1.6.17 · Physics › Oscillations & Waves
Jab do waves space ke ek hi region mein overlap karti hain, toh unke displacements har point pe aur har instant pe add hote hain. Yahi Principle of Superposition hai. Jahan crests, crests se milti hain wahan badi wave banti hai (constructive); jahan crest, trough se milti hai wahan woh cancel ho jaati hain (destructive). Interference bas superposition hi hai, lekin ek fixed pattern of points pe observe ki gayi.
YEH kyun important hai: Yeh waves ka fingerprint hai. Particles kabhi ek doosre ko cancel nahi karte — lekin waves karte hain. Bright/dark fringes, sound ke "dead spots", thin-film ke colours, noise-cancelling headphones — yeh sab ek hi idea hai.
Do (ya zyada) coherent waves ka superposition jo maximum aur minimum resultant amplitude ke points ka ek stationary pattern produce karta hai.
Coherent = do sources time mein constant phase difference rakhte hain (aur same frequency bhi). Iske bina pattern wash out ho jaata hai.
HUM kya add karte hain: displacements, na ki intensities. Intensity sirf end mein aati hai, jab humein resultant amplitude pata chal jaaye.
Do waves ko point P pe aate huye lo, same frequency ω , same amplitude a , lekin phase difference ϕ hai:
y 1 = a sin ( ω t ) , y 2 = a sin ( ω t + ϕ )
Intuition Phase difference aata kyun hai?
Do waves ne alag-alag distances travel ki hain (ya phir woh out of step shuru hue). Path difference Δ x ka matlab hai ek wave "peeche" hai — aur poore wavelength λ ki lag ek poore cycle 2 π ke phase ke barabar hai. Yahi ek sentence humein neeche diya master link deta hai.
Superpose karo (bas add karo):
y = y 1 + y 2 = a sin ω t + a sin ( ω t + ϕ )
sin A + sin B = 2 sin 2 A + B cos 2 A − B use karo jahan A = ω t , B = ω t + ϕ :
y = resultant amplitude A 2 a cos 2 ϕ sin ( ω t + 2 ϕ )
Yeh step kyun? Same frequency ke do sines ka sum abhi bhi same frequency ka sine hi hai — sirf uski amplitude aur phase change hoti hai. Amplitude woh bracketed factor hai.
I ∝ A 2 kyun? Ek wave ki energy (aur brightness/loudness) amplitude ke square ke saath scale hoti hai — yeh ek universal wave fact hai.
Poora subtopic isi mein hai — cos 2 ( ϕ /2 ) kab maximum hai aur kab zero.
Constructive — waves in step hain, amplitudes add hoti hain:
ϕ = 2 π n ⟹ cos 2 ϕ = ± 1 ⟹ A m a x = 2 a , I m a x = 4 I 0
Δ x = nλ n = 0 , ± 1 , ± 2 , …
Destructive — waves opposite hain, cancel ho jaati hain:
ϕ = ( 2 n + 1 ) π ⟹ cos 2 ϕ = 0 ⟹ A m i n = 0 , I m i n = 0
Δ x = ( n + 2 1 ) λ n = 0 , ± 1 , …
4 I 0 kyun aur 2 I 0 kyun nahi?
Energy overall conserved rehti hai! Constructive points ko ek source ka 4× milta hai; destructive points ko 0 milta hai. Pattern pe average = 2 I 0 = do individual intensities ka sum. Energy bas redistribute hoti hai, create nahi hoti.
Worked example Example 1 — Phase given hai
Do coherent waves, har ek ki intensity I 0 hai, phase difference ϕ = 6 0 ∘ ke saath milti hain. Resultant intensity nikalo.
I = 4 I 0 cos 2 ( 2 6 0 ∘ ) = 4 I 0 cos 2 3 0 ∘ = 4 I 0 ( 2 3 ) 2 = 3 I 0
Yeh step kyun? Hum ϕ directly derived intensity law mein daal dete hain; path ki koi info nahi chahiye.
Worked example Example 2 — Path difference given hai
Do speakers se sound waves (λ = 0.5 m) ek listener tak path difference 0.75 m ke saath pahunchi hain. Constructive hai ya destructive?
λ Δ x = 0.5 0.75 = 1.5 = ( 1 + 2 1 )
Yeh ek half-integer hai → destructive : ek quiet spot.
Yeh step kyun? Δ x ko λ se compare karo. Integer ⇒ loud, half-integer ⇒ silent.
Worked example Example 3 — Unequal amplitudes
Sources ke amplitudes 3 a aur 4 a hain, phase difference ϕ hai. General resultant:
A = ( 3 a ) 2 + ( 4 a ) 2 + 2 ( 3 a ) ( 4 a ) cos ϕ
Max (ϕ = 0 ): A = 7 a . Min (ϕ = π ): A = ∣4 a − 3 a ∣ = a (poori tarah cancel nahi ho sakta!).
