Oscillations: Class 11 Physics Chapter 13 Hinglish mein

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

Published 22 Oct 2026 · Updated 22 Oct 2026

80 min
24 practice questions
24 MCQs

Is chapter mein aap kya seekhenge

Periodic aur oscillatory motion mein farq samjho aur period, frequency aur angular frequency ko aapas mein jodo
Simple harmonic motion ko uske amplitude, angular frequency, phase aur phase constant se describe karo, aur tay karo ki time ka koi diya hua function simple harmonic hai, periodic hai ya dono mein se koi nahi
Reference circle ki madad se pata karo ki oscillate karte particle ki position, velocity aur acceleration time ke saath kaise badalti hain, aur unke phases compare karo
Force law F = −kx ko state karo aur use karo, spring–mass systems ka period nikalo, jismein vertical springs aur series ya parallel mein jude springs bhi shaamil hain
Simple harmonic oscillator ki kisi bhi position par kinetic, potential aur total energy calculate karo
Chhote swings ke liye simple pendulum ka period derive karo aur samjhao ki small-angle approximation kab sahi rehta hai
SHM par board aur entrance-level ke problems solve karo, jismein initial conditions aur effective gravity bhi shaamil hain

NCERT Physics Class 11, Chapter 13, pp. 259–277 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

13.1 Introduction

Kuch motions ek baar hote hain aur khatam ho jaate hain. Fielder ki taraf phenki gayi ball apna arc follow karke gir jaati hai; ek train Pune se Mumbai tak chalti hai aur ruk jaati hai. Is course ke pichhle chapters ke zyadatar motions isi ek-baar wale type ke the. Kuch doosre motions baar-baar wahin laut aate hain jahan se shuru hue the: Earth ke chaaron taraf ghoomta Moon, Sun ke chakkar lagata planet, ceiling fan ke blade ka tip. Aise motions periodic hote hain.

Yeh chapter baar-baar hone wale motions ki ek khaas family ko dekhta hai, jismein koi body ek beech wali position ke dono taraf aage-peeche hoti hai. Speed breaker par se gaadi chalao, to car apne springs par ek-do second tak upar-neeche hilti rehti hai. Diver springboard ke sire se kood jaata hai aur board kaanpta rehta hai. Tanpura ka taar chhedo, to vibrate karte hue woh dhundhla dikhta hai. Inmein se har ek oscillation hai (Fig. 13.1).

4/4Alag-alag objects, ek hi pattern: beech wali position ke aas-paas aage-peeche.

Fig. 13.1Teen oscillators jo tumne aaj dekhe honge. Har ek beech wali position ke dono taraf aage-peeche hota hai.

Teen panels, har ek mein dashed outlines do extreme positions dikhati hain aur ek double arrow aage-peeche ka motion. Step 1 mein ek car highlight hoti hai jiski body wheels ke upar lage springs par upar-neeche hilti hai. Step 2 mein ek end par fixed diving springboard highlight hota hai jiska free end upar-neeche mudta hai. Step 3 mein do khoontiyon ke beech taana hua taar highlight hota hai jo beech mein upar-neeche mudta hai. Step 4 mein teeno saath dikhte hain: har ek beech wali position ke dono taraf aage-peeche hota hai.

Physics ke kaafi bade hisse ke neeche oscillations chhupe hain. Jab bhi koi sound tumhare eardrum tak pahunchta hai, woh aage-peeche hota hai. Wristwatch mein time rakhne wala quartz crystal ek second mein dasiyon hazaar baar vibrate karta hai. India mein wall socket ka voltage har second $50$ poore cycles mein badhta aur ghatta hai. Aur ek bilkul shaant dikhne wale solid ke andar har atom crystal mein apni jagah ke aas-paas hilta rehta hai, aur solid jitna garam ho, utni zor se.

Inmein se kisi ko bhi describe karne ke liye humein ek chhoti, exact vocabulary chahiye: period, frequency, displacement, amplitude aur phase. Agla section yeh vocabulary banata hai, aur baaki chapter isse sabse simple aur sabse important oscillation, simple harmonic motion, ko study karta hai.

Quick check

2 mein se 1

Inmein se kaun sa motion khud ko repeat nahi karta?

fielder ki taraf ek baar phenki gayi ballEarth ke chaaron taraf ghoomta Moonghoomte fan blade ka tipjhoolta pendulum

Answer dekho

Answer (A) fielder ki taraf ek baar phenki gayi ball

Throw ek hi baar hota hai: ball apna arc follow karti hai aur pakad li jaati hai. Moon, fan ka tip aur pendulum teeno baar-baar usi state mein laut aate hain.

13.2 Periodic aur oscillatory motions

Car ke turn indicator ko dekho: on, off, on, off, har baar same timing ke saath. Ya wall clock ki second hand ko follow karo, jiska angle ek minute tak lagataar badhta hai aur phir twelve se dobara shuru hota hai. Agar tum inke behaviour ko time ke against plot karo, to graph ek hi tukde ko baar-baar copy karke bana hota hai (Fig. 13.2). Periodic motion ka test yahi hai: ek fixed interval ke baad motion khud ko bilkul waisa hi repeat karta hai.

3/4Complicated shape bhi periodic hai agar woh bilkul waise hi repeat ho.

Fig. 13.2Teen periodic motions aur ek jo periodic nahi hai. Har periodic graph mein

T

duration ka ek hi tukda end to end copy hota hai.NCERT Fig. 13.1 · p. 260

Har step same time axis par ek graph dikhata hai. Step 1: indicator light on aur off hoti hai, jisse ek jaise rectangular pulses ki row banti hai, ek repeat par T mark hai. Step 2: clock ki second hand ka angle seedhi line mein badhta hai aur wapas zero par girta hai, ek sawtooth, ek tooth par T mark hai. Step 3: ek heartbeat trace jismein ek chhota bump, ek lamba spike aur ek gol hump hai, bilkul waise hi repeat hota hai, ek repeat par T mark hai. Step 4: akela chhoda gaya jhoola ghatti height ke saath aage-peeche hota hai, isliye graph ka koi bhi hissa kabhi bilkul waisa repeat nahi hota.

Definition

Periodic motion

Aisa motion jo barabar time intervals ke baad khud ko bilkul waisa hi repeat karta hai.

ExampleEarth ka apne axis par ghoomna, jo har din repeat hota hai.

