Superposition principle: Worked examples — Hinglish deep dive

1.6.16 · D3 · Physics › Oscillations & Waves › Superposition principle

Kuch bhi shuru karne se pehle, yeh pakka kar lete hain ki symbols samajh aate hain.

Definition Amplitude

$A_{res}$ , angular frequency $ω$ , phase difference $ϕ$ , aur resultant phase $α$ — seedhi bhasha mein

Amplitude $A$ = ek akeli wave kisi point ko jo sabse bada dhakka de sakti hai, flat resting line se measure karke. Socho ek jhule ki sabse unchi height — woh height hi $A$ hai.
Angular frequency $ω$ = wave kitni tezi se cycle karti hai, radians per second mein. $y = A sin (ω t)$ mein, quantity $ω t$ woh angle (radians mein) hai jo wave ne $t = 0$ se ab tak ghumaya hai. Ek pura cycle tab hota hai jab $ω t$ mein $2 π$ ka izafa ho. Isse "angular" isliye kehte hain kyunki hum wave ko ek circle pe angle ki tarah track karte hain (neeche phasor idea dekho).
Phase difference $ϕ$ = ek wave dusri ke comparison mein apne cycle mein kitni aage ya peeche shifted hai, angle ki tarah measure karke. Ek pura cycle $= 36 0^{\circ} = 2 π rad$ hota hai. Agar wave 2 apna up-swing tab shuru kare jab wave 1 pehle se quarter cycle kar chuki ho, toh wave 2, $ϕ = 9 0^{\circ}$ se lead karti hai.
Resultant amplitude $A_{res}$ = do waves ko superpose karne ke baad jo ek combined wave milti hai uska sabse bada dhakka — summed wave ki height, kisi ek original ki nahin.
Resultant phase $α$ = combined wave ka direction (phase angle), wave 1 se measure karke. Jab aap do arrows ko add karke ek green arrow banate hain (Figure s01), $α$ hai woh green arrow wave 1 se kitna tilted hai. Positive $α$ matlab combined wave wave 1 se leads (pehle up-swing shuru kare); negative matlab woh lags.

Hum ek master formula use karenge, jo parent note se liya gaya hai aur un do arrows pe cosine rule se derive kiya gaya hai:

Intuition

$tan α = \frac{A _{2} sin ϕ}{A _{1} + A _{2} cos ϕ}$ kahan se aaya — khud banao Figure s01 dekho aur do arrows ko x–y axes par rakh do.

Wave 1 flat horizontal mein hai, toh uske components hain $(A_{1}, 0)$ .
Wave 2, $ϕ$ se upar tila hua hai, toh "adjacent = length· $cos$ , opposite = length· $sin$ " se uske components hain $(A_{2} cos ϕ, A_{2} sin ϕ)$ .
Arrows add karna = components add karna. Green resultant ke isliye:
- horizontal part $= A_{1} + A_{2} cos ϕ$ (dono arrows ki rightward push),
- vertical part $= A_{2} sin ϕ$ (sirf wave 2 ka vertical push hai).
Green arrow horizontal se angle $α$ banata hai, aur kisi bhi arrow ke liye $tan (tilt) = \frac{vertical part}{horizontal part}$ (opposite over adjacent). Yeh exactly wahi formula hai: $tan α = \frac{vertical}{horizontal} = \frac{A _{2} s i n ϕ}{A _{1} + A _{2} c o s ϕ} .$ Toh $tan α$ ek alag sawaal ka jawaab deta hai $A_{res}$ se: "green arrow kitna lamba hai?" nahin, balki "woh kis direction mein point kar raha hai?"

The scenario matrix

Har superposition problem jo aap miloge, in cells mein se ek mein aayegi. Neeche ke examples label karte hain woh kaunsi cell hit karte hain.

