Standing waves — formation, nodes, antinodes: Exercises

1.6.18 · D4 · Physics › Oscillations & Waves › Standing waves — formation, nodes, antinodes

Definition Jo symbols hum use karte rahenge (ek baar define karo, hamesha use karo)

$x$ — string ke saath position, metres mein. Socho "rope ke saath kitni door."
$t$ — time, seconds mein.
$λ$ (lambda) — wavelength: ek poori wiggle ki length. Ek complete hill+valley imagine karo.
$k$ — wave number: $k = \frac{2 π}{λ}$ . Yeh count karta hai ki pratyek metre mein "wave phase" ke kitne radians milte hain. Bada $k$ = tightly packed wiggles.
$T$ — period: kisi point ke ek poore up-down-up breath ke liye time (seconds mein).
$f$ — frequency: har second mein kitne poore breaths, hertz (Hz) mein. Bas $f = \frac{1}{T}$ hai.
$ω$ (omega) — angular frequency: $ω = \frac{2 π}{T} = 2 π f$ , breathing ke radians per second.
$A$ — har ek travelling wave ka amplitude; standing pattern $2 A$ tak swing karta hai.
$R (x)$ — position-dependent amplitude $2 A sin (k x)$ : har point $x$ par swing kitni badi hai.
$μ$ (mu) — string ka mass per unit length, kg/m mein (rope ka ek metre kitna bhaari hai). Yeh $v = F / μ$ ke through wave speed set karta hai; dekho Waves on a string.
$F$ — string mein tension (string ke saath pulling force, newtons mein).

Neeche wala picture is poore page ka map hai: yeh fixed envelope $R (x)$ (blue), time mein ek single snapshot (yellow dashed), red nodes jo kabhi nahi hilte, aur green antinodes jo sabse zyada swing karte hain, yeh sab dikhata hai. Labelled $λ /4$ gap notice karo ek node se agli antinode tak aur $λ /2$ gap neighbouring nodes ke beech — yeh do spacings is page par lagbhag har problem drive karte hain, toh jab bhi koi position calculate karo iske paas wapas dekho.

Figure — Standing waves — formation, nodes, antinodes

Figure 1 — Standing-wave envelope $R (x) = 2 A sin (k x)$ : red = nodes, green = antinodes, labelled $λ /4$ aur $λ /2$ spacings.

L1 · Recognition

Recall Solution 1

KYA dhundhna hai: ek standing wave ( $x$ mein kuch) $\times$ ( $t$ mein kuch) mein split ho jaata hai. $x$ aur $t$ kabhi ek hi bracket ke andar nahi baithte. Yeh exactly woh shape-times-breathing split hai jo Figure 1 mein draw ki gayi hai.

(a) $x$ aur $t$ ek bracket mein tangled hain $\to$ yeh travel karta hai. Standing nahi.
(b) $sin (3 x) \cdot cos (5 t)$ ke roop mein factored $\to$ shape $\times$ breathing $\to$ standing wave ✓.
(c) Do opposite travellers; sum-to-product se yeh $0.4 sin (3 x) cos (5 t)$ ban jaata hai $\to$ yeh bhi standing hai. Answer: (b) aur (c). (c) kyun count karta hai: Superposition principle hume unhe add karne deta hai, aur trig identity sum ko (b) jaise factored form mein collapse kar deti hai.

Recall Solution 2

$y = 2 A sin (k x) cos (ω t)$ se compare karo. Yahan $2 A = 0.2$ , $k = 3$ , $ω = 5$ .

$R (x) = 0.2 sin (3 x)$ metres.
Maximum swing $= 2 A = 0.2$ m, jahan bhi $∣ sin (3 x) ∣ = 1$ ho (green antinode points Figure 1 mein).

L2 · Application

Recall Solution 3

Wavelength: $k = 2 π$ , aur $k = \frac{2 π}{λ} \Rightarrow λ = \frac{2 π}{k} = 1$ m. Nodes: $R (x) = 0 \Rightarrow sin (2 π x) = 0 \Rightarrow 2 π x = nπ \Rightarrow x = \frac{n}{2}$ . $x > 0$ par pehle teen: $x = 0.5, 1.0, 1.5$ m. $sin$ zeros kyun? ek node kabhi nahi hilta, toh uska swing $R (x)$ exactly zero hona chahiye, aur $sin$ $π$ ke poore multiples par vanish karta hai. Spacing $= 0.5$ m $= λ /2$ ✓ — yeh exactly Figure 1 mein red-dot spacing hai ( $λ = 1$ m ke liye drawn).

