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PhysicsCh 14 Waves

Waves: Class 11 Physics Chapter 14 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

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  • 90 min
  • 26 practice questions
  • 26 MCQs

Is chapter mein aap kya seekhenge

  • Mechanical wave ko ek travelling disturbance ki tarah samjhao jo energy le jaata hai, matter nahi, aur transverse aur longitudinal waves mein farq karo, saath hi yeh bhi ki kaunsa medium kise carry karta hai
  • y(x, t) = a sin(kx − ωt + φ) likho aur iska matlab samjho: isse amplitude, phase, wavelength, angular wave number, period, frequency aur travel ki direction padho
  • v = ω/k = νλ derive karo aur string par speed v = √(T/μ) aur longitudinal waves ke liye v = √(B/ρ), √(Y/ρ) aur √(γP/ρ) nikaalo, Laplace correction ke saath
  • Pulses aur do harmonic waves par principle of superposition lagao, aur resultant amplitude 2a cos(φ/2) nikaalo
  • Rigid aur free boundaries par reflection samjhao aur standing wave y = 2a sin kx cos ωt ko uske nodes aur antinodes ke saath derive karo
  • Dono ends par fixed strings aur ek end par closed ya dono ends par open pipes ke normal modes aur harmonics nikaalo, aur resonance ke sawaal solve karo
  • Beats samjhao, beat frequency ν₁ − ν₂ derive karo aur beats ki madad se decide karo ki do sources mein se kiski frequency zyada hai

NCERT Physics Class 11, Chapter 14, pp. 278–299 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

14.1 Introduction

Socho, ek tense IPL chase ke dauraan cricket stadium khachakhach bhara hai. Ek end par koi haath upar karke khada hota hai aur phir baith jaata hai, agla insaan ek pal baad wahi karta hai, phir uske baad wala, aur khade hote fans ki ek lehar poore stands mein ghoom jaati hai. Koi bhi apni seat nahi chhodta. Har spectator sirf uthta aur baithta hai, phir bhi kuch to clearly travel kar raha hai: motion ka ek pattern (Fig. 14.1).

2/3Khada group beech tak aa gaya hai. Riya khadi hai, lekin usne apni seat nahi chhodi.

Fig. 14.1Crowd apni seats par hi rehti hai; sirf khade logon ka pattern row mein aage badhta hai.

Ek lambi bench par terah spectators baithe hain, aur ek arrow jis par the wave travels likha hai, right ki taraf point karta hai. Step 1 mein teesri spectator poori khadi hai aur uske dono taraf wale aadhe uthe hain. Step 2 mein khada group row ke beech mein aa gaya hai, aur highlighted spectator Riya poori khadi hai. Step 3 mein group right end ke paas pahunch gaya hai aur Riya phir se apni purani seat par baithi hai.

Nature mein yahi trick har jagah hai. Talaab mein tairti ek patti, girte pathar se bani ripples ke guzarne par upar-neeche hilti hai, lekin kinaare tak nahi pahunchti. Mandir ki ghanti apne aas-paas ki hawa ko kampa deti hai, aur awaaz aangan paar karke tum tak pahunchti hai, phir bhi ghanti se tumhare kaan tak hawa ka koi jhonka nahi aata. Is tarah ka travelling disturbance, jismein ek pattern medium se hokar aage badhta hai jabki medium khud sirf fixed positions ke aas-paas hilta hai, wave (तरंग) kehlaata hai.

Definition
Wave

Ek disturbance jo ek jagah se doosri jagah travel karta hai, energy aur information le jaata hai, jabki medium ka har hissa sirf apni mean position ke aas-paas move karta hai.

ExampleTalaab par ripples: tairti patti ke paas paani upar-neeche hota hai, lekin patti bahti nahi.

Waves energy aur information le jaati hain, matter nahi. Guwahati mein rehne wale cousin ko phone call ke baare mein socho. Tumhari awaaz sound waves banati hai; phone ka microphone unhe electrical signal mein badalta hai; phone us signal ko radio waves ke roop mein tower tak bhejta hai; phir woh optical fibre mein light pulses ki tarah travel kar sakta hai, aur doosre end par poori chain ulti chalti hai. Har step message ko ek tarah ki wave se doosri tarah ki wave ko saunp deta hai.

Waves ki teen badi families hain. Mechanical waves (ripples, strings par waves, sound, earthquake ke jhatke) ko material medium chahiye; yeh isliye travel karti hain kyunki medium ke paas-paas ke hisse ek doosre ko push aur pull karte hain, isliye inhe Newton ke laws aur medium ki elastic properties control karti hain. Electromagnetic waves (light, radio waves, X-rays) ko koi medium nahi chahiye: taaron ki roshni lagbhag khaali space ko sadiyon tak paar karke hum tak pahunchti hai. Vacuum mein yeh sab same speed se travel karti hain, jo 1983 se metre ki definition ke dwara exactly fix hai:

Vacuum mein light ki speed
c=299 792 458 m/s
(14.1)

Teesri family, matter waves, quantum physics ka hissa hai: electrons, protons aur poore atoms bhi wave jaisa behaviour dikhate hain. Yeh abstract lagta hai, lekin electron microscopes viruses aur materials ke atomic structure ki image banane ke liye electrons ki waves par hi depend karte hain. Electromagnetic waves se tumhari mulaqat Class 12 mein hogi; yeh chapter mechanical waves ke baare mein hai.

Ek disturbance medium mein aage kaise badhta hai? Morning assembly ki line mein kandhe se kandha milakar khade students ko socho. Pehle student ko ek halka dhakka do, aur har student agle ko takkar maarkar waapas apni jagah par jhool jaata hai: dhakka line mein aage daudta hai jabki har student apni jagah par rehta hai. Asli medium springs se jude chhote masses ki chain jaisa behave karta hai. Solid mein yeh springs crystal ke paas-paas ke atoms ke beech ke bonds hote hain. Ek atom ko displace karo, to uske bonds use waapas kheenchte hain aur saath hi uske neighbours ko bhi kheenchte hain.

Hawa mein bonds nahi hote, phir bhi usmein springiness hoti hai. Hawa ki ek patli slab ko dabao, to uski density thodi badhti hai, isliye uska pressure bhi badhta hai. Yeh extra pressure aage wali slab ko push karta hai, jo phir dab jaati hai, jabki pehli slab relax hoti hai aur thoda aage nikalkar ek patli, low-pressure state mein chali jaati hai. Is tarah pressure differences springs ka kaam karte hain, aur dabe hue regions, compressions, aur khinche hue regions, rarefactions, ka ek pattern hawa mein aage badhta hai. Energy medium ke paas-paas ke hisson ke beech in elastic links se aage pahunchti hai. Isliye elastic media mein waves Chapter 13 ke oscillators ki kareebi rishtedaar hain: medium ka har hissa oscillate karta hai, aur neighbours ke beech ka coupling oscillation ko travel karwata hai.

Quick check
3 mein se 1
Ek pathar se bani ripples shaant talaab par phailti hain aur ek tairti patti ke paas se guzarti hain. Patti kya karti hai?
Answer dekho

Answer (A) Yeh upar-neeche (aur thoda aage-peeche) hilti hai lekin lagbhag wahin rehti hai jahan thi

Patti ke paas ka paani sirf apni mean position ke aas-paas oscillate karta hai, isliye patti upar-neeche hoti hai jabki pattern bahar ki taraf travel karta hai. Bahar ki taraf bahne ka matlab hota ki paani khud flow kar raha hai, jiski wave ko zaroorat nahi.

14.2 Transverse aur longitudinal waves

Mechanical waves ko is basis par baanta jaata hai ki medium ke hisse wave ki travel direction ke comparison mein kaise move karte hain. Transverse wave (अनुप्रस्थ तरंग) mein har hissa travel ki direction ke right angles par move karta hai. Longitudinal wave (अनुदैर्ध्य तरंग) mein har hissa travel ki direction ke along aage-peeche move karta hai. Ek slinky spring dono dikhati hai (Fig. 14.2).

2/3Longitudinal: coils paas aakar compressions (C) aur door hokar rarefactions (R) banate hain; ribbon travel ke along aage-peeche move karta hai.

Fig. 14.2Har case mein wave right ki taraf travel karti hai. Farq is baat ka hai ki medium ka ek marked hissa kaise move karta hai.NCERT Figs 14.2–14.4 · p. 280

Step 1: ek slinky ko uske left end se sideways hilaaya jaata hai aur woh lehardaar curve banati hai; ek coil par bandha laal ribbon crest par hai, uske saath ek vertical double arrow hai jis par ribbon: up and down likha hai, aur neeche ek arrow kehta hai ki wave right ki taraf travel karti hai. Step 2: slinky seedhi hai lekin uske coils C wale regions mein paas aate hain aur R wale regions mein door hote hain; ribbon ke saath ek horizontal double arrow hai jis par ribbon: to and fro likha hai. Step 3: ek lehardaar water surface aur ek cork; ek dashed circle dikhata hai ki cork loop mein ghoomta hai jabki wave right ki taraf travel karti hai.

Gate se bandhi ek rassi transverse wave ko saaf dikhati hai. Apne end ko ek baar jhatke se upar aur waapas neeche karo, to ek akela kubad, ek pulse, gate ki taraf daud jaata hai. Apna haath pendulum bob ki tarah steadily upar-neeche karo, to tum ek continuous harmonic (sinusoidal) wave bhejte ho: barabar doori par crests aur troughs ki ek train. Tum kuch bhi karo, rassi mein bandhi gaanth sirf upar-neeche hoti hai; woh kabhi gate ki taraf nahi badhti.

Ek wave ko do bilkul alag tareekon se socha ja sakta hai, aur tumhe dono chahiye. Ek instant par li gayi photo poori rassi par phaila poora shape pakad leti hai. Ek akeli gaanth ka slow-motion video dikhata hai ki samay ke saath ek jagah kaise upar-neeche hoti hai. Section 14.3 dono pictures ko ek formula mein pack karta hai.

Sound roz ki longitudinal wave hai. Hawa ki ek lambi tube ke end par loudspeaker rakho (Fig. 14.3). Jab speaker cone aage uchhalta hai to woh paas ki hawa ko compression mein thoons deta hai; jab woh peeche hatta hai to ek rarefaction chhod deta hai. Ek steady tone tube mein compressions aur rarefactions ki ek regular train bhejti hai, aur unke guzarne par hawa ki har patli layer tube ke along aage-peeche move karti hai.

3/3Ek steady tone lagaataar compressions (C) aur rarefactions (R) banati rehti hai. Excess pressure Δp har C par sabse zyada hota hai.

Fig. 14.3Hawa ki har layer sirf tube ke along aage-peeche move karti hai; travel karta hai compressions aur rarefactions ka pattern.NCERT Fig. 14.4 · p. 281

Ek horizontal tube ke left end par loudspeaker hai aur andar patli vertical lines hawa ki layers dikhati hain; ek layer laal rang mein moti banayi gayi hai. Step 1: cone right ki taraf move hua hai aur uske theek aage ki layers paas-paas aa gayi hain, jis par compression likha hai. Step 2: compression tube mein aur aage chala gaya hai, laal layer ke paas se, jo thodi aage khisak gayi hai, aur cone ke paas ka region phaila hua hai, jis par rarefaction likha hai. Step 3: ek steady tone baari-baari se ghane aur phaile regions deti hai jin par C aur R likha hai; tube ke neeche excess pressure ka graph har C par sabse upar aur har R par sabse neeche hai, aur ek arrow ek C se agle C tak ek wavelength dikhata hai.