#	Case class	Kya khaas hai	Example
A	1-D signed sum, same sign	pulses reinforce, koi angles nahin	Ex 1
B	1-D signed sum, opposite sign	partial / total cancellation	Ex 1, Ex 2
C	Degenerate: equal & opposite	$A_{res} = 0$ , boundary case	Ex 2
D	In phase, $ϕ = 0$	limiting max, $A_{res} = A_{1} + A_{2}$	Ex 3
E	Out of phase, $ϕ = 18 0^{\circ}$	limiting min, $A_{\text{res}}=	A_1-A_2
F	Quarter phase, $ϕ = 9 0^{\circ}$	phasors perpendicular ⇒ Pythagoras	Ex 4
G	General angle $0 < ϕ < 18 0^{\circ}$	full cosine rule chahiye	Ex 5
H	Angle beyond $18 0^{\circ}$ ( $ϕ > π$ )	$cos$ ka sign palta, cos-half trap	Ex 6
I	Resultant phase (direction, size nahin)	$tan α$ / atan2 use hota hai	Ex 7
J	Real-world word problem	words → $ϕ$ translate karo	Ex 8
K	Exam twist: $A_{res}$ diya hai, $ϕ$ nikalo	formula ko invert karo	Ex 9
L	Negative phase $ϕ < 0$	lag; same amplitude, flipped $α$	Ex 10

Worked Examples

Cells A, B — 1-D signed sum

Worked example Example 1 — Do pulses milte hain (same-sign phir opposite-sign)

Point P par, ek instant mein pulse 1 deta hai $+ 6 cm$ aur pulse 2 deta hai $+ 2 cm$ . Kuch der baad pulse 1 deta hai $+ 6 cm$ aur pulse 2 deta hai $- 2 cm$ . Har instant par net displacement nikalo.

Forecast: Dono padhne se pehle guess karo — kya numbers bas add ho jaate hain?

Instant 1: $y = y_{1} + y_{2} = 6 + 2 = + 8 cm$ . Yeh step kyun? Ek dimension mein wave ka displacement ek signed number hota hai (upar = $+$ , neeche = $-$ ). Superposition kehta hai medium dono dhakke ek saath karta hai, toh hum signed numbers add karte hain — koi angles ki zaroorat nahin kyunki move karne ke liye sirf ek line hai.
Instant 2: $y = 6 + (- 2) = + 4 cm$ . Yeh step kyun? Same rule; pulse 2 ka negative sign matlab woh neeche dhakelta hai jabki pulse 1 upar, toh woh partially cancel karte hain.

Verify: Dono answers available total push $6 + 2 = 8$ ke chhote-ya-barabar hain. Same-sign ne max diya ( $8$ ), opposite-sign ne kam diya ( $4$ ). Units puri tarah cm mein. ✓

Cell C — the degenerate cancellation

Cells D, E — the two limiting phases

Worked example Example 3 — In phase vs out of phase (master formula ki limits)

Do waves, $A_{1} = 4$ , $A_{2} = 3$ (same units). $A_{res}$ nikalo (a) $ϕ = 0$ ke liye, (b) $ϕ = 18 0^{\circ}$ ke liye.

Forecast: $4$ aur $3$ saath milke sabse bada aur sabse chhota kya bana sakte hain?

(a) $ϕ = 0$ : convert: $0^{\circ} = 0 rad$ , toh $cos 0 = + 1$ , isliye $A_{res} = 4^{2} + 3^{2} + 2 (4) (3) (1) = 16 + 9 + 24 = 49 = 7.$ Yeh step kyun? $ϕ = 0$ par do arrows same direction mein point karte hain (Figure s01 mein socho red angle zero ho jaaye toh blue aur yellow ek line mein aa jaayein): jawaab simply $A_{1} + A_{2} = 7$ hai. Yeh maximum hai — constructive interference.
(b) $ϕ = 18 0^{\circ}$ : convert: $18 0^{\circ} = π rad$ , toh $cos π = - 1$ , isliye $A_{res} = 16 + 9 + 2 (4) (3) (- 1) = 25 - 24 = 1 = 1.$ Yeh step kyun? $ϕ = 18 0^{\circ}$ par arrows opposite directions mein point karte hain; jawaab $∣ A_{1} - A_{2} ∣ = ∣4 - 3∣ = 1$ hai. Yeh minimum hai — destructive interference. Yahan yeh zero nahin kyunki amplitudes unequal hain.