Kuch repeat hone wale motions aage-peeche kyun hote hain? Valley jaise shape ke track mein ek kancha daalo (Fig. 13.3). Sabse neeche wale point par dheere se rakho, to woh wahin tika rehta hai: wahan gravity aur track ka push cancel ho jaate hain, isliye yeh uski equilibrium position hai. Use kisi bhi slope par thoda upar le jao to balance toot jaata hai; bacha hua force hamesha neeche bottom ki taraf point karta hai. Yeh restoring force neeche aate waqt kanche ki speed badhata hai, woh aage nikal jaata hai, doosri taraf chadhta hai, phir se wapas mod diya jaata hai, aur ek oscillation shuru ho jaata hai.

2/4Right slope par upar, net force wapas bottom ki taraf point karta hai.

Fig. 13.3Valley mein kancha isliye oscillate karta hai kyunki har displacement ek aisa force laata hai jo bottom ki taraf wapas point karta hai. Hilltop par force door point karta hai, isliye koi oscillation nahi hota.

Valley jaisa ek curved track. Step 1: ek kancha sabse neeche wale point par ruka hai, label hai at rest, net force zero. Step 2: kancha right slope par upar hai aur track ke along ek arrow wapas neeche bottom ki taraf point karta hai, label hai pulled back. Step 3: left slope par bhi yahi, arrow right ki taraf point karta hai. Step 4: valley ki jagah ek hill aa jaati hai; top se thoda hatkar rakhe kanche par arrow top se aur door point karta hai, label hai pushed away, no oscillation.

Definition

Oscillatory motion

Kisi body ka equilibrium position ke dono taraf periodic aage-peeche wala motion, jo aise force se chalta hai jo hamesha usi position ki taraf wapas point karta hai.

ExampleBachche ka jhoola, ya spring par uchhalta weight.

Har oscillation periodic hota hai, lekin ulta sach nahi hai. Fan blade ka tip har chakkar mein apna path repeat karta hai, phir bhi woh kisi line ke along kabhi direction nahi badalta, isliye woh periodic hai par oscillatory nahi. Apne orbit mein ghoomte planet ke saath bhi yahi baat hai.

Physics mein oscillation aur vibration ka matlab ek hi hai. Kaun sa word use hoga, yeh aadat tay karti hai: toofan mein jhoomti oonchi building, har jhoom mein kuch seconds, ke liye kehte hain ki woh oscillate karti hai, jabki tuning fork ke prongs, jo ek second mein saikdon baar aage-peeche hote hain, ke liye kehte hain ki woh vibrate karte hain.

Jhoole ko akela chhod do to har arc pichhle se thoda neecha hota hai, jab tak woh ruk kar latak nahi jaata. Hooks par friction aur hawa ka drag uski energy kha jaate hain; is dheeme padne ko damping kehte hain. Har cycle mein sahi time par push dene wale parent khoi hui energy wapas de dete hain aur jhoola chalta rehta hai. Is chapter ke zyadatar hisse mein hum damping ko ignore karte hain aur sabse saaf case, simple harmonic motion, study karte hain, jismein restoring force equilibrium se doori ke bilkul proportion mein badhta hai. Ek asli, dheere-dheere damp hone wala oscillator aisa behave sirf utne time tak karta hai jo uske ruk jaane mein lagne wale time se bahut kam ho.

Oscillators shayad hi kabhi akele hote hain. Ek taane hue taar par beads ki row baandho, to pehle bead ko diya gaya jhatka agle bead tak, phir uske agle tak pahunch jaata hai. Jude hue oscillators ki chain mein is tarah aage badhne wala disturbance ek wave hai. Hawa mein sound, earthquake ke baad zameen ka hilna aur light, sab waves hain, aur yahi agle chapter ka topic hain.

13.2.1 Period aur frequency

Woh sabse chhota time jiske baad periodic motion repeat hota hai, uska period hai, jise $T$ likhte hain. Iski SI unit second hai. Jo motion $T$ ke baad repeat hota hai, woh $2 T$ , $3 T$ aur $T$ ke har whole-number multiple ke baad bhi repeat hota hai, lekin sirf sabse chhote time ko period kehte hain.

Nature mein periods ki range bahut badi hai. $512 H z$ mark wala tuning fork har vibration lagbhag $2$ milliseconds mein poora karta hai, jabki Sun par sunspots lagbhag $11$ saal ke cycle mein badhte-ghatte hain aur Saturn ko Sun ka ek chakkar lagane mein lagbhag $29$ saal lagte hain. Isliye aksar yeh batana aasaan hota hai ki har second mein kitne repeats hote hain. Is number ko frequency kehte hain, jise $ν$ (Greek letter nu) likhte hain:

Frequency aur period

ν = \frac{1}{T}

(13.1)

Frequency $s^{- 1},$ yaani repeats per second, mein naapi jaati hai. Is unit ka apna naam hai, hertz (Hz):

Hertz

1 H z = 1 s^{- 1}

(13.2)

Frequency ka whole number hona zaroori nahi. Jo pendulum $10 s$ mein $7$ swings poore karta hai, uske liye $ν = 0.7 H z;$ aisa koi niyam nahi ki ek poora swing theek second khatam hote hi poora ho.

Example 13.1NCERT Example 13.1 · p. 261

Ek fan aur ek crystal

(a) Ek ceiling fan $360$ revolutions per minute par steady ghoomta hai. Uske motion ki frequency aur period nikalo. (b) Wristwatch ka quartz crystal har second $32 768$ baar vibrate karta hai. Uska period nikalo.

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

13.2.2 Displacement

Ab tak displacement ka matlab position mein badlaav tha. Oscillations ke liye iska matlab thoda wide karna padta hai. Yahan displacement woh amount hai jitna koi repeat hone wali cheez kisi pal apni equilibrium value se alag hai. Equilibrium se naapna ek choice hai, lekin lagbhag hamesha sabse convenient choice.

To spring par rakhe block ka displacement uski woh doori $x$ hai jo spring ki relaxed position se hai, aur pendulum ke liye sabse natural displacement string aur vertical ke beech ka angle $θ$ hai. Yeh zaroori nahi ki yeh length ya angle hi ho. Mains supply ke liye yeh socket ke do pins ke beech ka voltage hai, jo positive aur negative mein jhoolta hai; tumhare kaan tak pahunchne wale sound ke liye yeh hai ki air pressure apni normal value se kitna upar uthi ya neeche giri hai (Fig. 13.4). Displacements ka sign hota hai, aur oscillation study karne ka matlab hai record karna ki displacement time ke saath kaise badalta hai.

5/5Alag-alag quantities, ek idea: equilibrium se displacement jo repeat hota hai.

