ek poori wavelength hold karti hai, $λ_{2} = L$ . Bottom (plum), $n = 3$ : teen humps aur do internal nodes, $λ_{3} = 2 L /3$ . Figure jo pattern obvious karta hai woh yeh hai: mode $n$ mein $n$ humps aur $n - 1$ internal nodes hote hain, aur har higher mode ek shorter wave pack karta hai — yeh exactly Resonance and normal modes ki staircase hai. In ke beech ki koi bhi wavelength ek end ko hilane ki zaroorat hogi, jo fixed bars forbid karte hain, toh woh string par simply reh nahi sakti.

Step 1. Dono ends fixed $\Rightarrow$ $x = 0$ par node aur $x = L$ par node. Node spacing $λ /2$ hai, toh $L$ mein half-wavelengths ka poora number fit hona chahiye (Figure 2 mein humps count karo): $L = n \frac{λ _{n}}{2} \Rightarrow λ_{n} = \frac{2 L}{n}, n = 1, 2, 3, \dots$ Yeh step kyun? Sirf isi ko maanne waali wavelengths boundary mein fit hoti hain; baaki har wave ko ends ko hilane ki zaroorat hogi, jo woh kar nahi sakte. Yeh Resonance and normal modes ki origin hai.

Step 2. Fundamental ( $n = 1$ , Figure 2 mein orange string): $λ_{1} = 2 L = 2 \times 0.65 = 1.3 m$ . Yeh step kyun? Sabse low note woh sabse lambi wave hai jo fit hoti hai — woh $n = 1$ hai, ek single hump.

Step 3. $v = f λ \Rightarrow f_{1} = \frac{v}{λ _{1}} = \frac{260}{1.3} = 200 Hz$ se fundamental frequency. Yeh step kyun? Wave speed $v$ string se fixed hai; ise wavelength se divide karke "metres per repeat" ko "repeats per second" mein convert kiya jaata hai, jo frequency hai.

Step 4. 3rd harmonic ( $n = 3$ , Figure 2 mein plum string): $λ_{3} = \frac{2 L}{3} = \frac{1.3}{3} \approx 0.4333 m$ , aur uski frequency $f_{3} = \frac{v}{λ _{3}} = 3 f_{1} = 600 Hz$ . Yeh step kyun? Harmonics fundamental ke integer multiples hain kyunki $n$ count karta hai ki kitne half-loops fit hote hain — teen loops matlab teen guna frequency.

Verify: $f_{1} = 200$ Hz, $λ_{1} = 1.3$ m dete hain $v = f_{1} λ_{1} = 260$ m/s ✓. $f_{3} = 600$ Hz, $λ_{3} = 0.4333$ m dete hain $600 \times 0.4333 = 260$ m/s ✓. Waves on a string harmonic pattern aur Figure 2 mein count kiye gaye humps se match karta hai.

Example 8 — Cell C8: exam twist (point ki speed; node mein strain)

Step 1. Pehla antinode: $sin (5 π x) = 1 \Rightarrow 5 π x = π /2 \Rightarrow x = 0.1 m$ . Wahan amplitude poora $0.02$ m hai. Yeh step kyun? Hame ek aisa point chahiye jo actually hilta ho; antinode maximum-swing point hai.

Step 2. Transverse velocity, fixed $x$ par $y$ ka time ke saath change hone ka rate hai: $v_{y} = \frac{\partial y}{\partial t} = - 0.02 sin (5 π x) \cdot (200 π) sin (200 π t) .$ Derivative kyun, aur $\partial / \partial t$ kyun? "Is point ki speed" poochh rahi hai ki $x$ ko fixed rakh ke per unit time mein uski height kaise badalti hai — yeh exactly constant position par time-derivative hai. Hum partial derivative use karte hain kyunki $x$ frozen hai (hum ek point dekh rahe hain, string par move nahi kar rahe). Note karo yeh transverse speed $v_{y}$ upar define ki gayi wave speed $v$ se alag cheez hai — ek batata hai ki material kitni tezi se upar-neeche jhakta hai, doosra batata hai ki pattern kitni tezi se travel karta.

Step 3. Antinode par $sin (5 π x) = 1$ hai, aur $∣ sin (200 π t) ∣$ 1 par peak karta hai, toh $v_{y, m a x} = 0.02 \times 200 π = 4 π \approx 12.57 m/s .$ Yeh step kyun? Dono sine factors apne max par hone se sabse badi possible speed milti hai — antinode ka sabse tez jhak.

Step 4. Node par force / strain: slope $\frac{\partial y}{\partial x} = 0.02 (5 π) cos (5 π x) cos (200 π t)$ hai. Node par, $sin (5 π x) = 0$ toh $cos (5 π x) = \pm 1$ — slope factor sabse bada exactly wahan hota hai jahan displacement zero hai. Bada slope matlab string wahan sabse zyada stretched/kinked hai, toh restoring tension node par peak karti hai. Yeh step kyun? Yeh parent note ka "node mein max strain hoti hai" wala fact hai, ab derive kiya gaya: displacement aur ek sine ki slope quarter-cycle out of step hain, toh ek ke zeros doosre ke peaks ke saath line up karte hain.

Verify: $v_{y, m a x} = 4 π \approx 12.566$ m/s ✓. Units: $[m] [rad/s] = m/s$ ✓. Node ka slope factor $cos (5 π x) = \pm 1$ jahan $sin = 0$ hai, nodes par max strain confirm karta hai ✓.

Active recall

Recall Yeh kaunsa cell hai? (jawab cover karo)

$y = 0.4 sin (3 x) cos (5 t)$ diya gaya hai, "origin ke baad 3rd node" ::: $sin (3 x) = 0 \Rightarrow x = nπ /3$ ; origin ke baad 3rd $n = 3 \Rightarrow x = π \approx 3.14$ m hai (Cell C3). Free end at $x = 0$ : shape factor kya hai? ::: $cos (k x)$ — origin ek displacement antinode hai (Cell C5). Both-fixed string, allowed wavelengths? ::: $λ_{n} = 2 L / n$ (Cell C7). String $t = T /4$ par flat kyun ho jaati hai? ::: $cos (ω t) = 0$ wahan; poori energy kinetic hai, gayi nahi hai (Cell C6). Amplitude $R$ wale antinode ki max transverse speed? ::: $R ω$ (Cell C8).

Connections

Superposition principle — "do waves ko add karo" wala har step.
Travelling waves — $y_{1}, y_{2}$ building blocks (Ex 2).
Reflection of waves at boundaries — isliye origin forced node/antinode hota hai (Ex 4–5).
Resonance and normal modes / Waves on a string / Sound in pipes — boundary-fit example (Ex 7).
Wave number k and wavelength — $k = 2 π / λ$ har extraction mein use ki gayi.
Parent: Standing waves — formation, nodes, antinodes.