Verify: $A_{res}$ hamesha band $[∣ A_{1} - A_{2} ∣, A_{1} + A_{2}] = [1, 7]$ mein pada rehta hai. Har doosra phase iske andar koi value deta hai. ✓

Cell F — quarter phase, perpendicular phasors

Worked example Example 4 — Ninety degrees ⇒ Pythagoras

$y_{1} = 4 sin ω t$ , $y_{2} = 3 sin (ω t + 9 0^{\circ})$ , jahan $ω$ shared angular frequency hai (dono waves same rate se cycle karti hain, toh unke phasors saath spin karte hain aur difference $ϕ = 9 0^{\circ}$ fixed rehta hai). $A_{res}$ nikalo.

Forecast: Arrows right angles par hain — yeh kaun si shape banata hai?

Phasors draw karo: arrow 1 (length 4) reference ke along, arrow 2 (length 3) usse $9 0^{\circ}$ ghuma hua. Yeh step kyun? $9 0^{\circ}$ phase difference matlab doosra arrow pehle ke relative seedha upar point karta hai — woh perpendicular hain. (Kyunki dono $ω$ share karte hain, poori picture ko spin karne se yeh right angle nahin badalti.)
Tip-to-tail mein woh right triangle banate hain legs 4 aur 3 ke saath. Convert: $9 0^{\circ} = \frac{π}{2} rad$ , aur $cos \frac{π}{2} = 0$ , toh cosine term gayab ho jaata hai: $A_{res} = 4^{2} + 3^{2} = 25 = 5.$ Yeh step kyun? Right angle ke liye, law of cosines apna cosine term kho deta hai ( $cos 9 0^{\circ} = 0$ ) aur plain Pythagoras ban jaata hai.

Verify: Master formula se check karo: $16 + 9 + 2 (4) (3) cos \frac{π}{2} = 25 + 24 (0) = 5$ . ✓ Band $[1, 7]$ ke andar. ✓

Cell G — a general in-between angle

Cell H — phase beyond $18 0^{\circ}$ (the cos-half trap)

Worked example Example 6 — Jab

$ϕ > 18 0^{\circ}$ Do equal waves, $A = 2$ , phase difference $ϕ = 24 0^{\circ}$ ke saath. $A_{res}$ nikalo.

Forecast: Log aksar $2 A cos (ϕ /2)$ mein plug karte hain aur negative number paate hain — kya yeh allowed hai?

Equal amplitudes, toh $A_{res} = 2 A cos (ϕ /2)$ use karo. Convert: $ϕ = 24 0^{\circ} = \frac{4 π}{3} rad$ , toh $ϕ /2 = 12 0^{\circ} = \frac{2 π}{3} rad$ , aur $cos \frac{2 π}{3} = - 0.5$ : $2 (2) cos (\frac{24 0 ^{\circ}}{2}) = 4 cos 12 0^{\circ} = 4 (- 0.5) = - 2.$ Yeh step kyun? Formula $- 2$ deta hai, lekin amplitude negative nahin ho sakta — length kabhi zero se neeche nahin jaati.
Fix — magnitude lo: physical amplitude hai $A_{res} = ∣ - 2∣ = 2$ . Yeh step kyun? $cos (ϕ /2)$ negative ho gaya kyunki $ϕ /2 = 12 0^{\circ}$ , $9 0^{\circ}$ se aage hai, jahan cosine $+$ se $-$ mein cross karta hai. Negative $cos (ϕ /2)$ ka matlab negative length nahin hai — length hamesha positive hoti hai. Minus sign asal mein encode karta hai ki resultant arrow doosri taraf point karne laga hai (resultant ke khud ke phase mein extra $18 0^{\circ}$ add ho gaya). Toh amplitude ko absolute value $∣2 A cos (ϕ /2) ∣$ ke roop mein padhte hain, aur direction ki flip alag se note karte hain agar direction ki zaroorat ho.
Safer route — sign-stable master formula use karo, jo kabhi spurious minus nahin deta kyunki poora expression square root ke neeche hai: $A_{res} = 2^{2} + 2^{2} + 2 (2) (2) cos \frac{4 π}{3} = 4 + 4 + 8 (- 0.5) = 8 - 4 = 4 = 2.$ Yeh step kyun? Yahan $cos ϕ$ (nahin $cos ϕ /2$ ) aata hai, aur square root automatically non-negative answer deta hai — toh $ϕ > 18 0^{\circ}$ ke liye yeh form prefer karo aur trap se side-step ho jaao.