Fig. 13.4Displacement koi bhi oscillate karti quantity hai jo apni equilibrium value se naapi jaaye: position, angle, voltage ya pressure.NCERT Fig. 13.2 · p. 261

Grid mein chaar panels. Upar left: deewar ke paas spring par ek block, dashed equilibrium line se right ki taraf x doori par displaced. Upar right: ek pendulum jiski string dashed vertical line ke saath theta angle banati hai. Neeche left: ek AC source jo wave sign wale circle ki tarah bana hai, saath mein uske badhte-ghatte voltage ka chhota graph. Neeche right: hawa ki layers ke liye vertical lines ki row, kuch jagah paas-paas sikudi hui aur kuch jagah phaili hui, label hai air pressure change. Steps 1 se 4 tak har step ek panel highlight karta hai; step 5 chaaron ko saath dikhata hai.

Maths mein, periodic displacement time ka ek periodic function hai. Sabse simple functions sine aur cosine hain. Lo

f (t) = A \cos ω t

(13.3a)

Cosine tab repeat hota hai jab uska argument $2 π$ se badhta hai. To $f$ usi value par tab laut aata hai jab $ω t$ ki value $2 π$ se badhti hai, yaani jab $t$ itna badhta hai

T = \frac{2 π}{ω}

(13.3b)

aur har $t$ ke liye $f (t + T) = f (t)$ . $A \sin ω t$ ke liye bhi yahi sach hai. Same $ω$ wale dono ka mixture bhi periodic hai, aur pata chalta hai ki woh ek single shifted sine curve hai.

DerivationSame frequency ke sine aur cosine ko jodna

Is combination se shuru karo

\begin{aligned} f (t) & = A \sin ω t \\ + B \cos ω t \end{aligned}

(13.3c)

Do naye constants $D$ aur $ϕ$ aise chuno ki $A = D \cos ϕ$ aur $B = D \sin ϕ$ . Tab $f (t)$ ban jaata hai

\begin{aligned} D \sin ω t \cos ϕ \\ + D \cos ω t \sin ϕ \end{aligned}

yeh $D$ guna hai is expansion ka, $\sin (ω t + ϕ) :$

f = D \sin (ω t + ϕ)

(13.3d)

$A = D \cos ϕ$ aur $B = D \sin ϕ$ ko square karke jodne se, aur ek ko doosre se divide karne se, naye constants milte hain:

D = \sqrt{A^{2} + B^{2}}

\tan ϕ = \frac{B}{A}

To sum same period $2 π / ω$ ka ek sine curve hai, jiski height $D$ zyada hai aur jo time axis ke along angle $ϕ$ se shifted hai (Fig. 13.5).

3/4Sum ek single sine curve hai: $D = \sqrt{3^{2} + 4^{2}} = 5$ aur $\tan ϕ = 4 / 3,$ isliye $ϕ \approx 53^{\circ} .$

Fig. 13.5

3 \sin ω t

aur

4 \cos ω t

ko jodne par milta hai

5 \sin (ω t + 53^{\circ}) :

ek sine curve, same period. Kai sines ko jodkar koi bhi periodic shape banaya ja sakta hai (Fourier).

Do cycles tak omega t ke against value ke axes, plus aur minus 5 par grid lines. Step 1: height 3 ka dashed blue sine curve. Step 2: height 4 ka dotted orange cosine curve joda jaata hai. Step 3: dono components fade hote hain aur ek moti vermillion curve, unka sum, height 5 ke saath dikhti hai, jo sine curve se thoda pehle peak karti hai, 53 degrees se shifted. Step 4: curves ki jagah ek dashed square wave aur ek green curve aati hai jo frequencies omega, 3 omega, 5 omega aur 7 omega wale chaar sines ko jodkar bani hai, aur jo pehle se hi square wave ko kaafi kareeb se follow karti hai.

Example 13.2NCERT Example 13.2 · p. 262

Periodic hai ya nahi?

Yahan $ω$ ek positive constant hai. Time ke inmein se kaun se functions periodic hain? Har periodic function ka period batao. (i) $3 \sin ω t + 4 \cos ω t$ (ii) $\cos 2 ω t + \sin 3 ω t$ (iii) $\sin ω t + \sin \sqrt{2} ω t$ (iv) $e^{- ω t} \sin ω t$

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

Quick check

3 mein se 1

Inmein se kaun sa motion periodic hai lekin oscillatory nahi?

spring par upar-neeche hilta weightaage-peeche jhoolta jhoolavibrate karta sitar ka taarEarth ke chaaron taraf ghoomta Moon

Answer dekho

Answer (D) Earth ke chaaron taraf ghoomta Moon

Moon apna orbit repeat karta hai, isliye uska motion periodic hai, lekin woh kisi equilibrium point ke aas-paas aage-peeche nahi hota. Baaki teeno beech wali position ke aas-paas direction badalte hain, isliye woh periodic hone ke saath oscillations bhi hain.

13.3 Simple harmonic motion

Ab chapter ke star ki baari. Equilibrium position ko $x$ -axis ke origin par rakho. Is axis ke along oscillate karti body simple harmonic motion mein hoti hai jab har pal uska displacement ek single cosine (ya sine) curve se describe ho:

SHM mein displacement

x (t) = A \cos (ω t + ϕ)

(13.4)

jahan

$x (t)$: time $t$ par equilibrium se displacement
$A$: amplitude
$ω$: angular frequency
$ω t + ϕ$: phase, jo time ke saath badalta hai
$ϕ$: phase constant, $t = 0$ par phase

Definition

Simple harmonic motion (SHM)

Ek line ke along aisa oscillation jismein equilibrium se displacement time ka sine ya cosine function hota hai, $x = A \cos (ω t + ϕ),$ jahan $A$ , $ω$ aur $ϕ$ constant hain.

Simple word isi shape ki wajah se mila hai. Kai motions repeat hote hain, lekin sirf woh qualify karte hain jinka $x$ – $t$ graph ek pure sine curve ho. Strobe photograph dikhata hai ki aisa motion kaisa dikhta hai. Spring par ek weight latkao, use upar kheencho, chhod do, aur barabar time intervals par flashes ke saath uski photo lo (Fig. 13.6). Images dono ends ke paas, jahan weight dheema hai, paas-paas jama ho jaati hain, aur beech mein, jahan woh tez hai, door-door phail jaati hain. Flashes ko time axis ke along phailao, to woh ek cosine curve banate hain.

3/3Ek period $T$ ke baad weight wapas $+ A$ par hai, aur dots ek cosine curve banate hain.

Fig. 13.6Barabar time ke flashes swing ke ends par paas-paas aur beech mein door-door hote hain. Time ke against plot karne par woh banate hain

x = A \cos ω t .