Ceiling fan ke neeche khade ho, to tumhe hawa lagti hai: hawa room mein ek jagah se doosri jagah le jaayi ja rahi hai. Fan ki bhinbhinahat bhi tum tak pahunchti hai, lekin wave ki tarah, bina kisi hawa ke safar kiye. Isi tarah nadi ka current paani ko neeche ki taraf le jaata hai, jabki shaant tank ko paar karti ripple koi paani nahi le jaati. Aisi waves, jinka pattern medium ke ek region se agle region tak aage badhta hai, progressive (travelling) waves (प्रगामी तरंग) hain, transverse ya longitudinal.

Ek taani hui string transverse waves carry kar sakti hai jabki uske aas-paas ki hawa nahi, aisa kyun? Medium ko patli layers ki tarah socho. Jab transverse wave guzarti hai, to har layer apne neighbours ke relative sideways ghaseeti jaati hai, jo ek shear hai (Chapter 8). Solid shear ke against waapas kheenchta hai, isliye woh displace hui layers ko waapas line mein push kar sakta hai. Liquid ya gas apni layers ko bas phisalne deta hai, isliye fluid ke andar transverse wave ke liye koi restoring force nahi hota aur woh turant khatam ho jaati hai. Dabana alag baat hai: solids, liquids aur gases sab compress hone par waapas push karte hain. Isiliye longitudinal waves har tarah ke medium se guzarti hain, isiliye copper rod dono tarah ki waves carry kar sakti hai, aur isiliye hawa mein sound purely longitudinal hai. Dono tarah ki waves alag-alag elastic constants par depend karti hain, isliye same solid mein bhi aam taur par unki speeds alag hoti hain.

Jheel par uchhalte cork ko dhyaan se dekho, to dikhega ki woh har crest par thoda aage aur har trough mein thoda peeche bhi khisakta hai, aur ek chhota loop banata hai (Fig. 14.2 ka teesra step). Isliye surface water waves transverse aur longitudinal motion ko milati hain, aur gehraai ke saath loops chhote hote jaate hain. Surface ko waapas level kaun kheenchta hai, yeh wave ke size par depend karta hai. Lagbhag do centimetre se chhoti tiny ripples ke liye surface tension jeetta hai; yeh capillary waves hain. Khule samudra ki lambi waves ke liye, lagbhag ek metre se lekar saikdon metre tak, Earth ki gravity yeh kaam karti hai; yeh gravity waves hain.

Example 14.1NCERT Example 14.1 · p. 281

Wave motions ko sort karna

Batao ki har wave transverse hai, longitudinal hai ya combination: (a) taani hui kapde sukhane ki rassi par bheja gaya ek jhatka; (b) lohe ki lambi railing mein suni gayi 'ping' jab hammer uske end ko uski length ke along thapthapata hai; (c) jheel mein tairte duck se phailti ripples; (d) ship dwara samudra ke paani mein bheja gaya sonar pulse; (e) chattan mein se guzarti earthquake ki S-wave.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check
3 mein se 1
Transverse wave mein medium ke particles travel ki direction ke move karte hain; longitudinal wave mein woh uske move karte hain.
Answer dekho

Answer (1) perpendicular, (2) parallel

Transverse: travel ki direction ke aar-paar (hilaayi gayi rassi). Longitudinal: uske along (sound).

14.3 Progressive wave mein displacement relation

Ek formula ko do kaam karne hain. Fixed time par padhne par use poori string ka shape dena chahiye; fixed jagah par padhne par use string ke us hisse ki motion deni chahiye. Isliye use is par depend karna chahiye ki tum kahan dekhte ho, x, aur kab dekhte ho, t. Maan lo string x-axis ke along padi hai aur y(x,t) uska sideways displacement hai. Sabse simple travelling wave, +x ki taraf jaati ek harmonic wave, hai

+x ke along harmonic wave
y=asin⁡(kx−ωt+ϕ)
(14.2)
jahan
y
position x aur time t par displacement
a
amplitude
k
angular wave number
ω
angular frequency
ϕ
shuruaati phase angle
DerivationPhase constant φ ki ijaazat kyun hai

θ=kx−ωt likho aur sum ke sine ko expand karo: sin⁡(θ+ϕ)=sin⁡θcos⁡ϕ+cos⁡θsin⁡ϕ. A=acos⁡ϕ aur B=asin⁡ϕ lene par yeh milta hai

y=Asin⁡θ+Bcos⁡θ
(14.3)

Isliye Eq. (14.2) bas same k aur ω wali ek sine wave aur ek cosine wave ka mixture hai. Square karke jodne aur divide karne se constants waapas mil jaate hain: a=A2+B2 aur tan⁡ϕ=B/A. Aisa koi bhi mixture khud ek harmonic wave hota hai.

Eq. (14.2) ko teen tarah padhne se dikhta hai ki yeh sach mein aisi hi wave hai.

  • Ek crest ko follow karo. Crest woh point hai jahan phase kx−ωt+ϕ ki ek khaas value hoti hai. Jaise-jaise t badhta hai, us value ko banaye rakhne ke liye x ko badhna padega, isliye crest +x ki taraf move karta hai.
  • Ghadi rok do. Kisi bhi fixed time t0 par formula asin⁡(kx+constant) hai: snapshot ek sine curve hai.
  • Ek jagah ko dekho. Kisi bhi fixed x0 par yeh asin⁡(constant−ωt) hai: string ka woh hissa simple harmonic motion karta hai.
  • ωt ke aage ka sign badal do, to crest ko ulti taraf move karna padega, aur −x ki taraf jaati wave milti hai:
−x ke along harmonic wave
y=asin⁡(kx+ωt+ϕ)
(14.4)

Table 14.1 mein saare symbols ek jagah hain, aur Fig. 14.4 Eq. (14.2) ko kaam karte hue dikhata hai: ek crest (orange marker) steadily right ki taraf move karta hai, jabki x=0 par bead apni jagah par hi sirf neeche, upar aur phir neeche move karta hai.

Table 14.1Harmonic wave ki quantitiesNCERT Fig. 14.5 · p. 282
SymbolNaamSI unit
y(x,t)position x, time t par displacementm
aamplitudem
kangular wave numberrad m−1
ωangular frequencyrad s−1
kx−ωt+ϕphaserad
ϕinitial phase angle (x = 0, t = 0 par phase)rad

4/4Sirf bead ko follow karne par y(0,t)=−asin⁡ωt milta hai: period T wali simple harmonic motion.

Fig. 14.4Crest x ke along move karta hai; x = 0 par bead sirf upar-neeche move karta hai. Ek period mein pattern ek wavelength aage badhta hai.NCERT Figs 14.6–14.8 · p. 282

Upar: y versus x ke axes ke saath ek neeli sine curve; ek orange triangle crest ko mark karta hai, ek bracket crests ke beech ek wavelength lambda ko mark karta hai, ek arrow amplitude a ko mark karta hai, aur ek laal bead curve par x equals 0 par baitha hai. Step 1, t equals 0: bead zero displacement par hai aur ek arrow dikhata hai ki woh neeche move karne wala hai. Step 2, t equals T over 4: poori curve quarter wavelength right khisak gayi hai, crest marker bhi uske saath, aur bead seedha neeche trough level tak aa gaya hai. Step 3, t equals T over 2: curve aadhi wavelength khisak gayi hai aur bead waapas zero par hai, upar move karta hua. Step 4: neeche ek doosra graph aata hai jo bead ka displacement time ke saath dikhata hai, ek sine curve jo pehle neeche jaati hai, aur ek bracket ek period T ko mark karta hai.

14.3.1 Amplitude aur phase

Sine ke aage ka constant a amplitude (आयाम) hai: medium ka koi bhi hissa apni rest position se zyada se zyada kitni door jaata hai. Kyunki sin⁡ ka size kabhi 1 se zyada nahi hota, y hamesha +a aur −a ke beech rehta hai. Hum sign convention aisa chunte hain ki a khud positive ho, jabki y axis ke upar positive aur neeche negative ho. Snapshot ke sabse upar wale points crests hain aur sabse neeche wale troughs.

String ke har hisse ko ghadi ki sui ki tarah ek cycle mein ghoomta hua socho. Angle kx−ωt+ϕ batata hai ki time t par x wala hissa apni cycle mein kitna aage hai; ise phase (कला) kehte hain aur ise radians mein naapte hain. Ek baar a pata ho, to displacement nikaalne ke liye bas phase chahiye. Jab ghadi zero dikhaati hai tab origin par iski value ϕ hai, yaani initial phase angle. x=0 kahan rakhein, ya ghadi kab shuru karein, yeh badalne se ϕ badal jaata hai; inhe sahi chunne se ϕ=0 ho jaata hai, aur ab se hum aam taur par yahi karenge.

14.3.2 Wavelength aur angular wave number

Do points jo same instant par apni cycle ke same stage par hon, in phase hote hain: jaise do paas-paas ke crests, ya do paas-paas ke troughs. In phase points ke beech ki sabse chhoti doori wavelength λ (तरंगदैर्ध्य) hai. Ise Eq. (14.2) se nikaalne ke liye ϕ=0 rakho aur t=0 par snapshot dekho:

y(x,0)=asin⁡kx
(14.5)

x ke along 2π/k aage badhne se angle mein 2π jud jaata hai aur sine waapas same value par aa jaata hai; 4π/k, 6π/k waghairah se bhi yahi hota hai. In repeat distances mein sabse chhoti wavelength hai:

λ=2πkork=2πλ
(14.6)

Hum k ko angular wave number kehte hain; kuch books ise propagation constant kehti hain. Iski SI unit radian per metre hai; kyunki radian ek ratio hai, ise aksar sirf m−1 likha jaata hai. k ko per metre phase change ki tarah padho: ek metre mein jitni wavelengths fit hoti hain, har ek ke liye 2π.

14.3.3 Period, angular frequency aur frequency

Snapshot ki jagah ab ek hi jagah ko dekhte raho, jaise Fig. 14.4 ke step 4 ke history graph mein. ϕ=0 ke saath, x=0 wala hissa y(0,t)=asin⁡(−ωt)=−asin⁡ωt ki tarah move karta hai. Iska period T ek poori cycle ka time hai, jiske baad motion bilkul repeat hoti hai: har t ke liye sin⁡ωt aur sin⁡(ωt+ωT) barabar hone chahiye. Aisa karne wala sabse chhota T, ωT ko 2π ke barabar banata hai:

ω=2πT
(14.7)

ω wave ki angular frequency hai, rad s−1 mein. Frequency ν (Greek 'nu') per second oscillations ginti hai, aur ise hertz (Hz) mein naapte hain:

ν=1T=ω2π
(14.8)

Upar ki koi bhi baat wave ke transverse hone par depend nahi karti thi. Longitudinal wave mein har hisse ka displacement travel ki direction ke along hota hai, aur hum aam taur par ise y ki jagah s kehte hain:

s=asin⁡(kx−ωt+ϕ)
(14.9)
jahan
s
time t par us layer ka x ke along displacement jiski rest position x hai
a
displacement amplitude

Baaki saare symbols ke matlab wahi rehte hain. Sound ke liye s(x,t) batata hai ki jo hawa ki layer normally x par hoti hai, woh kitni aage ya peeche khisak gayi hai.

Example 14.2NCERT Example 14.2 · p. 284

Wave equation padhna

Ek lambi string par wave y(x,t)=0.0040sin⁡(12.0x−90.0t+0.50) hai, saare constants SI units mein (y aur x metres mein, t seconds mein, phase radians mein). Nikaalo (a) amplitude, (b) wavelength, (c) period aur frequency, aur (d) t=0.050 s par x=20.0 cm par displacement.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check
4 mein se 1
Ek wave SI units mein y=0.020sin⁡(5.0x−40t) hai. Metres mein iski wavelength kya hai?
m
Answer dekho

Answer 1.257 m

λ=2π/k=2π/5.0=1.26 m. 40 to ω hai; yeh period set karta hai, wavelength nahi.