Verify: Dono routes agree karte hain: magnitude $2$ . ✓ Aur $2$ band $[∣2 - 2∣, 2 + 2] = [0, 4]$ ke andar hai. ✓

Cell I — resultant ki direction (phase), size nahin

Worked example Example 7 — Summed arrow kahan point karta hai?

$y_{1} = 4 sin ω t$ , $y_{2} = 3 sin (ω t + 9 0^{\circ})$ . Resultant phase $α$ nikalo (green arrow ka tilt wave 1 ke relative).

Forecast: Bada arrow (4) reference ke along point karta hai; chhota (3) use upar tilt karta hai — expect karo ek modest angle $4 5^{\circ}$ se kam.

$tan α = \frac{A _{2} sin ϕ}{A _{1} + A _{2} cos ϕ}$ use karo — woh formula jo humne upar components se banaya. Convert: $ϕ = 9 0^{\circ} = \frac{π}{2} rad$ , toh $sin \frac{π}{2} = 1$ aur $cos \frac{π}{2} = 0$ : $tan α = \frac{3 s i n 9 0 ^{\circ}}{4 + 3 c o s 9 0 ^{\circ}} = \frac{3 ( 1 )}{4 + 0} = \frac{3}{4} = 0.75.$ Yeh step kyun? Numerator = green arrow ka vertical part (sirf wave 2 upar contribute karta hai); denominator = horizontal part (wave 1 plus wave 2 ka sideways hissa). Unka ratio hai opposite-over-adjacent = tilt. Yeh "kis direction mein?" ka jawaab deta hai, Example 4 ke "kitna lamba?" se alag sawaal.
Yahan horizontal part $A_{1} + A_{2} cos ϕ = 4$ positive hai, toh plain $arctan$ safe hai (koi quadrant flip nahin). $α = arctan (0.75) \approx 36.8 7^{\circ}$ . Yeh step kyun? $arctan$ poochta hai "kaunse angle ka tangent $0.75$ hai?" — yeh tan ko undo karta hai. Dono parts positive ⇒ first quadrant ⇒ atan2(3,4) arctan(0.75) se agree karta hai, koi correction nahin chahiye.

Verify: Poori wave hai $y = 5 sin (ω t + 36.8 7^{\circ})$ . Check karo size Ex 4 ( $5$ ) se match kare aur angle $4 5^{\circ}$ se kam ho forecast ki tarah (kyunki wave 1 lamba arrow hai, woh direction "jeet" leti hai). ✓

Cell J — real-world word problem

Worked example Example 8 — Do loudspeakers (words →

$ϕ$ translate karo) Do loudspeakers same pure tone in phase bajate hain, har ek aapke kaan par amplitude $A = 1$ (arbitrary units) ki sound produce karta hai. Kyunki aap ek ke zyada paas hain, uski sound aap tak pahunchne se pehle ek extra half-wavelength ( $λ /2$ ) travel karti hai. Aap kitna amplitude sunenge?