Recall Solution 4

Antinodes: ek antinode sabse zyada swing karta hai, toh uska amplitude factor $∣ sin (2 π x) ∣$ apni largest possible value, $1$ , hit karna chahiye. $∣ sin ∣ = 1$ kyun? kisi angle ki sine ki size $1$ se zyada kabhi nahi hoti, toh swing $R (x) = 2 A sin (2 π x)$ exactly wahan sabse badi hoti hai jahan $sin$ apni peak $\pm 1$ tak pahunche — woh point poore $2 A$ ke saath flap karta hai. $∣ sin (2 π x) ∣ = 1 \Rightarrow 2 π x = (n + \frac{1}{2}) π \Rightarrow x = \frac{2 n + 1}{4} .$ Pehle do: $x = 0.25$ m aur $x = 0.75$ m. Max swing $= 2 A = 0.5$ m. Check karo: $x = 0$ par node, $x = 0.25$ par antinode; gap $= 0.25 = λ /4$ ✓ — Figure 1 mein yellow $λ /4$ arrow.

Recall Solution 5

Add karo page ke top par stated sum-to-product identity, $sin P + sin Q = 2 sin \frac{P + Q}{2} cos \frac{P - Q}{2}$ , $P = 4 x - 2 t$ aur $Q = 4 x + 2 t$ ke saath: $y = 2 (3) sin (4 x) cos (2 t) = 6 sin (4 x) cos (2 t) .$ Antinode: $sin (4 x) = 1 \Rightarrow 4 x = \frac{π}{2} \Rightarrow x = \frac{π}{8} \approx 0.3927$ m. Check karo $λ = \frac{2 π}{4} = \frac{π}{2}$ , toh $λ /4 = \frac{π}{8}$ ✓ (pehla antinode origin par node se ek quarter-wave ki doori par baithta hai).

L3 · Analysis

Recall Solution 6

Boundary kyun matter karta hai: fixed ends hil nahi sakte, toh dono ends nodes hone chahiye. Dono ends par nodes wala pattern tabhi fit hota hai jab $n$ half-waves ka poora number $L$ span kare (dekho Resonance and normal modes): $L = n \frac{λ}{2} \Rightarrow λ = \frac{2 L}{n} .$ (a) $λ = \frac{2 ( 1.2 )}{3} = 0.8$ m. (b) $n = 3$ mode mein $n + 1 = 4$ nodes hain (dono ends + 2 interior) aur $n = 3$ antinodes. Sketch check: teen "bellies" (antinodes) jo chaar nodes se separate hain — exactly neeche Figure 2 mein $n = 3$ picture.

Neeche wala figure precisely yahi $n = 3$ mode draw karta hai: white walls dono fixed ends (forced nodes) mark karte hain, blue curves $+ R (x)$ aur $- R (x)$ envelope hain jiske beech string swing karta hai, red dots chaar nodes hain aur green dots teen antinodes. Part (b) confirm karne ke liye picture se unhe count karo.

Figure 2 — Dono ends par fixed string par third harmonic ( $n = 3$ ): 4 red nodes (dono walls including) aur 3 green antinodes, $λ = 0.8$ m ke saath.

Recall Solution 7

$R (0.25) = 0.5 sin (2 π \cdot 0.25) = 0.5 sin (\frac{π}{2}) = 0.5$ m (yahan poora swing available hai). Breathing factor: $cos (100 π \cdot \frac{1}{300}) = cos (\frac{π}{3}) = 0.5$ . $y = 0.5 \times 0.5 = 0.25 m .$ Interpretation: antinode $0.5$ m tak reach kar sakta hai, lekin is instant breathing factor sirf $0.5$ hai, toh yeh $+ 0.25$ m par halfway up baitha hai aur neeche ki taraf ja raha hai.

Recall Solution 8

Derivative kyun: velocity "displacement time ke saath kitni tezi se change hoti hai," yaani $v = \frac{\partial y}{\partial t}$ . $sin (2 π x)$ factor time mein ek constant hai, toh sirf $cos (100 π t)$ differentiate hota hai: $v = \frac{\partial y}{\partial t} = - 0.5 sin (2 π x) (100 π) sin (100 π t) = - 50 π sin (2 π x) sin (100 π t) .$ $x = 0.25$ par, $sin (2 π \cdot 0.25) = 1$ .

$t = 0$ : $sin (0) = 0 \Rightarrow v = 0$ . (Antinode top par hai, momentarily still — max displacement, zero speed.)
$t = \frac{1}{200}$ : $100 π /200 = \frac{π}{2}$ , $sin (\frac{π}{2}) = 1 \Rightarrow v = - 50 π \approx - 157.1$ m/s. (Top speed par equilibrium se guzar raha hai.)