14.4 Travelling wave ki speed

Pattern kitni tezi se move karta hai? Wave ka koi bhi feature chuno, sabse aasaan ek crest, aur dekho ki woh ek chhote time Δt mein kitni door jaata hai. Fig. 14.4 mein poori curve rigidly right ki taraf khisakti hai, isliye fixed phase wala har point same speed v se move karta hai; agar alag phases alag speeds se move karte to shape bacha nahi rehta.

DerivationConstant phase wale point se speed

Wave par fixed displacement wala point, fixed phase wala point hota hai:

kx−ωt=const.
(14.10)

Agar yeh point time t par x par hai aur time t+Δt par x+Δx par, to k(x+Δx)−ω(t+Δt)=kx−ωt, isliye k Δx−ω Δt=0. Chhote intervals ki limit mein,

Wave speed
v=dxdt=ωk
(14.11)

ω=2π/T aur k=2π/λ substitute karne par:

v=λT=νλ
(14.12)

Eq. (14.12) ko timing ke baare mein ek statement ki tarah padho. Jab tak medium ka koi ek hissa ek cycle poori karta hai, tab tak jo crest uske upar tha woh wahan pahunch jaata hai jahan agla crest tha: ek wavelength. Yeh har travelling wave ke liye sach hai, chahe woh kisi bhi tarah ki ho. Lekin equation yeh nahi batati ki speed kitni hai. Mechanical waves ke liye yeh medium decide karta hai. Do properties muqabla karti hain: medium deform hone ka kitni sakhti se virodh karta hai (string mein tension, ya koi elastic modulus: Young's, bulk ya shear) aur uski inertia kitni hai (string ki mass per metre, ya density). Ek baar medium ne v fix kar diya, to frequency chunne se wavelength fix ho jaati hai. Medium shear aur compression ka alag-alag stiffness se virodh karta hai, isiliye usmein transverse aur longitudinal waves ki speed aam taur par alag hoti hai.

InteractiveTravelling-wave lab

String ke along beads ki ek row ya hawa ka ek column. Sliders amplitude, frequency aur wave speed set karte hain, aur buttons travel ki direction aur transverse ya longitudinal view chunte hain. Play wave ko slow motion mein animate karta hai: ek crest marker aage travel karta hai jabki ek highlighted bead apni jagah par oscillate karta hai, aur readouts wavelength, period, angular wave number, angular frequency aur periods mein beeta hua time dikhate hain.

Fig. 14.5Amplitude, frequency aur wave speed set karo, phir Play dabao. Marked bead aur crest marker ko dekho; same wave ke transverse aur longitudinal view ke beech switch karo.
Example

Equation se speed aur direction

Ek rassi par wave SI units mein y=0.025sin⁡(4.0x+60t) hai. Yeh kis direction mein travel karti hai, aur iski speed, wavelength aur frequency kya hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

14.4.1 Taani hui string par transverse wave ki speed

String par wave ki speed kis par depend kar sakti hai? Mudi hui string ko seedha karne wala khinchaav tension T hai, isliye zyada taani hui string waves ko tez le jaani chahiye. Jise move karana hai woh string ka apna mass hai, jise linear mass density μ=m/L, yaani uski mass per metre, se naapte hain, isliye bhaari string dheemi honi chahiye. (Is subsection mein T ka matlab tension hai; chapter mein baaki jagah yeh period hai, aur context se hamesha saaf hota hai ki kaunsa.) Newton ke second law se poori calculation is book ke level se aage hai, lekin T aur μ ke dimensions lagbhag baat tay kar dete hain.

DerivationDimensional analysis se string ka formula

Tension ek force hai, [MLT−2], aur μ mass per length hai, [ML−1]. Divide karne se mass hat jaata hai:

[Tμ]=[MLT−2][ML−1]=[L2T−2]

jo ek speed ka square hai. Agar sirf T aur μ hi matter karte hain,

v=CTμ
(14.13)

jahan C ek pure number hai jise dimensional analysis kabhi nahi nikaal sakta. Poora Newtonian treatment dikhata hai ki C=1:

String par transverse waves
v=Tμ
(14.14)

Eq. (14.14) ke baare mein do baatein underline karne layak hain (Fig. 14.6). Pehli, ismein sirf string aur uska tension hai; tension bahar se set hota hai, sitar ki khootiyon ya latakte weight se, akele material se nahi. Doosri, ismein na wavelength hai na frequency, isliye ek given string par har frequency ki waves same speed se travel karti hain. Frequency woh chunta hai jo string ko hilaata hai, aur phir string wavelength ko fit hone par majboor karti hai:

λ=vν
(14.15)

Har medium itna sahyogi nahi hota. Jahan speed khud frequency ke saath badalti hai, jaise glass mein light ke liye ya gehre paani par waves ke liye, us medium ko dispersive kehte hain; tab ek pulse, jo kai frequencies ka mel hai, chalte-chalte phail jaata hai.

2/3Halki string ka pulse do guna door hai: μ→4μ speed ko aadha kar deta hai.

Fig. 14.6Chaar guna bhaari string pulses ko aadhi speed se le jaati hai; chaar guna tension isko poora kar deta hai, kyunki v=T/μ.

Do horizontal strings left par clamp hain: upar mass per length mu wali patli halki string aur neeche mass per length 4 mu wali moti bhaari string. Step 1: ek jaise pulses same dashed line se shuru hote hain, dono mein same tension ke saath. Step 2: same time ke baad halki string ka pulse bhaari string ke pulse se do guna door pahunch gaya hai; un par v aur v over 2 likha hai. Step 3: bhaari string ka tension chaar guna kar diya jaata hai aur ab uska pulse halki string ke pulse ke saath-saath chalta hai, dono par v likha hai.

Example 14.3NCERT Example 14.3 · p. 285

Guitar string par wave ki speed

Ek steel guitar string ki vibrating length 0.65 m hai aur us length ka mass 1.3 g hai. Ise 80 N ke tension par tune kiya gaya hai. Is par transverse waves ki speed nikaalo, aur woh time bhi jo ek pulse ko bridge se nut tak daudne mein lagta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

14.4.2 Longitudinal wave ki speed (sound ki speed)

Longitudinal wave mein medium ki layers baari-baari se dabti aur khinchti hain, isliye jo elastic property matter karti hai woh yeh hai ki medium volume ke change ka kitni zor se virodh karta hai. Yeh Chapter 8 ka bulk modulus hai:

B=−ΔPΔV/V
(14.16)
jahan
ΔP
pressure mein change
ΔV/V
isse bana volume strain

Minus sign B ko positive banata hai, kyunki pressure badhne se volume ghatta hai. B ke dimensions pressure wale hain aur ise pascals mein naapte hain. Ab inertia density ρ hai, [ML−3].

DerivationDimensional analysis se sound ki speed

Modulus ke dimensions ko density ke dimensions se divide karo:

[Bρ]=[ML−1T−2][ML−3]=[L2T−2]
(14.17)

Phir se ek speed ka square, isliye ek unknown number C ke saath:

v=CBρ
(14.18)

Exact theory C=1 deti hai, isliye kisi bhi medium mein longitudinal waves is speed se travel karti hain

Longitudinal waves (general)
v=Bρ
(14.19)

Ek lambi patli bar ke sides thoda phoolne ya sikudne ke liye free hote hain, isliye bar effectively ek simple longitudinal stress aur strain mein hoti hai, aur relevant modulus Young's modulus Y hai. Wahi dimensional argument aur exact theory dete hain

Patli bar mein longitudinal waves
v=Yρ
(14.20)

Table 14.2 mein typical speeds di gayi hain. Rod wali values seedhe Eq. (14.20) se handbook ki Y aur ρ values lekar aati hain; usi metal ke bade block mein, jahan material side mein phool nahi sakta, longitudinal speed thodi zyada hoti hai.

Table 14.2Kuch media mein sound ki speed (lagbhag)NCERT Table 14.1 · p. 286
MediumConditionsSpeed (m/s)
Carbon dioxidegas, 0 °C259
Hawagas, 0 °C331
Hawagas, 30 °C349
Heliumgas, 20 °C1007
Mercuryliquid, 20 °C1450
Paaniliquid, 25 °C1497
Samudra ka paaniaam ocean≈ 1530
Body ka naram tissueultrasound scans mein use hota hai1540
Leadpatli rod, √(Y/ρ)≈ 1190
Copperpatli rod, √(Y/ρ)≈ 3660
Steelpatli rod, √(Y/ρ)≈ 5050
Aluminiumpatli rod, √(Y/ρ)≈ 5090

Liquids aur solids gases se kahin zyada dense hain, to unmein sound tez kyun hai? Kyunki unhe compress karna aur bhi mushkil hai. Paani aur hawa ko compare karo. Paani ka bulk modulus, lagbhag 2.2×109 Pa, hawa ke effective bulk modulus ka lagbhag 15 000 guna hai, jabki paani sirf lagbhag 780 guna zyada dense hai. Isliye ratio B/ρ lagbhag 20 guna bada hai, aur speed lagbhag 20≈4.5 guna zyada: lagbhag 1480 m/s versus 331 m/s.

Gas ke liye hum bulk modulus predict kar sakte hain. Newton ne pehli koshish ki, aur woh jis tarah galat hui, usse kaafi kuch seekhne ko milta hai.

DerivationNewton ka formula: constant temperature par compressions

Absolute temperature T par volume V mein N molecules wali ideal gas yeh follow karti hai (Chapter 12)

PV=NkBT
(14.21)

Newton ne maana ki sound wave mein dabna aur khinchna constant temperature par (isothermally) hota hai. Tab PV constant rehta hai, isliye ek chhota change P ΔV+V ΔP=0 deta hai, yaani −ΔP/(ΔV/V)=P. Bulk modulus bas pressure hai, B=P, aur Eq. (14.19) ban jaati hai

Newton ka formula
v=Pρ
(14.22)

Ise 0 °C aur normal pressure, P=1.013×105 Pa, par sookhi hawa par try karo. Hawa ke molar mass, M=2.897×10−2 kg/mol, ke saath ideal gas law ρ=PM/RT=1.29 kg/m3 deta hai, aur Newton ka formula predict karta hai

v=1.013×1051.29≈280 m/s
(14.23)

Dhyaan se kiye gaye measurements 331 m/s dete hain. Newton ki value lagbhag 15 % kam hai, jo experimental error se kahin zyada hai. Galti assumption mein thi, algebra mein nahi (Fig. 14.7). Sound wave ke compressions aur rarefactions ek second mein saikdon baar badalte hain, aur involved regions centimetres se metres tak ke hote hain. Itne kam time mein heat na to dabe hue region se bahar nikal sakti hai, na rarefied region mein andar ja sakti hai. Isliye compressions gas ko thoda garam karte hain aur rarefactions use thanda: changes adiabatic hain, isothermal nahi. Adiabatically dabayi gayi gas zyada zor se virodh karti hai, kyunki uska badhta temperature pressure ke rise mein aur jod deta hai.

3/3Tezi se dabane par, jaise sound wave mein, gas garam hoti hai aur pressure zyada badhta hai: adiabatic, B=γP.

Fig. 14.7Wahi dabaav pressure ko zyada badhata hai jab woh itna tez ho ki heat bahar na nikal sake, isliye sound ke liye gas Newton ke maane se zyada stiff hai.