Forecast: Half wavelength extra — zyada loud ya dead-silent?

Path difference ko phase mein convert karo: ek poora wavelength $λ$ ek poora cycle hai $= 36 0^{\circ} = 2 π rad$ . Toh extra $λ /2$ ka matlab hai $ϕ = \frac{λ /2}{λ} \times 2 π rad = π rad = 18 0^{\circ} .$ Yeh step kyun? Phase difference $ϕ = \frac{2 π}{λ} \times (path difference)$ — extra distance travelled matlab extra time travelled matlab extra phase accumulated. Yeh bridge hai geometry (metres) se phase (radians) tak.
Equal amplitudes, $ϕ = π rad$ : $A_{res} = 2 A cos (π /2) = 2 (1) (0) = 0.$ Yeh step kyun? $cos (π /2) = cos 9 0^{\circ} = 0$ , toh total cancellation — ek "dead spot".

Verify: Master formula: $1 + 1 + 2 (1) (1) cos π = 2 - 2 = 0$ . ✓ Wahan aap silence sunoge — ek real, sunaai dene wala effect Interference of waves mein. Energy khoyi nahin; kahin aur zyada loud hai.

Cell K — exam twist: formula ko invert karo

Worked example Example 9 — Jawaab diya hai, phase nikalo

Do equal waves of amplitude $A = 3$ milte hain aur $A_{res} = 3$ dete hain. Phase difference $ϕ$ kya hai (sabse chhoti positive value)?

Forecast: Result ek akeli wave ke amplitude ke barabar hai — dono thoda lad rahe honge. Guess karo $ϕ$ $9 0^{\circ}$ se kam hai ya zyada.

Equal-amplitude formula ko diye gaye result ke barabar set karo: $2 A cos (\frac{ϕ}{2}) = A_{res} \Rightarrow 2 (3) cos (\frac{ϕ}{2}) = 3 \Rightarrow cos (\frac{ϕ}{2}) = 0.5.$ Yeh step kyun? Output pata hai aur ek input pata hai; hum backwards solve karte hain $ϕ$ ke liye. Isliye $arccos$ exist karta hai — yeh poochta hai "kaunse angle ka yeh cosine hai?"
$\frac{ϕ}{2} = arccos (0.5) = 6 0^{\circ} = \frac{π}{3} rad \Rightarrow ϕ = 12 0^{\circ} = \frac{2 π}{3} rad .$ Yeh step kyun? 2 se wapas multiply karo. Kyunki $cos$ positive hai aur hum sabse chhota positive angle chahte hain, $6 0^{\circ}$ principal value hai.

Verify: Back plug karo: $2 (3) cos 6 0^{\circ} = 6 (0.5) = 3 = A_{res}$ . ✓ Aur $12 0^{\circ} > 9 0^{\circ}$ , forecast se match karta hai ki waves thodi oppose karte hain.

Cell L — negative phase difference

Worked example Example 10 — Wave 2 lags (

$ϕ < 0$ ) $y_{1} = 4 sin ω t$ , $y_{2} = 3 sin (ω t - 9 0^{\circ})$ , toh $ϕ = - 9 0^{\circ}$ (wave 2, wave 1 se lag karti hai). $A_{res}$ aur resultant phase $α$ nikalo.

Forecast: Example 7 ( $+ 9 0^{\circ}$ lead) ke comparison mein, kya loudness badlegi? Kya direction badlegi?