L4 · Synthesis

Recall Solution 9

Fundamental: $L = \frac{λ}{2} \Rightarrow λ = 2 L = 4$ m $\Rightarrow k = \frac{2 π}{λ} = \frac{π}{2}$ rad/m. Time part: $ω = 2 π f = 2 π (25) = 50 π$ rad/s. Amplitude: max swing $2 A = 0.04 \Rightarrow A = 0.02$ m. Sab dalo ( $x = 0$ par node hai toh $sin$ use karte hain): $y = 0.04 sin ((\frac{π}{2}) x) cos (50 π t) m .$ Check karo: $x = 1$ m par (beech mein, $L /2$ ), $sin ((\frac{π}{2}) \cdot 1) = sin (\frac{π}{2}) = 1$ $\to$ poora swing $0.04$ m ✓ (yeh fundamental ka single antinode hai).

Recall Solution 10

$v = f λ$ kyun: speed = (wiggles per second) $\times$ (length per wiggle). Standing pattern inhi speed ke travellers se banta hai (dekho Waves on a string). (a) $v = f λ = 25 \times 4 = 100$ m/s. (b) $v = F / μ \Rightarrow F = μ v^{2} = 5 \times 1 0^{- 3} \times 10 0^{2} = 50$ N.

L5 · Mastery

Recall Solution 11

Mode $n$ mein $λ_{n} = \frac{2 L}{n}$ hai, toh $k_{n} = \frac{2 π}{λ _{n}} = \frac{nπ}{L}$ aur $R_{n} (x) = 2 A sin (\frac{nπ x}{L})$ . $n = 2$ antinode: $sin (\frac{2 π x}{L}) = \pm 1 \Rightarrow \frac{2 π x}{L} = \frac{π}{2} \Rightarrow x = \frac{L}{4}$ . $n = 4$ node us $x$ par? $sin (\frac{4 π}{L} \cdot \frac{L}{4}) = sin (π) = 0$ ✓ — haan, yeh ek node hai. Answer: $x = \frac{L}{4}$ $n = 2$ ka ek antinode hai aur $n = 4$ ka ek node. (Symmetry se $x = \frac{3 L}{4}$ bhi kaam karta hai.) Yeh kyun kaam karta hai: $n = 4$ ki spatial frequency $n = 2$ se double hai, toh har $n = 2$ antinode ek $n = 4$ node par land karta hai.

Recall Solution 12

Pehle, yahan use hone wale do symbols recall karo (dono top mein panel mein hain): $F$ string ka tension newtons mein hai — wahi $F$ jo Problem 10 se hai — aur $P$ instantaneous power hai, woh rate (watts mein) jis par energy point $x_{0}$ se kisi given instant par flow karta hai. Positive $P$ matlab energy $+ x$ direction mein move ho rahi hai, negative matlab $- x$ . Formula $P = - F \frac{\partial y}{\partial x} \frac{\partial y}{\partial t}$ string ki transverse motion ke liye "force times velocity" hai.

Do slopes compute karo: $\frac{\partial y}{\partial x} = 2 A k cos (k x) cos (ω t), \frac{\partial y}{\partial t} = - 2 A ω sin (k x) sin (ω t) .$ Multiply karo: $P = - F [2 A k cos (k x) cos (ω t)] [- 2 A ω sin (k x) sin (ω t)]$ $= 4 F A^{2} k ω sin (k x) cos (k x) sin (ω t) cos (ω t) .$ Time part hai $sin (ω t) cos (ω t) = \frac{1}{2} sin (2 ω t)$ . Ek full cycle par uska average zero hai (ek sine average karke zero hoti hai). Isliye $⟨ P ⟩ = 0$ har $x_{0}$ par. Meaning: power aage-peeche oscillate karta hai lekin kabhi net out nahi hota — energy trapped rehti hai, kinetic aur potential ke beech swap karti hoti hai, exactly jaisa parent note ne claim kiya tha.

Recall Solution 13

Mode $n$ ke antinodes $x = \frac{( 2 m - 1 ) L}{2 n}$ par baithte hain $m = 1, 2, \dots$ ke liye (yaani $\frac{L}{2 n}$ ke odd multiples). Require karo $\frac{( 2 m - 1 ) ( 1.2 )}{2 n} = 0.3 \Rightarrow (2 m - 1) = \frac{0.3 \cdot 2 n}{1.2} = 0.5 n$ . Left side ek positive odd integer hai, toh $0.5 n$ ek positive odd integer hona chahiye: $n = 2 (\to 0.5 n = 1)$ , $n = 6 (\to 3)$ , $n = 10 (\to 5)$ , … Lowest hai $n = 2$ . Tab $λ = \frac{2 L}{n} = \frac{2 ( 1.2 )}{2} = 1.2$ m. Check karo: $n = 2$ ke antinodes $x = \frac{( 2 m - 1 ) L}{2 n} = \frac{( 2 m - 1 ) ( 1.2 )}{4}$ par hain, $x = 0.3$ m ( $m = 1$ ) aur $x = 0.9$ m ( $m = 2$ ) dete hain ✓. Toh lowest harmonic jo $x = 0.3$ m par antinode rakhta hai woh $n = 2$ hai, $λ = 1.2$ m ke saath.