Pistons wale gas ke do ek jaise cylinders, har ek par ek pressure dial aur ek thermometer. Step 1: dono pistons upar hain aur dono dials aur thermometers same reading dikhate hain. Step 2: left piston ko dheere neeche push kiya jaata hai; heat out likhe lehardaar arrows cylinder se bahar nikalte hain, thermometer steady rehta hai aur dial ki sui thodi ghoomti hai; label par B equals P likha hai. Step 3: right piston ko utna hi tezi se neeche push kiya jaata hai; uska thermometer badhta hai, jis par gas warms up likha hai, aur uske dial ki sui left wale se zyada ghoomti hai; label par B equals gamma P likha hai.

DerivationLaplace ka correction: bina heat flow ke compressions

Adiabatic change ke liye ideal gas PVγ=constant follow karti hai (Chapter 11), jahan γ=Cp/Cv specific heats ka ratio hai. Ek chhota change deta hai

γPVγ−1 ΔV+Vγ ΔP=0

Vγ se divide karne par dikhta hai ki adiabatic bulk modulus Bad=−ΔP/(ΔV/V)=γP hai. Ise Eq. (14.19) mein rakhne par:

Ideal gas mein sound ki speed (Laplace)
v=γPρ
(14.24)

Hawa zyaadatar diatomic hai, γ=1.40 ke saath, isliye Laplace ka formula Newton ke estimate ko 1.40=1.18 se multiply karta hai: 280 m/s×1.18≈331 m/s, jo experiment se bahut achhi tarah match karta hai.

Example 14.4NCERT Example 14.4 · p. 287

Helium ke liye Newton aur Laplace

Helium ek monatomic gas hai (γ=5/3) jiska molar mass 4.00×10−3 kg/mol hai. 0 °C aur 1.013×105 Pa par helium mein sound ki speed estimate karo (a) Newton ke formula se aur (b) Laplace ke formula se. R=8.314 J mol−1 K−1 lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check
4 mein se 1
0.125 kg/m linear mass density wali ek rassi 50 N ke tension mein hai. Is par transverse waves m/s mein kitni tezi se travel karti hain?
m/s
Answer dekho

Answer 20 m/s

v=T/μ=50/0.125=400=20 m/s.

14.5 Waves ka principle of superposition

Ek lambi slinky ke opposite ends par khade do dost ek hi pal mein ek-ek pulse bhejte hain. Beech mein pulses milte hain. Tum shayad takkar expect karo, jismein dono pulses waapas uchhal jaayein ya ek doosre ko bigaad dein, lekin woh bas ek doosre ke aar-paar nikal jaate hain aur aage badhte rehte hain, har ek apne original shape aur speed ke saath. Jab tak woh overlap karte hain, slinky ek naya shape leti hai, aur woh shape predict karna aasaan hai: har point par dono pulses ke alag-alag displacements ko jod do (Fig. 14.8). Yeh rule principle of superposition hai.

2/3Jahan woh overlap karte hain, wahan displacement dono alag-alag displacements (dashed) ka sum hai.

Fig. 14.8Overlap karte pulses point by point jud jaate hain, phir aise aage badhte hain jaise kabhi mile hi na hon.NCERT Fig. 14.9 · p. 287

Ek horizontal string par do upward pulses hain: right ki taraf jaata ek lamba nukeela pulse A aur left ki taraf jaata ek neecha gol pulse B. Step 1: woh opposite sides se ek doosre ki taraf aate hain. Step 2: woh beech mein overlap karte hain; dashed outlines har pulse ko akele dikhate hain aur ek solid curve, jis par sum likha hai, string dikhati hai, jo dono mein se kisi bhi akele pulse se oonchi uthti hai. Step 3: pulses ek doosre ke aar-paar nikal gaye hain aur apni original directions mein apne original shapes ke saath aage badhte hain, ab A right par aur B left par.

PrincipleSuperposition ka principle

Jab do ya zyada waves medium ke same region se guzarti hain, to kisi bhi point aur instant par resultant displacement un displacements ka algebraic sum hota hai jo har wave akele wahan paida karti. Waves ke guzar jaane ke baad har wave doosron se bina prabhavit hue aage badhti rehti hai.

Sum mein signs matter karte hain. Agar ek crest-shaped pulse ulti taraf se aate ek jaise trough-shaped pulse se milta hai, to ek pal aata hai jab woh bilkul overlap karte hain aur string poori seedhi hoti hai. Energy gaayab nahi hui: us instant par string ka har hissa tezi se move kar raha hai, isliye saari energy kinetic hai. Ek pal baad pulses phir dikhte hain aur door chale jaate hain. Yeh principle interference ki chaabi hai: overlap karti waves ka combined effect.

Ab do sources ko same frequency ki harmonic waves same string par bhejne do, jismein ek doosre se cycle ke kuch hisse baad shuru hota hai. Tab dono waves ka same ω, same k aur isliye same wavelength aur speed hai, aur hum unke amplitudes a barabar lete hain. Farq sirf ek constant phase difference ϕ ka hai. String kya karti hai?

DerivationPhase mein alag do waves ko jodna

Short mein θ=kx−ωt likhte hue, do waves hain

y1=asin⁡θ
(14.27)
y2=asin⁡(θ+ϕ)
(14.28)

Superposition se,

y=asin⁡θ+asin⁡(θ+ϕ)
(14.29)

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2 use karo:

y=2asin⁡(θ+ϕ2)×cos⁡ϕ2
(14.30)

Factors ka order badalne par,

y=(2acos⁡ϕ2)×sin⁡(θ+ϕ2)
(14.31)

Doosra factor +x ki taraf jaati ek normal harmonic wave hai, original frequency aur wavelength ke saath aur ϕ/2 phase constant ke saath. Pehla factor uske amplitude ka kaam karta hai, aur yeh poori tarah is par depend karta hai ki dono waves ek doosre se kitni out of step hain:

Resultant amplitude
A(ϕ)=2acos⁡ϕ2
(14.32)

(Strictly, amplitude magnitude ∣2acos⁡(ϕ/2)∣ hai; negative value sirf phase ko π se shift karti hai.)

Do special cases khaas hain (Fig. 14.9). Jab waves in phase hon, ϕ=0 (ya 2π ka koi poora multiple), to crest crest se milta hai aur

y=2asin⁡(kx−ωt)
(14.33)

sabse bada possible amplitude, 2a: constructive interference. Jab woh bilkul out of phase hon, ϕ=π, to har jagah crest trough se milta hai aur

y=0
(14.34)

saare points par aur har time: destructive interference. In do extremes ke beech amplitude 2a se lekar zero tak har value leta hai; jaise ϕ=2π/3 se 2acos⁡60∘=a milta hai.

2/3ϕ=2π/3: aanshik reinforcement, amplitude 2acos⁡60∘=a.

Fig. 14.9Resultant same wavelength wali ek wave hai jiska amplitude 2acos⁡(ϕ/2) hai.NCERT Fig. 14.10 · p. 288

Upar: barabar amplitude wali ek solid neeli wave y1 aur ek dashed orange wave y2. Neeche: hara sum y1 plus y2, plus aur minus 2a par dashed guide lines ke saath. Step 1, phi equals 0: dono waves ek doosre ke upar padi hain aur sum ka amplitude 2a hai, jo guide lines ko chhoota hai. Step 2, phi equals two pi over three: dashed wave khisak gayi hai aur sum ka amplitude a hai, har wave ke barabar. Step 3, phi equals pi: crest trough se milta hai aur sum zero amplitude wali ek flat line hai.

Example

Do waves ek doosre ko kitna majboot karti hain?

3.0 mm amplitude wali do waves, same frequency aur wavelength ke saath, ek string par ek hi taraf travel karti hain. Jab unka phase difference π/3 ho tab resultant amplitude nikaalo, aur woh phase difference bhi jo resultant amplitude ko 3.0 mm bana de.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check
3 mein se 1
5.0 mm amplitude wali do waves, jo π/2 ke phase difference ke alawa bilkul ek jaisi hain, ek string par saath travel karti hain. Resultant wave ka amplitude mm mein kya hai?
mm
Answer dekho

Answer 7.07 mm

A=2acos⁡(ϕ/2)=2(5.0)cos⁡45∘=7.07 mm. Seedha jodkar 10 mm lena sirf in phase waves ke liye sahi hai.

14.6 Waves ka reflection

Asli media kahin na kahin khatam hote hain. Ghaati ke paar ek cliff ki taraf chillao, to cliff tumhari awaaz echo ki tarah waapas bhejti hai: rigid boundary se reflect hua sound. Ek halki string ko bhaari rassi se baandho aur string ko hilao: har wave ka kuch hissa rassi mein aage chala jaata hai aur kuch string ke along waapas aata hai. Aam taur par do media ke beech ki boundary wave ka kuch hissa reflect karti hai aur baaki transmit karti hai. Jab wave boundary se kisi angle par takraati hai, to transmitted (refracted) wave Snell's law ke hisaab se mudti hai aur reflected wave laws of reflection follow karti hai.

Ek string ko deewar se mazbooti se baandho aur uske along ek upward pulse bhejo (Fig. 14.10, upar). Agar deewar uski koi energy absorb nahi karti, to pulse apne shape ke saath waapas aata hai lekin ulta ho kar. Hum kehte hain ki uska π (yaani 180∘) ka phase change hua hai. Pehle forces ke baare mein socho. Pulse pahunchne par string ka aakhri hissa upar uthta hai, jo deewar ko upar kheenchta hai; Newton ke third law se deewar string ko neeche kheenchti hai, aur yahi neeche ka khinchaav woh ulta pulse hai jo waapas daudta hai. Superposition se bhi yahi jawab milta hai: bandha hua end kabhi move nahi kar sakta, isliye jo bhi waapas aaye use wahan har instant par aane wale displacement ko cancel karna chahiye, jo sirf ek ulti copy hi kar sakti hai.

3/3Deewar se pulse ulta lautta hai; free ring se seedha lautta hai.

Fig. 14.10Fixed end π ke phase change ke saath reflect karta hai; free end bina kisi phase change ke reflect karta hai.NCERT Fig. 14.11 · p. 289

Do strings right ki taraf jaati hain. Upar wali ek hatched deewar se bandhi hai, jis par fixed end likha hai. Neeche wali ek ring par khatam hoti hai jo ek chikni vertical rod par slide kar sakti hai, jis par free end likha hai. Step 1: har string par ek upward pulse right ki taraf travel karta hai. Step 2: pulses ends tak pahunchte hain; deewar par string zero par pinned hai aur pulse kuch had tak cancel ho jaata hai, jabki rod par ring pulse ki height ke lagbhag do guna tak upar uchhal jaata hai. Step 3: dono pulses waapas left ki taraf travel karte hain; deewar wala ab downward pulse hai, aur rod wala abhi bhi upward hai.

Iska ulta case woh end hai jo poori tarah free ho. Textbooks ise ek halki ring ki tarah model karti hain jo chikni vertical rod par bina friction ke slide karti hai (Fig. 14.10, neeche). Ring ko waapas neeche koi nahi kheenchta, isliye pulse use apni height ke lagbhag do guna tak upar phenk deta hai, aur reflection seedha waapas aata hai: koi phase change nahi. Sound ke liye flute ya organ pipe ka open end hawa ke displacement ke liye free end jaisa behave karta hai.