NCERT Figs. 13.4, 13.5 · p. 263

Left mein ek weight spring se latka hai, saath mein uski positions ke liye tick marks ka ek column hai. Ek dashed line equilibrium dikhati hai, upar plus A aur neeche minus A. Step 1: weight plus A par pakda hua hai aur ek flash use time zero par ek dot ki tarah record karta hai. Step 2: aadhe cycle ke baad weight minus A par hai; saat flashes record ho chuke hain, aur tick marks dono ends ke paas paas-paas jama hain aur beech mein phaile hue hain; time graph par dots aadhe cosine curve ko follow karte hain. Step 3: terah dots ka ek poora cycle, jo dashed cosine curve se jude hain, ek period T ke baad wapas plus A par khatam hota hai.

Eq. (13.4) ke har constant ka ek kaam hai. $A$ se shuru karo, yeh amplitude hai. Cosine kabhi $- 1$ se $+ 1$ ki range se bahar nahi jaata, isliye body centre se kisi bhi taraf $A$ se zyada door nahi jaati: $A$ sabse door ki doori hai, aur convention se hum ise positive rakhte hain. Sirf $A$ badlo to motion bas chauda ya patla ho jaata hai, aur turning points phir bhi usi pal par aate hain.

Ab cosine ke andar wala angle. $ω t + ϕ$ ko ek pointer ki tarah socho jo batata hai ki body apne cycle mein kitna aage pahunchi hai: woh turning point par hai ya centre cross kar rahi hai, aur kis direction mein. Yeh angle phase hai, aur yeh time ke saath steady badhta hai. Clock shuru hote waqt pointer jahan tha, $ϕ$ , woh phase constant hai (ise phase angle bhi kehte hain). Yeh is baat se tay hota hai ki motion kaise shuru hua: starting displacement $A \cos ϕ$ se do possible angles bachte hain, aur starting direction unmein se ek chunti hai. Do bodies ko same $A$ aur $ω$ lekin alag $ϕ$ do, to unke graphs ek hi curve hote hain jo time axis ke along khiska diya gaya ho (Fig. 13.7).

3/4 $ϕ = - π / 3$ poore curve ko $T / 6$ baad mein khiska deta hai.

Fig. 13.7Teeno constants curve ka alag-alag feature badalte hain:

A

uski height,

ϕ

time axis ke along uski position,

ω

woh kitna kasa hua packed hai.NCERT Figs. 13.7, 13.8 · p. 263

Ek dashed reference curve x equals A cos omega t do periods tak chalta hai, T aur 2T par marks hain. Step 1: sirf reference curve. Step 2: 1.6 guna amplitude wala ek blue curve, jo same times par zero cross karta hai. Step 3: same height ka ek green curve jo period ke chhathe hisse jitna baad mein shifted hai, phase constant minus pi by 3. Step 4: double angular frequency wala ek purple curve, jo reference ke har period mein do cycles poore karta hai.

Teesra constant, $ω$ , period tay karta hai.

DerivationAngular frequency aur period

Time us pal se naapo jab phase zero ho, taaki $ϕ = 0$ :

x (t) = A \cos ω t

(13.5)

Ek period ke baad displacement phir se same hota hai, $x (t + T) = x (t) :$

\begin{aligned} A \cos ω t \\ = A \cos ω (t + T) \end{aligned}

(13.6)

Cosine pehli baar tab repeat hota hai jab uska argument $2 π$ se badh jaaye, isliye $ω (t + T) = ω t + 2 π,$ jisse milta hai

ω = \frac{2 π}{T} = 2 π ν

(13.7)

To $ω$ , yaani angular frequency, woh rate hai jisse phase pointer ghoomta hai, radians per second mein. Ek poora cycle $2 π$ radians hai, isiliye $ω$ hamesha cycles per second ka $2 π$ guna hota hai. Tez pointer same time mein zyada cycles bhar deta hai (Fig. 13.7 ka step 4).

Example 13.3NCERT Example 13.3 · p. 264

SHM hai ya sirf periodic?

Tay karo ki time ka har function (a) simple harmonic motion hai ya (b) aisa periodic motion jo simple harmonic nahi hai, aur har ek ka period batao: (1) $\sqrt{3} \cos ω t + \sin ω t$ (2) $\sin ω t \cos ω t$ (3) $\cos ω t + \cos 2 ω t$

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

Quick check

3 mein se 1

Ek particle SI units mein

x = 0.05 \cos (4 π t + π / 6)

ke hisaab se move karta hai. Uska period kya hai?

4 π s

0.25 s

2 s

0.5 s

Answer dekho

Answer (D) $0.5 s$

ω = 4 π r a d / s,

isliye

T = 2 π / ω = 0.5 s .

4 π

ko hi period maan lene se,

π / ω

use karne se, ya frequency (

2 H z

) ko time ki tarah bata dene se baaki options milte hain.

13.4 Simple harmonic motion aur uniform circular motion

Uniform circular motion aur SHM bilkul alag dikhte hain, phir bhi ek doosre ke andar chhupa hai. Steady ghoomte wheel ko edge-on dekho, yaani aankh wheel ke plane mein ho, to rim par ka point ghoomta hua nahi lagta. Woh ek seedhi line ke along upar-neeche khisakta hai, ends par dheema hota hai aur beech mein tezi se nikalta hai, bilkul ek oscillator ki tarah. Neeche wali activity tumhe ek bicycle se yeh dikhati hai.

Try karke dekho

Wheel mein chhupa ek oscillator

Tumhe chahiye

ek bicycle
kapde sukhane wali ek clip
phone ki torch
pedals ghumane ke liye ek dost

Pehle guess karo

Edge-on dekhne par clip kahan sabse tez move karti dikhegi?

apne path ke top aur bottom parapne path ke beech meinhar jagah same speed sewoh bilkul move karti nahi dikhegi

Bicycle ko ulta kar do taaki woh seat aur handlebar par tiki rahe. Clip ko rear wheel ke rim ke paas ek spoke par mazbooti se laga do.
Wheel ke paas aise baitho ki tumhari aankh axle ke level par aur wheel ke plane mein ho. Apne dost se kaho ki pedals ko haath se dheere aur steady ghumaye.
Sirf clip ko dekho. Phir room mein andhera karo aur torch ko wheel ke aar-paar, uske plane ke parallel, chamkao taaki clip ki shadow deewar par pade.
Gino ki wheel ke $10$ chakkar lagane tak shadow kitni baar upar-neeche jaati hai.