Recall Solution 14

$P = k x - ω_{1} t$ , $Q = k x + ω_{2} t$ ke saath sum-to-product: $y = 2 A sin (k x + \frac{ω _{2} - ω _{1}}{2} t) cos (\frac{ω _{1} + ω _{2}}{2} t) .$ Pehle factor mein ab dono $x$ aur $t$ hain, toh yeh ek fixed shape nahi hai — space aur time poori tarah separate nahi hue. Us factor ke zeros wahan baithte hain jahan $k x + \frac{ω _{2} - ω _{1}}{2} t = nπ$ , yaani $x = \frac{nπ}{k} - \frac{ω _{2} - ω _{1}}{2 k} t,$ jo speed se drift karta hai $v_{drift} = \frac{ω _{1} - ω _{2}}{2 k} .$ Direction — sign carefully padho:

Agar $ω_{1} > ω_{2}$ (right-mover faster hai), $v_{drift} > 0$ : nodes ka poora pattern $+ x$ ki taraf drift karta hai (faster wave ki direction mein). Yeh physically sense banata hai — faster wave "jeet" jaati hai aur interference pattern ko apne saath drag kar leti hai.
Agar $ω_{1} < ω_{2}$ , $v_{drift} < 0$ : nodes $- x$ ki taraf crawl karte hain, (ab faster) left-mover ke peeche.
Agar $ω_{1} = ω_{2}$ , $v_{drift} = 0$ : drift vanish ho jaata hai aur nodes lock ho jaate hain — hum true standing wave recover kar lete hain. Yeh exactly isliye hai ki ek genuine standing wave ke liye equal frequencies required hain: tabhi drift speed zero hoti hai.

Active recall

Recall Rapid-fire self-test (answers cover karo)

$k = 2 π$ ho toh $λ$ kya hoga? ::: $λ = 1$ m. $sin (2 π x)$ ke node positions kya hain? ::: $x = n /2$ (yaani $0, 0.5, 1.0, \dots$ ). $x > 0$ par $sin (2 π x)$ ka pehla antinode? ::: $x = 0.25$ m. $3 sin (4 x - 2 t) + 3 sin (4 x + 2 t)$ se standing wave kya hogi? ::: $y = 6 sin (4 x) cos (2 t)$ . $6 sin (4 x) cos (2 t)$ ke liye origin ke sabse paas antinode? ::: $x = π /8 \approx 0.39$ m. Dono ends par fixed $L = 1.2$ m string par 3rd harmonic ka $λ$ ? ::: $0.8$ m. Us $n = 3$ mode mein kitne nodes aur antinodes hain (ends included)? ::: 4 nodes, 3 antinodes. Dono ends par fixed $L = 2$ m string ka fundamental wavelength? ::: $4$ m. $L = 2$ m, $2 A = 0.04$ m, $f = 25$ Hz wali string ka poora standing wave? ::: $y = 0.04 sin ((\frac{π}{2}) x) cos (50 π t)$ m. Us string ( $μ = 5 \times 1 0^{- 3}$ ) ka wave speed aur tension? ::: $v = 100$ m/s, $F = 50$ N. Ek real standing wave zero net power kyun transport karta hai? ::: time factor $sin ω t cos ω t = \frac{1}{2} sin 2 ω t$ ek cycle mein average karke zero ho jaata hai. Do opposite waves kab true standing wave banana fail karte hain? ::: jab unki frequencies different hon ( $ω_{1} \neq = ω_{2}$ ), toh nodes $v = (ω_{1} - ω_{2}) /2 k$ par drift karte hain.

Connections

Standing waves — formation, nodes, antinodes — woh parent note jinhe yeh exercises drill karti hain.
Superposition principle — har derivation mein do waves add karne ke liye use hota hai.
Travelling waves — $y_{1}, y_{2}$ building blocks.
Reflection of waves at boundaries — isliye ek end ek node (ya antinode) fix karta hai.
Resonance and normal modes — L3–L5 mein use hone wala $L = nλ /2$ mode-fitting.
Waves on a string — $v = f λ$ aur $v = F / μ$ links.
Sound in pipes — boundary trap ka free-end/open-end variant.
Wave number k and wavelength — throughout $k = 2 π / λ$ .