Ise likhne ke liye, ek harmonic wave yi=asin⁡(ωt−kx) (+x ke along ek wave) ko left se x=0 par ek boundary tak aane do. Reflected wave −x ke along travel karti hai aur, do tarah ke ends ke liye,

Rigid boundary
yr=asin⁡(ωt+kx+π)=−asin⁡(ωt+kx)
(14.35)
Free boundary
yr=asin⁡(ωt+kx)
(14.36)

x=0 par rigid case har time yi+yr=asin⁡ωt−asin⁡ωt=0 deta hai, bilkul fixed end ki condition. Free case 2asin⁡ωt deta hai: end do guna amplitude se oscillate karta hai. (Books inhe alag sign conventions ke saath likhti hain; physics same hai: rigid end par sum zero hota hai aur free end par do guna.)

Quick check
2 mein se 1
Ek pulse halki string ke along us point ki taraf travel karta hai jahan woh ek bhaari lohe ke khambe se mazbooti se bandhi hai. Reflected pulse hai
Answer dekho

Answer (B) ulta, same shape ke saath

Rigid end fixed rehta hai, isliye reflection mein π ka phase change hota hai: pulse ulta waapas aata hai. Do guna, seedha displacement free end ki baat hai.

14.6.1 Standing waves aur normal modes

Guitar string par, ya flute ke andar ki hawa mein, do boundaries hoti hain, aur ek wave unke beech kai baar aage-peeche reflect hoti hai. Second ke ek chhote hisse mein string par opposite directions mein jaati do wave trains hoti hain, aur unka sum ek steady pattern mein tik jaata hai jo kahin jaata hua nahi lagta: ek standing wave ya stationary wave (अप्रगामी तरंग).

DerivationDo opposite waves ek standing wave banati hain

Ek wave ko +x ke along chalne do jabki uska reflection, jo amplitude aur wavelength mein uske barabar hai, −x ke along chalta hai (Eqs. (14.2) aur (14.4) ϕ=0 ke saath). Jaisa humne upar dekha, yeh jodi bilkul wahi hai jo x=0 par ek rigid end paida karta hai:

y1=asin⁡(kx−ωt)y2=asin⁡(kx+ωt)

Inhe jodo, aur Section 14.5 mein use kiye gaye sin⁡P+sin⁡Q ke formula se do sines ke sum ko ek product mein badlo. Result neeche Eq. (14.37) hai.

Standing wave
y=2asin⁡kx cos⁡ωt
(14.37)

Eq. (14.37) mein position aur time alag ho gaye hain: ek factor sirf x par depend karta hai, doosra sirf t par, aur moving combination kx−ωt gaayab ho gaya hai. Ise ek oscillation cos⁡ωt ki tarah padho, jo string ke har hisse ke liye same hai, aur jise ek amplitude ∣2asin⁡kx∣ se multiply kiya gaya hai jo har hisse ke liye fixed hai lekin string ke along badalta hai. Kuch points, jahan sin⁡kx=0, kabhi move hi nahi karte: yeh nodes (निस्पंद) hain. Unke theek beech mein swing sabse bada hota hai: antinodes (प्रस्पंद). Do paas-paas ke nodes ke beech ke saare hisse ek saath uthte aur girte hain, jabki string ka agla section ulti taraf move karta hai, kyunki node par sin⁡kx ka sign badal jaata hai. Crests aage nahi badhte, isiliye 'standing'. Aur kyunki node kabhi move nahi karta, uske aar-paar koi work nahi hota: energy har loop ke andar aage-peeche chhalakti hai lekin string ke along aage nahi jaati.

InteractiveStanding-wave lab

Do views. Pehle mein right ki taraf jaati ek wave aur left ki taraf jaata uska reflection do rangon mein bane hain, aur unka sum teesre rang mein; Play dikhata hai ki sum fixed nodes ke saath apni jagah par oscillate karta hai. Doosre mein tum dono ends par fixed string, ek end par closed pipe ya dono ends par open pipe chunte ho, ek harmonic number chunte ho, aur length aur wave speed set karte ho; standing wave ka envelope uske nodes aur antinodes ke marks ke saath banta hai, aur readouts mode ki wavelength aur frequency aur allowed frequencies ki list dete hain.

Fig. 14.11Do opposite travelling waves ko ek standing wave banate dekho, phir ek string ya pipe chuno aur uske harmonics ko ek-ek karke dekho. Animate karne ke liye Play dabao.

Nodes wahan hain jahan sin⁡kx=0, yaani kx=nπ. k=2π/λ ke saath:

Nodes
x=nλ2n=0,1,2,…
(14.38)

Antinodes wahan hain jahan ∣sin⁡kx∣=1, yaani kx=(n+12)π:

Antinodes
x=(n+12)λ2n=0,1,2,…
(14.39)

Paas-paas ke nodes ek doosre se λ/2 door hote hain, paas-paas ke antinodes λ/2 door hote hain, aur har antinode do nodes ke theek beech mein hota hai, har ek se λ/4 door.

Ab asli fayda dekho. Ek kabhi na khatam hone wali string par tum jo chaho us frequency ki waves bhej sakte ho. Lekin x=0 aur x=L par bandhi string kahin zyada choosy hoti hai, kyunki dono ends nodes hone chahiye. x=0 wala end Eq. (14.37) mein pehle se hi node hai. x=L ko bhi node banane ke liye Eq. (14.38) kehti hai ki length mein half-wavelengths ki poori sankhya fit honi chahiye:

L=nλ2n=1,2,3,…
(14.40)

Isliye allowed wavelengths aur frequencies (ν=v/λ, aur n=1,2,3,… ke saath) hain

λn=2Ln
(14.41)
Dono ends par fixed string ke normal modes
νn=nv2L
(14.42)

Yeh natural frequencies string ke normal modes hain. Sabse neechi, ν1=v/2L, fundamental mode ya first harmonic hai; n=2 second harmonic deta hai, n=3 third, aur aage bhi isi tarah, har ek fundamental ka whole-number multiple (Fig. 14.12). Yahan v=T/μ string se fix hota hai, isiliye string ko kasne (zyada T) ya ungli se chhota karne (kam L) se uski har frequency badh jaati hai.

3/4Third harmonic: teen loops, L=3λ/2.

Fig. 14.12Dono ends par fixed string mein half-wavelengths ki poori sankhya fit hoti hai: νn=nν1.NCERT Fig. 14.13 · p. 291

Length L ki ek string dono ends par clamp hai. Har step string ki do extreme positions dikhata hai, ek solid aur ek dashed, neeche N se nodes aur upar A se antinodes marked hain, upar left mein wavelength aur upar right mein frequency. Step 1: ek loop, nodes sirf ends par, lambda equals 2L, nu 1 equals v over 2L, fundamental. Step 2: do loops aur beech mein ek node, lambda equals L, frequency 2 nu 1. Step 3: teen loops, lambda equals 2L over 3, frequency 3 nu 1. Step 4: chaar loops, lambda equals L over 2, frequency 4 nu 1.

Sitar ki string chhedo, to tumhe ek pure mode nahi milta. Ek saath kai milte hain, fundamental aur harmonics ki ek family, har ek ki apni strength ke saath. Kaunse harmonics strong honge, yeh is par depend karta hai ki string ko kahan chheda ya bow kiya gaya: jis harmonic ka us point par node hai, woh mushkil se excite hota hai. Players jaan-boojh kar iska use karte hain, aur harmonics ka yeh mel hi kaafi had tak woh cheez hai jo same note par veena, violin aur sitar ki awaaz ko alag banati hai.

Example

Veena string ko tune karna

Ek veena string ki vibrating length 0.90 m aur linear mass density 1.6 g/m hai. Kaunsa tension ise 220 Hz ki fundamental frequency dega, aur tab uske second harmonic ki frequency kya hogi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Air columns ke bhi normal modes hote hain. Test tube ya patli bottle ke munh par phoonko, to tumhe andar ka air column sunai deta hai, jo neeche closed aur upar open hai. Closed end par hawa glass se chipki rehti hai, isliye wahan displacement hamesha zero hota hai (displacement node); lekin woh layer upar ki hawa se kisi bhi aur jagah se zyada dabti aur khinchti hai, isliye pressure change wahan sabse zyada hota hai (pressure antinode). Open end par ulta hota hai: hawa free move karti hai aur pressure room ke pressure ke kareeb rehta hai, yaani displacement antinode aur pressure node. Closed end x=0 par lene se ek node pehle se hi wahan hai, aur phir Eq. (14.39) x=L par ek antinode maangti hai:

L=(n+12)λ2n=0,1,2,…

isliye allowed wavelengths aur normal-mode frequencies, n=0,1,2,… ke liye, hain

λ=2Ln+12
(14.43)
Ek end par closed pipe
ν=(n+12)v2L
(14.44)

Fundamental (n=0) v/4L hai: tube mein ek quarter-wavelength fit hoti hai. Upar ke modes 3v/4L, 5v/4L waghairah hain, sirf odd harmonics, jinhe third, fifth, … harmonics kehte hain. Dono ends par open pipe ke dono ends par displacement antinodes hote hain; unke beech half-wavelengths ki poori sankhya fit hoti hai, isliye usmein har harmonic hota hai, νn=nv/2L, bilkul string ki tarah (Fig. 14.13).

2/4Agle mode mein teen quarter-wavelengths fit hoti hain: ν=3v/4L. Koi second harmonic nahi hota.

Fig. 14.13Closed end displacement node hai, open end displacement antinode. Closed pipes mein sirf odd harmonics hote hain; open pipes mein saare.NCERT Figs 14.14–14.15 · p. 292

Length L ka ek horizontal pipe, jismein neeli solid aur dashed curves andar ki hawa ke displacement ka envelope dikhati hain, pipe ke neeche N se nodes aur A se antinodes marked hain, upar left mein length condition aur upar right mein frequency. Step 1: left end par closed; closed end par node aur open end par antinode, L equals lambda over 4, nu equals v over 4L. Step 2: closed pipe, third harmonic, open end se ek-tihaai doori par ek node, L equals 3 lambda over 4, nu equals 3v over 4L. Step 3: dono ends par open; dono ends par antinodes aur beech mein ek node, L equals lambda over 2, nu equals v over 2L. Step 4: open pipe, second harmonic, dono ends aur beech mein antinodes, quarter points par nodes, L equals lambda, nu equals 2v over 2L.

Strings aur air columns ko ek bahari periodic force se bhi chalaya ja sakta hai, Chapter 13 ke forced oscillators ki tarah. Jab driving frequency kisi natural frequency se lagbhag match karti hai, to amplitude bahut bada ho jaata hai aur system resonate karta hai. Ek vibrate karta tuning fork ek aisi tube ke upar pakdo jismein paani ka level dheere-dheere neeche kiya ja raha ho, to air column kuch khaas lengths par zor se goonjta hai; yahi laboratory ka resonance-tube experiment hai. Drum string ka two-dimensional version hai. Uski khaal poore rim par kas kar pakdi hoti hai, isliye har mode ka poore rim ke along node hona chahiye; frequencies nikaalna string se mushkil hai, lekin idea same hai.

Compare

Progressive aur standing waves

Progressive wave

  • Pattern speed v=ω/k se travel karta hai
  • Har point ka amplitude same a hota hai
  • Phase point se point steadily badalta hai
  • Medium ke along energy le jaati hai
  • Koi bhi frequency allowed hai

Standing wave

  • Pattern apni jagah par rehta hai
  • Amplitude ∣2asin⁡kx∣ node se antinode tak badalta hai
  • Ek loop ke saare points in phase; paas-paas ke loops ulte
  • Node ke aar-paar koi energy flow nahi hoti
  • Ends fixed hon to sirf normal-mode frequencies
Example 14.5NCERT Example 14.5 · p. 292

Kaunsa harmonic resonate karta hai?