Yeh step kyun? Do vectors (phasors) ko alag-alag length ke saath add karna: law of cosines. Poori cancellation ke liye equal amplitudes chahiye.
Common mistake "Intensities add kar do:
I = 2 I 0 har jagah."
Yeh sahi kyun lagta hai: energy "ko" bas sum ho jaana chahiye. Fix: Intensities tabhi add hoti hain jab sources incoherent hoon. Coherent waves ke liye pehle amplitudes add karo, phir square karo. 2 I 0 sirf spatial average ke roop mein aata hai, kisi ek point pe nahi.
Common mistake "Constructive ke liye
Δ x = λ chahiye, toh Δ x = 0 destructive hai."
Yeh sahi kyun lagta hai: λ "the" condition lagti hai. Fix: n = 0 (Δ x = 0 ) sabse bright point hai — central maximum. Zero path difference = perfectly in step.
Common mistake Phase aur path conditions mein confusion.
Fix: Constructive: ϕ = 2 nπ ya Δ x = nλ . Destructive: ϕ = ( 2 n + 1 ) π ya Δ x = ( n + 2 1 ) λ . Hamesha check karo ki problem kaunsi quantity de rahi hai.
Recall Flip me
Q: Do coherent waves, phase diff ϕ , har ek ki intensity I 0 — resultant I ?
A: I = 4 I 0 cos 2 ( ϕ /2 ) .
Q: Destructive interference ke liye path difference?
A: Δ x = ( n + 2 1 ) λ .
Q: I m a x = 4 I 0 kyun hai agar har ek I 0 hai?
A: Amplitudes add hoti hain → 2 a → I ∝ ( 2 a ) 2 = 4 I 0 ; energy redistribute hoti hai, average abhi bhi 2 I 0 rehta hai.
Recall Feynman: ek 12-saal ke bacche ko samjhao
Socho do bachche ek hi jhooley ko dhakka de rahe hain. Agar woh sahi moment pe saath mein dhakka dein, toh jhoola bahut oopar jaata hai — yahi constructive hai. Agar ek aage dhakka deta hai exactly tab jab doosra peeche kheench raha hota hai, toh jhoola muskil se hilta hai — yahi destructive hai. Waves bhi yahi karti hain: jo dhakke line up hote hain woh badi wave banate hain, jo dhakke lad jaate hain woh kuch nahi banate. Yeh "sahi moment" isse set hota hai ki har wave kitni door tak travel ki.
Mnemonic Conditions yaad karo
"Whole = Wow, Half = Hush."
Pura number of wavelengths (nλ ) → loud/bright (Wow, constructive). Half-extra (( n + 2 1 ) λ ) → silent/dark (Hush, destructive).
Worked example Pehle predict karo!
Calculate karne se pehle: agar ϕ = π /2 ho, toh kya I average 2 I 0 se upar hoga ya neeche?
Forecast: cos 2 ( π /4 ) = 1/2 , toh I = 4 I 0 ⋅ 2 1 = 2 I 0 — exactly average. 9 0 ∘ phase "neutral" point hai. ✓
Principle of Superposition — woh parent rule jis par interference tika hai.
Young's Double Slit Experiment — Δ x = d sin θ ko in conditions par apply karta hai.
Standing Waves — opposite directions mein travel karti waves ka interference.
Beats — alag frequencies ka time mein interference.
Coherence and Path Difference — kyun steady ϕ zaroori hai.
Energy in Waves — kyun I ∝ A 2 aur fringes mein energy conservation.
Phase ϕ wali do equal-a waves ka resultant amplitude A = 2 a cos ( ϕ /2 )
Phase ϕ par resultant intensity (har ek I 0 ) I = 4 I 0 cos 2 ( ϕ /2 )
Constructive condition (path) Δ x = nλ , n = 0 , ± 1 , …
Destructive condition (path) Δ x = ( n + 2 1 ) λ
Path diff aur phase diff ka link ϕ = ( 2 π / λ ) Δ x
I m a x = 4 I 0 kyun hai 2 I 0 nahiamplitudes add hokar 2 a bante hain; I ∝ A 2 = ( 2 a ) 2 = 4 I 0
Stable interference ke liye sources kaisi honi chahiye coherent (constant phase difference, same frequency)
Kya interference mein energy conserve hoti hai haan — redistribute hoti hai; spatial average intensity 2 I 0 rehti hai
Amplitudes a 1 , a 2 ka general resultant Equal sources ke liye ϕ = π /2 par intensity 2 I 0 (average value)
requires coherent sources
master link phi = 2pi/lambda times dx
dx = n lambda, phi = 2 pi n
dx = n+half lambda, phi = 2n+1 pi
Superposition of displacements
Constant phase difference
Resultant amplitude A = 2a cos phi/2
Intensity I = 4 I0 cos squared phi/2