Ise algebra mein badalne ke liye teen numbers wheel ko describe karte hain: point $P$ jo circle banata hai uska radius $A$ , uski steady angular speed $ω$ (maan lo anticlockwise), aur woh angle $ϕ$ jo radius $O P$ ne clock shuru hote waqt positive $x$ -axis ke saath banaya tha. Angle constant rate se badhta hai, isliye time $t$ ke baad woh $ω t + ϕ$ ho jaata hai (Fig. 13.8).

2/4 $O P$ ghoom kar pahunch gaya hai $ω t + ϕ,$ aur $P^{'}$ yahan hai: $A \cos (ω t + ϕ) .$

Fig. 13.8Jaise-jaise

P

reference circle par ghoomta hai, diameter par uska foot

P^{'}

SHM mein move karta hai.

y

-axis par projection bhi SHM hai, bas ek quarter-cycle aage-peeche.NCERT Fig. 13.10 · p. 265

O par centred radius A ka ek circle, x aur y axes ke saath. Step 1: time zero par radius OP x-axis ke saath angle phi banata hai; P se ek dashed line x-axis par uske foot P prime tak girti hai. Step 2: P anticlockwise ghoom kar bade angle, omega t plus phi, par pahunch gaya hai, aur P prime axis ki negative side par khisak gaya hai. Step 3: P axis ke neeche hai aur diameter ke along ek double-headed arrow dikhata hai ki P prime sirf minus A aur plus A ke beech aage-peeche hota hai, x equals A cos of omega t plus phi. Step 4: ek horizontal dashed line P ki height ko right wale graph tak le jaati hai, jahan height time ke against ek sine curve banati hai, y equals A sin of omega t plus phi.

$P$ se $x$ -axis par ek perpendicular girao aur uske foot ko $P^{'}$ kaho. $P^{'}$ ki position $O P$ ka $x$ -component hai:

x (t) = A \cos (ω t + ϕ)

Yeh bilkul Eq. (13.4) hai. Edge-on dekhne par, jab sirf ek coordinate maayne rakhta hai, circle par steady motion simple harmonic motion hai, jismein circle ka radius amplitude hai, angular speed angular frequency hai aur starting angle phase constant hai. Particle $P$ ko reference particle kehte hain aur uske circle ko reference circle.

$x$ -axis mein kuch khaas nahi hai. $y$ -axis par shadow yeh follow karti hai $y (t) = A \sin (ω t + ϕ) :$ same swing aur same timing, lekin hamesha $x$ -shadow se ek quarter cycle ( $π / 2$ ) door. Ise ulta karo to ek kaam ki baat milti hai: perpendicular lines ke along do SHMs, barabar amplitudes aur $π / 2$ ke phase difference ke saath, milkar uniform circular motion banate hain.

Lekin geometry aur physics ko alag rakho. Reference circle SHM ka description hai, uska cause nahi. Line ke along SHM mein body us line ke along ek force se push hoti hai jo centre se uski doori ke saath badhta hai. Uniform circular motion wali body ko constant size ka centripetal force chahiye jo ghoomta rahe taaki centre ki taraf point kare. Dono forces bilkul alag hain.

Example 13.4NCERT Example 13.4 · p. 265

Do circular motions ki shadows

Har case mein particle ke $x$ -projection ka SHM nikalo. (a) Radius $5 c m$ , period $2 s$ , anticlockwise; $t = 0$ par radius positive $x$ -axis ke saath $60^{\circ}$ banata hai, axis ke upar. (b) Radius $3 c m$ , period $8 s$ , clockwise; $t = 0$ par radius positive $x$ -axis ke saath $135^{\circ}$ banata hai (upar aur left ki taraf point karta hua).

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

13.5 Simple harmonic motion mein velocity aur acceleration

Reference circle oscillator ki velocity aur acceleration lagbhag muft mein de deta hai. Reference particle $P$ radius $A$ ke circle par angular speed $ω$ se ghoomta hai, isliye uski speed hai

v = ω A

(13.8)

Uski velocity circle ke tangent ke along point karti hai. Oscillator $P^{'}$ hamesha $x$ -axis par $P$ ke theek neeche ya upar hota hai, isliye $P^{'}$ ki velocity bas $P$ ki velocity ka $x$ -component hai (Fig. 13.9). Jab $O P$ angle $ω t + ϕ,$ par hota hai, to tangent vertical ke saath wahi angle banata hai, aur $x$ -component hota hai $- ω A \sin (ω t + ϕ) .$ Minus sign batata hai ki jab $P$ circle ke upar wale half mein hai, tab $P^{'}$ negative $x$ ki taraf move karta hai.

2/3Centripetal acceleration $ω^{2} A$ ko project karo: $a = - ω^{2} A \cos (ω t + ϕ) = - ω^{2} x .$

Fig. 13.9Oscillator ki velocity aur acceleration reference particle ki velocity (tangent) aur centripetal acceleration (

O

ki taraf) ke

x

-components hain.NCERT Figs. 13.11, 13.12 · p. 266

Ek reference circle jismein P 60 degrees par hai aur x-axis par uska foot P prime hai. Step 1: P par tangent ke along omega A length ka ek green velocity arrow, aur uska horizontal component, left ki taraf point karta hua, P prime par bana hai. Step 2: P se centre ki taraf omega squared A length ka ek purple acceleration arrow, aur uska horizontal component, yeh bhi left ki taraf, P prime par bana hai. Step 3: do khaas positions. Jab P horizontal diameter ke end par hota hai to uski velocity vertical hoti hai, isliye x ke along P prime ki velocity zero hoti hai lekin acceleration sabse zyada, centre ki taraf. Jab P circle ke top par hota hai, to P prime centre par hota hai, sabse zyada leftward velocity aur zero acceleration ke saath.

DerivationSHM mein velocity aur acceleration

Velocity. $P$ ki tangent velocity ko project karne se milta hai

\begin{aligned} v (t) & = - ω A \\ \times \sin (ω t + ϕ) \end{aligned}

(13.9)

Yahi result bina kisi geometry ke Eq. (13.4) ko differentiate karke bhi milta hai, kyunki velocity displacement ke badalne ka rate hai:

v (t) = \frac{d}{d t} x (t)

(13.10)

Acceleration. Reference particle ka centripetal acceleration hai, size $v^{2} / A = ω^{2} A,$ jo $P$ se $O$ ki taraf point karta hai. Iska $x$ -component $P^{'}$ ka acceleration hai:

\begin{aligned} a (t) & = - ω^{2} A \\ \times \cos (ω t + ϕ) \\ = - ω^{2} x (t) \end{aligned}

(13.11)

Velocity ko ek baar aur differentiate karne se yahi milta hai:

a (t) = \frac{d}{d t} v (t)

(13.12)

Eq. (13.11) ko dheere se padho, kyunki yeh SHM ka dil hai: acceleration hamesha displacement ka ek fixed negative number guna hota hai, yaani $- ω^{2}$ guna. Centre se doori double karo to acceleration double ho jaata hai; doosri side jao to woh palat jaata hai. Isliye swing ke har point par body ko $x = 0$ ki taraf wapas dhakela jaata hai, turning points par sabse zor se aur centre par bilkul nahi.