25.0 cm lamba ek pipe ek end par closed hai. Uske open end ke paas ek loudspeaker steady 1020 Hz tone bajata hai. Sound ki speed 340 m/s lo. Kya pipe resonate karta hai, aur kis mode mein? Agar closed end khol diya jaaye to kya woh tab bhi resonate karega?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check
4 mein se 1
60 cm wavelength wali standing wave mein paas-paas ke nodes cm mein kitni door hote hain?
cm
Answer dekho

Answer 30 cm

Nodes λ/2=30 cm door hote hain. Ek node se sabse paas ke antinode ki doori λ/4=15 cm hai.

14.7 Beats

256 Hz aur 260 Hz ke do tuning forks ek saath bajao aur suno. Tumhe do notes sunai nahi dete. Ek sunai deta hai, beech ki pitch par, aur woh dhadakta hai: tez, dheema, tez, dheema, har second chaar baar. Yeh regular dhadkan, jo tab sunai deti hai jab barabar loudness aur thodi alag frequency ki do awaazein milti hain, beats (विस्पंद) kehlaati hai.

DerivationBeat frequency

Ek jagah, x=0, par dhyaan do aur barabar amplitude aur lagbhag barabar angular frequencies ω1>ω2 wali do sound waves ko wahan aane do. Har ek ke liye Eq. (14.9) mein phase constant ϕ=π/2 chunne par hawa ke displacements hain

s1=acos⁡ω1ts2=acos⁡ω2t
(14.45)

(Hum s likhte hain kyunki sound longitudinal hai.) Superposition aur cos⁡A+cos⁡B=2cos⁡A−B2cos⁡A+B2 se,

s=2acos⁡(ω1−ω22t)×cos⁡(ω1+ω22t)
(14.46)

ωb=(ω1−ω2)/2 aur ωa=(ω1+ω2)/2 likho:

s=[2acos⁡ωbt]×cos⁡ωat
(14.47)

Jab ∣ω1−ω2∣≪ω1, to ωa, ωb se bahut bada hota hai. Tab hawa average angular frequency ωa (jo pitch tum sunte ho) par tezi se oscillate karti hai, ek dheere-dheere badalte amplitude 2acos⁡ωbt ke andar. Awaaz tab sabse tez hoti hai jab cos⁡ωbt, +1 ya −1 ho, jo cos⁡ωbt ki har cycle mein do baar hota hai. Isliye loudness ke peaks angular frequency 2ωb=ω1−ω2 par dohraate hain, aur ω=2πν ke saath,

Beat frequency
νbeat=ν1−ν2
(14.48)

Fig. 14.14 ise 12 Hz aur 15 Hz ki do waves ke liye dikhata hai (sunne ki range se bahut neeche, lekin banana aasaan): unka sum ek second mein teen baar badhta aur ghatta hai. Musicians bilkul isi tarah kaan se tune karte hain: do sitars, ya tanpura aur harmonium, ko tab tak adjust kiya jaata hai jab tak dhadkan dheemi hokar ruk na jaaye. Kyunki formula ek difference deta hai, akeli beat frequency se yeh pata nahi chalta ki kaunsa source upar hai; Example 14.6 dikhata hai ki yeh kaise pata karein.

3/3Sum ek second mein 15−12=3 baar badhta aur ghatta hai: 3 beats per second.

Fig. 14.14Do paas-paas ki frequencies judkar aisi wave banati hain jiski loudness ν1−ν2 par badhti aur ghatti hai.NCERT Fig. 14.16 · p. 294

Ek second ke liye time ke saath teen graphs. Step 1: sirf upar wala graph, constant amplitude wali ek 12 hertz wave. Step 2: uske neeche same amplitude wali ek 15 hertz wave aati hai. Step 3: sabse neeche wala graph unka sum dikhata hai, jiska amplitude us second ke dauraan teen baar ek wave ke amplitude ke do guna tak badhta hai aur zero tak ghatta hai, uske chaaron taraf ek dashed envelope bana hai aur 0, 0.5 aur 1 second par time marks hain.

InteractiveBeats lab

Do sliders 200 aur 280 hertz ke beech do tone frequencies set karte hain. Ek graph chhote time window mein dono waves aur unka sum dikhata hai, beat envelope dashed bana hai, aur ek readout beat frequency aur average frequency deta hai. Play dono tones ko device speaker se halki volume par ek saath bajata hai, taaki awaaz ka badhna aur ghatna dekhne ke saath sunai bhi de; frequencies barabar hone par beats gaayab ho jaati hain.

Fig. 14.15Do frequencies chuno, unhe ek saath sunne ke liye Play dabao (halki volume rakho) aur unke sum ka envelope dekho. Ek frequency ko tab tak slide karo jab tak beats ruk na jaayein.
Example 14.6NCERT Example 14.6 · p. 294

Violin sharp hai ya flat?

Ek violinist apni A string ko 440 Hz ke harmonium reed ke saath tune karti hai aur 4 beats per second sunti hai. Woh string ko bahut thoda kasti hai aur beats tez hokar 6 per second ho jaati hain. Kasne se pehle string ki frequency kya thi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check
3 mein se 1
256 Hz aur 260 Hz ke tuning forks ek saath bajaaye jaate hain. Per second kitni beats sunai deti hain?
beats/s
Answer dekho

Answer 4 beats/s

νbeat=260−256=4 per second. Sunai dene wali pitch lagbhag average hai, 258 Hz.
Recap

Summary aur key points

  1. Mechanical wave ek disturbance hai jo material medium se travel karta hai, energy aur information le jaata hai lekin matter nahi; ise Newton ke laws aur medium ki elasticity control karte hain (Section 14.1).
  2. Transverse wave mein particles travel ki direction ke right angles par move karte hain; longitudinal wave mein uske along. Transverse waves ko aisa medium chahiye jo shear ka virodh kare; longitudinal waves solids, liquids aur gases mein travel karti hain (Section 14.2).
  3. +x ke along harmonic wave y(x,t)=asin⁡(kx−ωt+ϕ) hai, jismein amplitude a, angular wave number k, angular frequency ω aur phase kx−ωt+ϕ hai; kx+ωt se −x ke along wave milti hai (Section 14.3).
  4. λ=2π/k, T=2π/ω aur ν=1/T=ω/2π (Section 14.3.2, Section 14.3.3).
  5. Wave speed: v=ω/k=λ/T=νλ; medium v fix karta hai aur source ν fix karta hai (Section 14.4).
  6. Taani hui string par v=T/μ, jo frequency par depend nahi karti (Section 14.4.1).
  7. Longitudinal waves: fluid mein v=B/ρ aur patli bar mein v=Y/ρ; ideal gas mein v=γP/ρ (Laplace), Newton ka P/ρ nahi (Section 14.4.2).
  8. Superposition: overlap karti waves apne displacements ko algebraically jodti hain, y=∑fi(x−vt), aur bina badle aage nikal jaati hain (Section 14.5).
  9. Phase mein ϕ se alag do barabar waves 2acos⁡(ϕ/2) amplitude ki wave deti hain: ϕ=0,2π,… ke liye constructive, ϕ=π ke liye destructive (Section 14.5).
  10. Rigid end par reflection wave ko ulat deta hai (phase change π); free end par koi phase change nahi hota (Section 14.6).
  11. Opposite directions mein jaati do ek jaisi waves ek standing wave y=2asin⁡kxcos⁡ωt banati hain, jismein nodes aur antinodes λ/2 door hote hain (Section 14.6.1).
  12. Dono ends par fixed string, ya dono ends par open pipe, ke normal modes νn=nv/2L hain (saare harmonics); ek end par closed pipe ke ν=(n+12)v/2L hain (odd harmonics, fundamental v/4L). Har ek resonant frequency hai (Section 14.6.1).
  13. Thodi alag frequencies wali do waves νbeat=ν1−ν2 par beats banati hain, jo loudness ke badhne aur ghatne ki tarah sunai deti hain (Section 14.7).
Practice · 26 questions

Practice questions answers ke saath (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercise 14.1

    1.20 kg mass ki 15.0 m lambi nylon rassi do khambon ke beech 320 N ke tension se taani gayi hai. Ek end ko tez sideways jhatka diya jaata hai. Pulse ko doosre end tak pahunchne mein kitne seconds lagte hain?

    s
    Solution dikhao

    μ=1.20/15.0=0.0800 kg/m, isliye v=320/0.0800=63.2 m/s.

    t=15.0/63.2=0.237 s.

  2. NumericalMediumNCERT Exercise 14.2

    Ek pathar ko rest se ek kuein mein giraaya jaata hai jiski paani ki surface rim se 45.0 m neeche hai. Chhodne ke kitne seconds baad rim par chhapaak sunai deta hai? g=9.8 m/s2 aur sound ki speed 340 m/s lo.

    s
    Solution dikhao

    Girna: t1=2h/g=90.0/9.8=3.030 s.

    Sound: t2=45.0/340=0.132 s.

    Total: t=3.03+0.13=3.16 s. Sound ke travel time ko ignore karne se jawab lagbhag 4 % galat hota.

  3. NumericalEasyNCERT Exercise 14.3

    8.0 m lambi steel cable ka mass 4.0 kg hai. Kitne newtons ka tension is par transverse waves ko 340 m/s se travel karwayega, jo garam din par hawa mein sound ki speed hai?

    N
    Solution dikhao

    μ=4.0/8.0=0.50 kg/m.

    T=μv2=0.50×3402=5.78×104 N, lagbhag 58 kN.

  4. Short answerMediumNCERT Exercise 14.4

    In sawaalon ka jawab dene ke liye ideal gas ke liye Laplace ka formula v=γP/ρ use karo.

    1. Ek dost kehti hai ki hill station par sound tez travel karna chahiye kyunki wahan air pressure kam hai. Agar temperature same ho, to kya woh sahi hai?

      Solution dikhao

      AnswerNahi: fixed temperature par v pressure par depend nahi karta.

      ρ=PM/RT ke saath formula v=γRT/M ban jaata hai: pressure cancel ho jaata hai. Kam pressure ka matlab bilkul usi proportion mein kam density, isliye same temperature par speed nahi badalti.

    2. 0 °C par hawa mein sound 331 m/s se travel karta hai. 35 °C wali garmi ki dopahar mein iski speed m/s mein estimate karo.

      m/s
      Solution dikhao

      v∝T (absolute temperature), isliye v=331308/273=352 m/s.

    3. Same temperature aur pressure par sound sookhi hawa ki tulna mein nam hawa mein thoda tez kyun travel karta hai?

      Solution dikhao

      AnswerWater vapour average molar mass ko kam karti hai, isliye density ghat jaati hai.

      Water molecule (M=18 g/mol) average air molecule (29 g/mol) se halka hai. Kuch hawa ki jagah vapour aane se same P aur T par M aur isliye ρ kam hota hai, isliye v=γRT/M badhta hai (γ ka chhota change kam matter karta hai).

  5. Short answerHardNCERT Exercise 14.5

    Har one-dimensional travelling wave ko f(x−vt) ya f(x+vt) ki tarah likha ja sakta hai. Lekin aisa har function asli wave nahi ho sakta: asli medium ka displacement har jagah, har time finite bhi rehna chahiye. Decide karo ki har function, jismein A, b, k aur v positive constants hain, travelling wave ko represent kar sakta hai ya nahi.