Amplitude: convert: $- 9 0^{\circ} = - \frac{π}{2} rad$ , aur kyunki $cos$ even hai, $cos (- \frac{π}{2}) = cos \frac{π}{2} = 0$ , toh $A_{res} = 4^{2} + 3^{2} + 2 (4) (3) (0) = 5.$ Yeh step kyun? $ϕ$ ka sign paltane se $cos ϕ$ nahin badalti, toh amplitude Example 7 se same hai — lag bilkul lead jitna loud lagta hai.
Direction: convert: $sin (- \frac{π}{2}) = - 1$ , $cos (- \frac{π}{2}) = 0$ , toh $tan α = \frac{3 s i n ( - \frac{π}{2} )}{4 + 3 c o s ( - \frac{π}{2} )} = \frac{3 ( - 1 )}{4 + 0} = - 0.75.$ Horizontal part $4$ positive hai, toh atan2(-3, 4) deta hai $α = - 36.8 7^{\circ}$ . Yeh step kyun? $sin$ odd hai ( $sin (- \frac{π}{2}) = - sin \frac{π}{2}$ ), toh vertical part ka sign palta hai jabki horizontal part same rehta hai — resultant upar ki jagah neeche tilt hota hai. Same size, mirrored direction.

Verify: $A_{res} = 5$ , Example 7 se match karta hai. Phase $α = - 36.8 7^{\circ}$ exactly Example 7 ke $+ 36.8 7^{\circ}$ ka negative hai — mirror image, jaisa $ϕ \to - ϕ$ ke liye expected hai. ✓

Forecast-then-Verify

Recall Forecast: amplitude 5 aur 5 ki waves,

$ϕ = 12 0^{\circ}$ ke saath. $A_{res}$ predict karo, phir check karo. Forecast: thoda opposing, toh 0 aur 10 ke beech, 5 se kam. Verify: convert $12 0^{\circ} = \frac{2 π}{3}$ , toh $2 (5) cos 6 0^{\circ} = 10 (0.5) = 5$ . Exactly 5 — crossover point. ✓

Active Recall

Recall Kaunsi cell ko koi angles bilkul nahin chahiye?

Cells A, B, C — pure 1-D signed addition of displacements.

Path difference of

λ /2

ka kya phase difference hota hai?

18 0^{\circ} = π rad

(half a cycle).

2 A cos (ϕ /2)

negative kyun aa sakta hai, aur tab kya karte hain?

Kyunki

cos

9 0^{\circ}

ke aage negative ho jaata hai; minus sign ek

18 0^{\circ}

flip in direction flag karta hai, toh amplitude ke liye absolute value lo (ya sign-safe

form use karo).

ϕ = 9 0^{\circ}

ke liye amplitude formula kya ban jaata hai?

Pythagoras,

A_{res} = A_{1}^{2} + A_{2}^{2}

, kyunki

cos 9 0^{\circ} = 0

A_{res}

hamesha kis band mein rehna chahiye?

∣ A_{1} - A_{2} ∣

aur

A_{1} + A_{2}

ke beech.

Resultant ki direction kaunsa formula deta hai, aur woh kahan se aata hai?

tan α = \frac{A _{2} sin ϕ}{A _{1} + A _{2} cos ϕ}

, summed arrow ke vertical-over-horizontal components se; jab denominator negative ho tab atan2 use karo.

ϕ

ka sign palat ne se loudness badlti hai kya?

Nahin —

cos

even hai, toh

A_{res}

nahin badalti; sirf

α

ka sign palta hai (kyunki

sin

odd hai).

y = A sin (ω t)

mein

ω

kya hai?

Angular frequency (rad/s);

ω t

woh angle hai jo sweep hua, har

2 π

par ek cycle.

Resultant phase

α

physically kya matlab rakhta hai?

Combined wave ka tilt/direction wave 1 ke relative — positive matlab woh leads karta hai, negative matlab woh lags karta hai.

Connections

Interference of waves — Ex 8 ka dead-spot, interference in real life hai.
Phasor method — yahan har cell ek arrow-sum picture hai.
Beats — tab kya hota hai jab $ϕ$ time mein drift kare (frequencies alag hon).
Standing waves — $ϕ = 18 0^{\circ}$ cancellation, space mein fixed.
Simple Harmonic Motion — ek point par har wave ek SHM hi hai; humne SHMs add ki.
Superposition principle — parent rule jin examples se exercise kiya.