Eqs. (13.4) aur (13.9) ke beech time ko hata dene se kisi bhi position par speed milti hai, jo problems mein baar-baar use hota hai. Kyunki $\sin^{2} + \cos^{2} = 1,$

Displacement x par speed

v = \pm ω \sqrt{A^{2} - x^{2}}

Speed centre par sabse zyada, $ω A$ , hoti hai aur dono ends par zero, bilkul jaisa strobe photograph ne dikhaya tha. $ϕ = 0$ ke saath teeno quantities hain $x = A \cos ω t,$ $v = - ω A \sin ω t$ aur $a = - ω^{2} A \cos ω t .$ Inke graphs ko ek time axis par ek ke upar ek rakho (Fig. 13.10) aur ek cursor ko saath mein khiskao. Teeno ka period $T$ same hai; inki peak values $A$ , $ω A$ aur $ω^{2} A$ hain. Har velocity peak matching displacement peak se ek quarter-cycle pehle aata hai ( $π / 2$ ka phase lead), jabki acceleration hamesha displacement ki mirror image hota hai ( $π$ ka phase difference).

2/4 $x = 0$ , $v = - ω A$ , $a = 0$ : centre se tezi se guzarta hua.

Fig. 13.10SHM mein displacement, velocity aur acceleration ka period

T

same hai. Velocity displacement se ek quarter-cycle aage hai; acceleration aadha cycle door hai, hamesha

x

ke ulta.NCERT Fig. 13.13 · p. 267

Teen graphs ka ek hi time axis hai jo T tak jaata hai. Upar: displacement, height A ka cosine. Beech mein: velocity, height omega A ka negative sine. Neeche: acceleration, height omega squared A ka negative cosine. Ek dashed vertical cursor pal ko mark karta hai, har curve par ek dot ke saath. Step 1, t equals 0: x plus A par hai, v zero hai aur a minus omega squared A par hai. Step 2, t equals T by 4: x zero hai, v sabse zyada negative hai aur a zero hai. Step 3, t equals T by 2: x minus A par hai, v zero hai aur a plus omega squared A par hai. Step 4, t equals 3T by 4: x zero hai, v sabse zyada positive hai aur a zero hai.

InteractiveSHM explorer

Ek reference circle jis par ek particle ghoomta hai, horizontal diameter par uska projection, aur uske neeche spring par ek block jo projection ke saath move karta hai. Time ke against displacement, velocity aur acceleration ke graphs ek moving cursor share karte hain. Sliders amplitude, period aur phase constant set karte hain, aur ek mass slider zaroori spring constant set karta hai. Readouts phase, x, v, a, kinetic aur potential energy, aur unka constant total dikhate hain. Play motion shuru karta hai; ek step button ise period ke barahwen hisse jitna aage badhata hai.

Fig. 13.11Amplitude, period aur phase constant set karo, phir Play dabao. Dekho ki reference particle, spring par block aur teeno graphs kaise saath-saath move karte hain.

Example 13.5NCERT Example 13.5 · p. 267

Upar-neeche hilta float

Machhli pakadne ka ek float upar-neeche aise hilta hai ki apne resting level se uska displacement $y = 0.04 \cos (5 π t - \frac{π}{3}),$ metres mein hai, jahan $t$ seconds mein hai. $t = 0.3 s$ par nikalo (a) displacement, (b) velocity aur (c) acceleration. (d) Sabse zyada speed aur sabse zyada acceleration kya hain?

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

Quick check

3 mein se 1

SHM ke kis point par speed aur acceleration ka size dono sabse zyada hote hain?

mean position parextreme positions parmean aur extreme ke theek beech meinkisi point par nahi: jab ek sabse zyada hota hai to doosra zero hota hai

Answer dekho

Answer (D) kisi point par nahi: jab ek sabse zyada hota hai to doosra zero hota hai

Speed centre par sabse zyada hoti hai, jahan

a = - ω^{2} x = 0;

acceleration ends par sabse zyada hota hai, jahan particle ek pal ke liye rest mein hota hai. Mean position chunna badi speed ko bade acceleration se confuse karna hai.

13.6 Simple harmonic motion ka force law

Newton ka second law Eq. (13.11) ke acceleration ko force mein badal deta hai. SHM mein mass $m$ ke particle ke liye,

F = m a = - m ω^{2} x

Positive constant $m ω^{2}$ ko $k$ likhne par,

SHM ka force law

F (t) = - k x (t)

(13.13)

k = m ω^{2}

(13.14a)

ω = \sqrt{\frac{k}{m}}

(13.14b)

isliye spring constant $k$ wale spring se jude mass $m$ ka period hai

Spring–mass oscillator ka period

T = 2 π \sqrt{\frac{m}{k}}

LawSHM ka force law

Ek particle theek tab simple harmonic motion mein move karta hai jab us par net force equilibrium se uske displacement ke proportional ho aur ulti direction mein point kare: $F = - k x$ .

Kyunki yeh force body ko hamesha mean position ki taraf wapas kheenchta hai, ise restoring force kehte hain. Dhyaan do ki ab SHM ko do tareekon se pehchana ja sakta hai. Tumhe motion diya ja sakta hai, $x = A \cos (ω t + ϕ),$ ya force diya ja sakta hai, $F = - k x$ . Dono mein same information hai: do baar differentiate karne se motion se force milta hai, aur do baar integrate karne se (yeh skill baad mein aayegi) wapas motion milta hai, jismein $A$ aur $ϕ$ integration ke do constants ki tarah aate hain. Problems mein force wala raasta aksar jaldi hota hai: chhote displacement ke liye net force nikalo, aur agar woh aaye $- (constant) \times (displacement),$ to motion SHM hai.