    1. y=A e−(x−vt)2/b2

      Answer dekho

      Answer Sahi

      x−vt ka ek function jo 0 aur A ke beech rehta hai: +x ke along move karta ek bell-shaped pulse.
    2. y=A(x+vt)3

      Answer dekho

      Answer Galat

      Yeh x+vt ka function hai, lekin x+vt badhne par yeh bina kisi limit ke badhta jaata hai, isliye koi asli medium ise carry nahi kar sakta.
    3. y=Ab2b2+(x−vt)2

      Answer dekho

      Answer Sahi

      x−vt ka ek function jiska maximum A hai aur koi infinity nahi: +x ke along move karta ek pulse.
    4. y=Asin⁡[k(x2−v2t2)]

      Answer dekho

      Answer Galat

      x2−v2t2=(x−vt)(x+vt) mein dono combinations hain, isliye shape bas speed v se aage nahi khisakta.
  6. NumericalEasyNCERT Exercise 14.6

    Ek dolphin samudra ke paani mein 100 kHz ka click nikaalti hai, jahan sound 1530 m/s se travel karta hai. Click samudra ki surface tak pahunchta hai; kuch hissa waapas paani mein reflect hota hai aur kuch upar hawa mein chala jaata hai, jahan sound 343 m/s se travel karta hai.

    1. Reflected sound ki wavelength cm mein kya hai?

      cm
      Solution dikhao

      Reflected wave abhi bhi samudra ke paani mein hai, isliye λ=1530/1.00×105, jo 0.0153 m yaani 1.53 cm hai.

    2. Hawa mein transmit hue sound ki wavelength mm mein kya hai?

      mm
      Solution dikhao

      Same frequency, kam speed: λ=343/1.00×105, jo 3.43 mm hai.

  7. NumericalEasyNCERT Exercise 14.7

    Ek clinic ka ultrasound scanner 3.5 MHz par kaam karta hai. Soft tissue mein sound 1540 m/s se travel karta hai.

    1. Tissue mein ultrasound ki wavelength mm mein kya hai?

      mm
      Solution dikhao

      λ=1540/3.5×106, jo 4.4×10−4 m yaani 0.44 mm hai.

    2. Is wavelength ko aadha karne ke liye (zyada sharp image ke liye) MHz mein kitni frequency chahiye hogi?

      MHz
      Solution dikhao

      Same tissue mein λ∝1/ν, isliye frequency do guna karo: 7.0 MHz. (Zyada frequencies zyada absorb hoti hain, isliye woh utni gehraai tak nahi pahunch paati.)

  8. Short answerMediumNCERT Exercise 14.8

    String par ek transverse wave y(x,t)=4.0sin⁡(60t+0.050x−π/6) hai, jismein x aur y cm mein aur t s mein hain. Maan lo x right ki taraf badhta hai.

    1. Dikhao ki wave travel karti hai, aur nikaalo ki woh kitni tezi se (m/s mein) aur kis taraf move karti hai.

      Solution dikhao

      AnswerTravelling, 12 m/s, right se left.

      x aur t, 0.050x+60t ke roop mein aate hain, isliye yeh −x ke along, right se left, travelling wave hai.

      v=ω/k=60/0.050=1200 cm/s=12 m/s.

    2. Iska amplitude aur frequency kya hain?

      Solution dikhao

      Answer4.0 cm; 9.55 Hz

      a=4.0 cm; ν=ω/2π=60/2π=9.55 Hz.

    3. t=0 par x=0 par wave ka phase kya hai?

      Solution dikhao

      Answer−π/6

      x=0, t=0 par phase −π/6≈−0.52 rad hai.

    4. Do paas-paas ke crests ke beech sabse chhoti doori cm mein kya hai?

      cm
      Solution dikhao

      λ=2π/k=2π/0.050=126 cm, lagbhag 1.26 m.

  9. Short answerMediumNCERT Exercise 14.9

    Question 8 ki wave ke liye x=0, x=10 cm aur x=20 cm wale points ke displacement–time graphs sketch karo (ya describe karo).

    1. Graphs ka shape kya hai, aur unke amplitudes aur periods kya hain?

      Solution dikhao

      AnswerSine curves, har ek ka amplitude 4.0 cm, period 0.105 s.

      Fixed x par y=4.0sin⁡(60t+ϕx): har point SHM karta hai. Teeno ka amplitude 4.0 cm aur period 2π/60=0.105 s hai.

    2. Point se point motion kis baat mein alag hai: amplitude, frequency ya phase? Kitna?

      Solution dikhao

      AnswerSirf phase: har 10 cm par 0.50 rad.

      ϕx=0.050x−π/6: 0, 10 aur 20 cm par −0.52 rad, −0.02 rad aur +0.48 rad.

      Isliye sirf phase alag hai, har 10 cm par 0.50 rad, jo har graph ko time mein 0.50/60=8.3 ms side mein khisaka deta hai.

  10. NumericalMediumNCERT Exercise 14.10

    Ek travelling harmonic wave y(x,t)=3.0cos⁡2π(25t−0.0125x+0.20) hai, jismein x aur y cm mein aur t s mein hain. String par do points ek doosre se d doori par hain. Unke oscillations ke beech phase difference radians mein nikaalo, jab d ho

    1. 1.0 m

      rad
      Solution dikhao

      λ=1/0.0125=80 cm. Δϕ=2π(100/80)=2.5π=7.85 rad (π/2 wali hi oscillation state).

    2. 0.20 m

      rad
      Solution dikhao

      Δϕ=2π(20/80)=π/2=1.57 rad.

    3. λ/3

      rad
      Solution dikhao

      Δϕ=2π/3=2.09 rad.

    4. 3λ/2

      rad
      Solution dikhao

      Δϕ=3π=9.42 rad: dono points hamesha opposite directions mein move karte hain.

  11. NumericalMediumNCERT Exercise 14.11

    Dono ends par clamp ki gayi ek string ka transverse displacement y(x,t)=0.030sin⁡(2.5πx)cos⁡(200πt) hai, jismein x aur y m mein aur t s mein hain. String 1.2 m lambi hai aur uska mass 2.4×10−2 kg hai.

    1. Yeh kis tarah ki wave hai?

      Answer dekho

      Answer (B) standing

      x aur t alag factors mein aate hain, sin⁡kx aur cos⁡ωt: ek standing wave.
    2. Ise opposite directions mein jaati do waves mein todo, aur har ek ka amplitude, wavelength, frequency aur speed batao.

      Solution dikhao

      Answer0.015 m, 0.80 m, 100 Hz, 80 m/s

      2sin⁡Acos⁡B=sin⁡(A−B)+sin⁡(A+B) use karne par displacement y1+y2 hai, jahan y1=0.015sin⁡(2.5πx−200πt) aur y2=0.015sin⁡(2.5πx+200πt).

      Har ek ke liye a=0.015 m, λ=2π/2.5π=0.80 m, ν=200π/2π=100 Hz aur v=νλ=80 m/s.

    3. String kitni taani hui hai? Tension newtons mein batao.

      N
      Solution dikhao

      μ=2.4×10−2/1.2=0.020 kg/m; T=μv2=0.020×802=128 N.

  12. Short answerMediumNCERT Exercise 14.12

    Question 11 ki string ke liye:

    1. Alag-alag points (nodes ko chhodkar) jin frequencies par oscillate karte hain, unhe compare karo.

      Solution dikhao

      AnswerHaan, 100 Hz.

      Haan: har point mein factor cos⁡(200πt) hai, isliye sab 100 Hz par oscillate karte hain. x=0, 0.40, 0.80 aur 1.2 m par nodes bilkul move nahi karte.

    2. Kisi given instant par unke phases compare karo.

      Solution dikhao

      AnswerEk loop ke andar in phase; paas-paas ke loops mein opposite phase.

      Ek loop ke andar (paas-paas ke nodes ke beech) ke points apne extremes par ek saath pahunchte hain. Paas-paas ke loops ke points ulta move karte hain, kyunki wahan sin⁡(2.5πx) ke signs ulte hain.

    3. Unke amplitudes compare karo.

      Solution dikhao

      AnswerNahi; yeh nodes par 0 se lekar antinodes par 0.030 m tak badalta hai.

      Nahi: amplitude ∣0.030sin⁡(2.5πx)∣ nodes par zero hai aur antinodes x=0.20, 0.60, 1.0 m par 0.030 m hai.

    4. Ek end se 0.10 m door wala point apni rest position se zyada se zyada kitne cm door jaata hai?

      cm
      Solution dikhao

      0.030sin⁡(0.25π)=0.030×0.707, jo 0.0212 m hai, lagbhag 2.1 cm.

  13. Quick answerMediumNCERT Exercise 14.13

    Har function ek elastic medium ka displacement deta hai (constants suitable units mein). Har ek ko travelling wave, standing wave, ya dono mein se koi nahi, mein classify karo.

    1. y=0.5cos⁡(4x)sin⁡(12t)

      Answer dekho

      Answer (B) standing

      cos⁡4x aur sin⁡12t ka product: ek standing wave.
    2. y=3sin⁡(2x+7t)+2cos⁡(2x+7t)

      Answer dekho

      Answer (A) travelling

      Dono terms sirf 2x+7t par depend karte hain; milkar yeh 13≈3.6 amplitude ki ek harmonic wave hain jo −x ke along 3.5 units se travel karti hai.
    3. y=(x−2t)2+1

      Answer dekho

      Answer (C) dono mein se koi nahi

      Yeh x−2t ka function hai, lekin yeh bina limit ke badhta jaata hai, isliye yeh asli wave ko describe nahi kar sakta.
    4. y=sin⁡(3x)+cos⁡(2t)

      Answer dekho

      Answer (C) dono mein se koi nahi

      Ek fixed shape aur poore medium ka ek uniform oscillation: kuch bhi travel nahi karta, aur yeh standing-wave form ka bhi nahi hai.
  14. NumericalMediumNCERT Exercise 14.14

    Ek sonometer wire ki vibrating length apna fundamental 120 Hz par bajati hai. Wire ki us length ka mass 6.0 g hai, aur wire ka mass per metre 5.0 g hai.

    1. Wire ke along transverse waves m/s mein kitni tezi se travel karti hain?

      m/s
      Solution dikhao

      L=6.0/5.0=1.20 m, isliye λ=2.40 m aur v=νλ=120×2.40=288 m/s.

    2. Wire mein tension newtons mein kitna hai?

      N
      Solution dikhao

      T=μv2=(5.0×10−3)(288)2=415 N.

  15. NumericalMediumNCERT Exercise 14.15

    Ek resonance-tube experiment mein, ek lambi tube mein paani ke upar ka air column 512 Hz ke tuning fork ke saath tab resonate karta hai jab column 16.0 cm lamba hota hai, aur phir se jab woh 49.6 cm lamba hota hai.

    1. Hawa mein sound ki speed m/s mein estimate karo.

      m/s
      Solution dikhao

      λ/2=49.6−16.0=33.6 cm, isliye λ=67.2 cm.

      v=νλ=512×0.672=344 m/s. Difference lene se end effect cancel ho jaata hai.

    2. Doosri length pehli ki bilkul teen guna nahi hai. Data se end correction (open end ke aage ki extra effective length) cm mein nikaalo.

      cm
      Solution dikhao

      Pehli resonance ke liye L1+e=λ/4=16.8 cm, isliye e=0.8 cm. Check: L2+e=50.4 cm=3λ/4.

  16. NumericalMediumNCERT Exercise 14.16

    Aluminium rod ko rosin lage kapde se lambai mein ragadne par woh goonjne lagti hai. Rod 1.50 m lambi hai aur apne midpoint par ek clamp se pakdi hui hai, aur uska sabse neecha longitudinal note 1.70 kHz hai. Aluminium mein sound ki speed m/s mein nikaalo.

    m/s
    Solution dikhao

    Fundamental mein rod ke ends par antinodes aur centre par ek node hota hai, isliye usmein aadhi wavelength fit hoti hai: λ=2L=3.00 m.

    v=νλ=1700×3.00=5100 m/s, jo Table 14.2 mein aluminium ke Y/ρ se milta hai.