Sirf yahi force law SHM deta hai. Floor par uchhalti ball apna motion repeat karti hai, lekin us par force, steady gravity aur har bounce par ek tez jhatka, ka $- k x$ se koi lena-dena nahi, isliye motion periodic hai par simple harmonic nahi. Jo body $x$ ke bilkul proportional force ke under move karti hai, use linear harmonic oscillator kehte hain. Yeh ek case itna common kyun hai? Lagbhag kisi bhi potential-energy valley ke bottom ke paas force ko $- k x$ plus $x^{2}$ , $x^{3}$ wagairah ke chhote terms ki tarah likha ja sakta hai. Chhote swings ke liye yeh extra terms mushkil se farq daalte hain, isliye crystal mein atoms, tairti naav aur jhoolta lamp, sab kaafi achhe approximation tak linear oscillators ki tarah behave karte hain. Jab extra terms farq daalte hain, tab oscillator ko non-linear kehte hain.

Is formula mein ek kamaal ki baat dekho, $T = 2 π \sqrt{m / k} :$ amplitude isme aata hi nahi. Spring oscillator chaude swing mein bhi utna hi time leta hai jitna patle swing mein, kyunki chauda swing proportionally bada restoring force laata hai. Planetary orbits bilkul alag behave karte hain. Kepler ke third law se bada orbit zyada time leta hai, jahan $T^{2}$ orbit ke size ke cube ke proportional hota hai.

Horizontal spring par block standard example hai, lekin force law vertically latakte springs par bhi laagu hota hai. Tab gravity SHM ko kharab nahi karti; woh sirf equilibrium point ko khiska deti hai (Fig. 13.12).

3/3 $y$ jitna aur kheenchne se ek spring force $k y$ judta hai jise weight balance nahi karta: $F_{net} = - k y .$

Fig. 13.12Spring par mass latkane se equilibrium sirf

d = m g / k

neeche khisakta hai. Naye equilibrium se naapne par net force

- k y

hai: same

k

ke saath SHM.NCERT Figs. 13.14, 13.15 · p. 268

Teen springs ceiling se agal-bagal latke hain. Ek dashed line bina load wale spring ka end mark karti hai, label hai natural length. Step 1: sirf bina load wala spring. Step 2: beech wale spring par ek block hai aur woh d tak khinch kar naye equilibrium par pahunch gaya hai, jo ek doosri dashed line se mark hai; upar ki taraf k d label wala green arrow aur neeche ki taraf m g label wala vermillion arrow balance karte hain, k d equals m g. Step 3: right wale block ko naye equilibrium se y doori aur neeche kheencha gaya hai; upar ki taraf spring force k times d plus y ab m g se bada hai, aur net force minus k y hai.

Example 13.6NCERT Example 13.6 · p. 267

Vertical spring par mass

Ek halke spring se latkaya gaya $250 g$ mass use $6.0 c m$ kheench deta hai aur wahin ruk jaata hai. Phir use thoda aur neeche kheench kar chhoda jaata hai. Dikhao ki woh apni resting position ke aas-paas SHM mein oscillate karta hai, aur period nikalo. Lo $g = 9.8 m / s^{2} .$

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

Do springs ek hi mass share kar sakte hain. Agal-bagal (parallel mein) hon to dono same amount khinchte hain aur unke forces jud jaate hain. End to end (series mein) hon to dono par same force hota hai aur unke stretches jud jaate hain. Har arrangement par force law laagu karne se woh single spring constant milta hai jo wahi kaam karega:

k_{parallel} = k_{1} + k_{2}

\frac{1}{k_{series}} = \frac{1}{k_{1}} + \frac{1}{k_{2}}

Jaise, $300 N / m$ aur $600 N / m$ ke springs se parallel mein juda $2.0 k g$ ka block is period se oscillate karta hai: $T = 2 π \sqrt{2.0 / 900} \approx 0.30 s .$ Yahi springs series mein hon to $k = 200 N / m$ aur $T \approx 0.63 s .$ Aamne-saamne ki deewaron se jude do springs ke beech rakha block parallel arrangement hai: displacement ek spring ko kheenchta hai aur doosre ko dabata hai, aur dono forces beech ki taraf wapas point karte hain.

13.7 Simple harmonic motion mein energy

Oscillator lagataar ek tarah ki energy ko doosri mein badalta rehta hai. Swing ke ends par woh ek pal ke liye rukta hai, isliye uske paas koi kinetic energy nahi hoti; centre par woh sabse tez hota hai. Isi beech spring ends par sabse zyada khincha ya daba hota hai aur centre par relaxed. Har tarah ki energy zero se peak tak badhti hai aur phir wapas girti hai, jabki unka sum kabhi nahi badalta.

DerivationKinetic, potential aur total energy

Kinetic energy. Eq. (13.9) se $v$ ko $K = \frac{1}{2} m v^{2}$ mein rakho aur $m ω^{2} = k$ use karo:

\begin{aligned} K & = \frac{1}{2} k A^{2} \\ \times \sin^{2} (ω t + ϕ) \end{aligned}

(13.15)

Square karne se $v$ ka sign hat jaata hai, isliye $K$ har oscillation mein do baar repeat hota hai: iska period $T / 2$ hai.

Potential energy. Potential energy sirf conservative force ke liye define ki ja sakti hai (Chapter 5). Restoring force $- k x$ is shart ko poora karta hai, aur body ko centre se, jahan hum $U = 0$ rakhte hain, $x$ tak le jaane mein iske against kiya gaya work is tarah store hota hai

U = \frac{1}{2} k x^{2}

(13.16)

Eq. (13.4) se $x$ rakhne par yeh ban jaata hai

\begin{aligned} U & = \frac{1}{2} k A^{2} \\ \times \cos^{2} (ω t + ϕ) \end{aligned}

(13.17)

Total energy. Dono ko jodne par bracket $\sin^{2} + \cos^{2}$ ki value $1$ hoti hai:

\begin{aligned} E & = K + U \\ = \frac{1}{2} k A^{2} \end{aligned}

(13.18)

Eq. (13.18) mein koi $t$ nahi bachta: friction na ho to koi bhi cheez energy kam ya zyada nahi kar sakti, isliye chhodne ke pal se hi total lock ho jaata hai. Yeh amplitude ke square ke saath badhta hai, isliye swing double karne se chaar guna energy store hoti hai. Eqs. (13.16) aur (13.18) ko milane se kisi bhi position par kinetic energy bhi milti hai, $K = \frac{1}{2} k (A^{2} - x^{2}),$ jo Section 13.5 ke speed formula se match karti hai.

$x$ ke against plot karne par potential energy ek parabola hai, ek katori jiska kinaara total-energy line ko theek $x = \pm A$ par chhoota hai (Fig. 13.13). Kisi bhi position par katori aur line ke beech ka gap kinetic energy hai. Particle kabhi $\pm A$ se aage nahi ja sakta, kyunki wahan $U$ ki value $E$ se zyada ho jaata aur $K$ ko negative hona padta, jo impossible hai. Time ke against plot karne par $K$ aur $U$ do humps hain jo baari-baari aate hain, har ek period mein do baar peak karta hai.