  17. NumericalMediumNCERT Exercise 14.17

    50 cm lamba ek pipe dono ends par open hai. Sound ki speed 340 m/s lo.

    1. 1020 Hz ka source pipe ka kaunsa harmonic excite karta hai?

      Answer dekho

      Answer (C) teesra

      ν1=v/2L=340 Hz aur saare harmonics hote hain, isliye 1020 Hz=3ν1 teesra harmonic hai.
    2. Agar ab ek end band kar diya jaaye, to kya wahi source phir bhi resonance paida karega? Samjhao.

      Solution dikhao

      AnswerNahi.

      Ek end par closed: ν1=v/4L=170 Hz aur sirf odd multiples hote hain: 170, 510, 850, 1190 Hz, … 1020 Hz=6×170 Hz ek even multiple hai, isliye koi resonance nahi hota.

  18. NumericalMediumNCERT Exercise 14.18

    Ek tabla player apne drum ko 280 Hz ke harmonium note ke saath tune karti hai aur 4 beats per second sunti hai. Woh drumhead ko thoda dheela karti hai aur beats dheemi hokar 2 per second ho jaati hain. Dheela karne se pehle drum ki frequency Hz mein kya thi?

    Hz
    Solution dikhao

    Pehle: 276 ya 284 Hz. Dheela karne se drum ki pitch kam hoti hai. Beats dheemi huin, isliye woh 280 Hz ki taraf gaya, jiska matlab hai ki woh upar se shuru hua tha: 284 Hz (ab 282 Hz). Agar woh 276 Hz hota, to dheela karne se beats tez ho jaatin.

  19. Long answerMediumNCERT Exercise 14.19 (a)–(c)

    Samjhao:

    1. sound wave mein zero displacement wala point sabse zyada pressure variation wala point kyun hota hai (aur ulta bhi);

      Solution dikhao

      Displacement node par dono taraf ki layers ek saath uski taraf move karti hain (wahan hawa ko compress karti hain) aur phir ek saath door jaati hain (use rarefy karti hain), isliye pressure sabse zyada jhoolta hai. Displacement antinode par paas-paas ki layers ek saath same direction mein move karti hain, isliye wahan hawa mushkil se dabti hai: pressure mein thoda sa change.

    2. chamgaadad andhere mein kisi rukawat ki doori, direction, size aur nature ka andaaza kaise laga sakta hai;

      Solution dikhao

      Chamgaadad chhote ultrasonic pulses bhejta hai aur echoes sunta hai (echolocation). Delay se doori milti hai (d=vt/2); uske do kaanon par pahunchne ke time aur loudness ka farq direction deta hai; echo kitna strong hai aur frequency ke saath kaise badalta hai, isse size aur texture pata chalte hain. Ultrasound ki bahut chhoti wavelengths (kuch mm) chhote insects se bhi achhi tarah reflect hoti hain.

    3. same note bajate flute aur shehnai ko alag kaise pehchana ja sakta hai.

      Solution dikhao

      Dono same fundamental paida karte hain, jo pitch fix karta hai, lekin har ek alag strengths ke saath harmonics ka alag mix bhi paida karta hai. Yeh mix (tone quality ya timbre) sound wave ke shape ko alag banata hai, aur kaan ise pehchaan leta hai.

  20. Short answerMediumNCERT Exercise 14.19 (d)–(e)

    Samjhao:

    1. solids longitudinal aur transverse dono waves kyun carry karte hain, jabki gases sirf longitudinal waves;

      Solution dikhao

      Solid compression aur shear dono ka virodh karta hai, isliye woh dono tarah ke strain ke against waapas push kar sakta hai. Gas compression ka virodh karti hai (uska bulk modulus hota hai) lekin uska koi shear modulus nahi hota: uski layers ek doosre ke paas se free hokar phisalti hain, isliye transverse motion ke liye koi restoring force nahi hota.

    2. dispersive medium se travel karte waqt pulse ka shape kyun badalta hai.

      Solution dikhao

      Pulse alag-alag frequencies ki kai harmonic waves ka superposition hai. Dispersive medium mein yeh alag speeds se travel karti hain, isliye woh step se bahar ho jaati hain aur sum (pulse ka shape) phail kar badal jaata hai.

  21. Quick answerEasy

    Taani hui string ka tension 44 % badhaya jaata hai. Us par transverse waves ki speed kitne percent badhti hai?

    Answer dekho

    Answer (B) 20 %

    1.44=1.20, yaani 20 % ki badhat. Percentage ko aadha karna (22 %) sirf chhote change ka approximation hai, aur 44 % square root bhool jaata hai.
  22. Quick answerMedium

    Assertion (A): String par standing wave mein node ke aar-paar koi energy nahi jaati. Reason (R): Node kabhi move nahi karta, isliye wahan ek taraf ki string doosri taraf ki string par koi work nahi karti.

    Answer dekho

    Answer (A) A aur R dono true hain, aur R, A ko samjhata hai

    Kisi point ke aar-paar work karne ki rate force guna velocity hai; node par velocity hamesha zero hoti hai, isliye koi energy use paar nahi karti. R hi A ka reason hai.
  23. Quick answerEasy

    Ek end par closed sabse chhota pipe kaunsa hai jiski fundamental frequency 425 Hz ho? v=340 m/s lo aur length cm mein batao.

    cm
    Answer dekho

    Answer 20 cm

    L=v/4ν=340/1700=0.20 m=20 cm.
  24. Quick answerEasy

    Har quantity ko uski SI unit se match karo.

    Left mein koi item tap karo, phir right mein uska match.

    Answer dekho
    Answer
    • angular wave number : rad m−1
    • angular frequency : rad s−1
    • frequency : Hz
    • linear mass density : kg m−1
    k per metre phase hai, ω per second phase hai, ν per second oscillations hai, aur μ per metre mass hai.
  25. Quick answerMedium

    Same frequency aur wavelength wali do harmonic waves, jinke amplitudes 3a aur a hain, ek string par saath travel karti hain. Sabse bade aur sabse chhote possible resultant amplitude ka ratio kya hai?

    Answer dekho

    Answer (A) 2

    Sabse bada 3a+a=4a (in phase), sabse chhota 3a−a=2a (opposite phase): ratio 2. 4 ka ratio in amplitudes ke squares ka ratio hota.
  26. Quick answerMedium

    Hawa ko ideal gas maankar usmein sound ke liye kaunse statements true hain?

    Saare sahi options chuno
    Answer dekho
    Answer
    • (A) Yeh temperature ke saath badhti hai
    • (B) Constant temperature par pressure badalne se yeh nahi badalti
    • (D) Nam hawa mein yeh thodi zyada hoti hai
    v=γRT/M, T ke saath badhti hai, ismein P nahi hai, aur M ghatne par (nam hawa) badhti hai. Hawa sound ke liye dispersive nahi hai, isliye saari sunai dene wali frequencies same speed se travel karti hain; warna door ka music gadbad hokar pahunchta.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook meinYahan practice question
NCERT Exercise 14.1Practice Q1
NCERT Exercise 14.2Practice Q2
NCERT Exercise 14.3Practice Q3
NCERT Exercise 14.4Practice Q4
NCERT Exercise 14.5Practice Q5
NCERT Exercise 14.6Practice Q6
NCERT Exercise 14.7Practice Q7
NCERT Exercise 14.8Practice Q8
NCERT Exercise 14.9Practice Q9
NCERT Exercise 14.10Practice Q10
NCERT Exercise 14.11Practice Q11
NCERT Exercise 14.12Practice Q12
NCERT Exercise 14.13Practice Q13
NCERT Exercise 14.14Practice Q14
NCERT Exercise 14.15Practice Q15
NCERT Exercise 14.16Practice Q16
NCERT Exercise 14.17Practice Q17
NCERT Exercise 14.18Practice Q18
NCERT Exercise 14.19 (a)–(c)Practice Q19
NCERT Exercise 14.19 (d)–(e)Practice Q20

Quick check: MCQs

  1. 14.1 Introduction3 questions
  2. 14.2 Transverse aur longitudinal waves3 questions
  3. 14.3.3 Period, angular frequency aur frequency4 questions
  4. 14.4.2 Longitudinal wave ki speed (sound ki speed)4 questions
  5. 14.5 Waves ka principle of superposition3 questions
  6. 14.6 Waves ka reflection2 questions
  7. 14.6.1 Standing waves aur normal modes4 questions
  8. 14.7 Beats3 questions
Glossary · 15

Key terms (English + Hindi)

amplitudeआयामaayaam
Medium ke kisi particle ka apni equilibrium position se sabse bada displacement; hamesha positive.Chapter mein pehli baar yahan aaya →
angular wave numberकोणीय तरंग संख्याkoniya taranga sankhya
k=2π/λ, per unit length phase change; SI unit rad m−1.Chapter mein pehli baar yahan aaya →
antinodeप्रस्पंदpraspand
Standing wave ka sabse bade amplitude wala point, do nodes ke theek beech mein.Chapter mein pehli baar yahan aaya →
beatsविस्पंदvispand
Thodi alag frequencies ki do awaazein milne par sunai dene wala loudness ka regular badhna aur ghatna, ν1−ν2 per second.Chapter mein pehli baar yahan aaya →
interferenceव्यतिकरणvyatikaran
Overlap karti waves ka combined effect: in phase hone par majbooti, opposite phase mein hone par cancellation.Chapter mein pehli baar yahan aaya →
longitudinal waveअनुदैर्ध्य तरंगanudairghya taranga
Aisi wave jismein particles travel ki direction ke along move karte hain, aur compressions aur rarefactions banate hain.Chapter mein pehli baar yahan aaya →
nodeनिस्पंदnispand
Standing wave ka woh point jo kabhi move nahi karta (zero amplitude).Chapter mein pehli baar yahan aaya →
normal mode
Ek natural frequency jis par ek bounded system (string, air column) standing wave mein vibrate kar sakta hai; sabse neechi fundamental hai.Chapter mein pehli baar yahan aaya →
phaseकलाkalaa
Harmonic wave ka argument kx−ωt+ϕ, jo kisi given jagah aur time par displacement fix karta hai.Chapter mein pehli baar yahan aaya →
principle of superpositionअध्यारोपण का सिद्धांतadhyaropan ka siddhant
Jahan waves overlap karti hain, wahan har point par displacement un displacements ke sum (signs ke saath) ke barabar hota hai jo har wave akele paida karti.Chapter mein pehli baar yahan aaya →
progressive waveप्रगामी तरंगpragaami taranga
Aisi wave jiska pattern medium mein aage badhta hai aur apne saath energy le jaata hai.Chapter mein pehli baar yahan aaya →
standing waveअप्रगामी तरंगapragaami taranga
Opposite directions mein travel karti do ek jaisi waves se bana pattern 2asin⁡kxcos⁡ωt; yeh medium ke along aage nahi badhta.Chapter mein pehli baar yahan aaya →
transverse waveअनुप्रस्थ तरंगanuprastha taranga
Aisi wave jismein medium ke particles travel ki direction ke right angles par move karte hain.Chapter mein pehli baar yahan aaya →
waveतरंगtaranga
Ek travelling disturbance jo energy ko ek jagah se doosri jagah le jaata hai, jabki medium sirf fixed positions ke aas-paas oscillate karta hai.Chapter mein pehli baar yahan aaya →
wavelengthतरंगदैर्ध्यtarangdairghya
Ek instant par same phase wale do points ke beech sabse chhoti doori, λ=2π/k.Chapter mein pehli baar yahan aaya →

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