2/4Yahan $U = E / 4$ aur $K = 3 E / 4$ : bar ek quarter aur teen quarters mein bant jaata hai.

Fig. 13.13SHM mein energy: katori

U = \frac{1}{2} k x^{2}

aur flat line

E = \frac{1}{2} k A^{2} .

Kinetic energy gap bharti hai. Time mein

K

aur

U

har period mein do baar jagah badalte hain.NCERT Fig. 13.16 · p. 269

Step 1: ek blue parabola U equals half k x squared, x equals 0 par zero se uthkar minus A aur plus A par horizontal vermillion line E equals half k A squared se milta hai. Step 2: x equals A by 2 par axis se E tak ek vertical bar do hisson mein bant jaata hai, parabola ke neeche blue hissa U, bar ka ek quarter, aur uske upar vermillion hissa K, teen quarters. Step 3: plus A par poora bar U hai; centre par poora bar K hai. Step 4: ek period ka time graph: blue curve U E se shuru hokar T by 4 par zero tak girta hai, dashed vermillion curve K iska ulta karta hai, aur dono hamesha milkar flat line E banate hain; har ek period mein do baar peak karta hai.

Dono graphs mein kuch bhi zero se neeche nahi jaata. $K$ ke liye yeh apne aap hota hai, kyunki yeh $v^{2}$ par depend karta hai. $U$ ke liye yeh mean position par zero chunne ka nateeja hai: sirf $U$ ke changes physical hain, aur zero kahin aur rakhne se katori upar ya neeche khisak jaati, motion nahi badalta.

Example 13.7NCERT Example 13.7 · p. 269

Khisakte block ki energy

Smooth table par rakha $0.50 k g$ ka block $200 N / m$ spring constant wale spring se juda hai. Use equilibrium se $6.0 c m$ kheench kar rest se chhoda jaata hai. Jab block equilibrium se $2.0 c m$ door ho, tab potential, kinetic aur total energy nikalo, aur wahan block ki speed bhi. Kinetic energy potential energy ke barabar kahan hoti hai?

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

Quick check

3 mein se 1

Spring oscillator ka amplitude double kar diya jaata hai. Uski total energy

double ho jaati haisame rehti haichaar guna ho jaati haiaadhi ho jaati hai

Answer dekho

Answer (C) chaar guna ho jaati hai

E = \frac{1}{2} k A^{2}

amplitude ke square ke saath badhta hai, isliye

A

double karne se

E

ki value

4

guna ho jaata hai. 'Double' square bhool jaata hai.

13.8 Simple pendulum

Physicists ek asli pendulum ko ek ideal roop mein badal dete hain, simple pendulum: saara mass $m$ ek itne chhote bob mein hai ki use point maana ja sake, length $L$ ki string na khinchti hai na uska koi weight hai, aur support move nahi karta. Ek taraf se chhodne par yeh ek hi vertical plane mein aage-peeche jhoolta hai. Dhaage par bandha ek nut iske kaafi kareeb aa jaata hai.

Try karke dekhoNCERT 13.8 · p. 270

Apne pendulum ka time naapo

Tumhe chahiye

lagbhag $1.2 m$ mazboot dhaaga
do bhaari steel nuts ya washers
ek door frame ya hook
phone stopwatch aur ek metre scale

Pehle guess karo

Agar dhaage ko chaar guna lamba kar do, to period

aadha ho jaayegasame rahegadouble ho jaayegachaar guna lamba ho jaayega

Nut ko dhaage se mazbooti se baandho aur support se aise latkao ki upar ki gaanth se nut ke beech tak length $80 c m$ ho.
Nut ko lagbhag $5 c m$ ek taraf kheencho aur bina dhakka diye chhod do. Jab woh sabse neeche wale point se guzre tab timer shuru karo, aur $20$ poore swings (jaana aur aana) ke baad rok do. $T$ paane ke liye $20$ se divide karo.
$40 c m$ aur $20 c m$ lengths ke saath dohrao. Phir wapas $80 c m$ par jao aur do nuts saath baandh kar try karo, aur aakhir mein $15 c m$ ka pull-aside.
Table bharo aur $T^{2}$ ko length se compare karo.

Table 13.1

Set-up	20 swings ka time (s)	Period T (s)	T² (s²)
80 cm, ek nut
40 cm, ek nut
20 cm, ek nut
80 cm, do nuts
80 cm, chauda swing

Tumhari entries apne aap save hoti hain.

$80 c m$ par lagbhag $1.8 s$ , $40 c m$ par $1.3 s$ aur $20 c m$ par $0.9 s$ ki ummeed rakho, yaani $T^{2}$ length ke proportional hai, aur length ka ek quarter period ko aadha kar deta hai. Mass double karne se kuch bhi aisa nahi badalta jo tum naap sako, aur $15 c m$ ka pull-aside period ko ek percent se bhi kaafi kam badhata hai. Ek ki jagah $20$ swings ka time lene se tumhare reaction time ka asar bees guna kam ho jaata hai.

Chhote swings simple harmonic kyun hote hain? Bob ko ek taraf kheencho taaki string vertical ke saath angle $θ$ banaye, aur dekho us par kya act karta hai: uska weight $m g$ , seedha neeche, aur string ka pull $F_{T}$ , support ki taraf (Fig. 13.14). Inmein se kaun bob ko wapas la sakta hai? Weight ko do hisson mein todo, string ki line ke along $m g \cos θ$ aur swing ke arc ke along $m g \sin θ$ .

3/4String ke along $m g \cos θ$ , arc ke along $m g \sin θ$ .

Fig. 13.14String ke along

F_{T} - m g \cos θ

centripetal force deta hai. Arc ke along

m g \sin θ

bob ko mean position ki taraf wapas kheenchta hai aur restoring torque deta hai.NCERT Fig. 13.17 · p. 270

Ek bob hatched support se length L ki string se latka hai, dashed vertical line ke left ki taraf kheencha hua, taaki string uske saath angle theta banaye. Ek dotted arc bob ka path dikhata hai. Step 1: sirf displaced pendulum, L aur theta label ke saath. Step 2: string ke along support ki taraf ek green arrow, tension ke liye F T label, aur seedha neeche ek vermillion arrow, label m g. Step 3: weight ek dashed component m g cos theta mein bant jaata hai jo string ke along support se door point karta hai, aur ek purple component m g sin theta mein jo arc ke along mean position ki taraf hai. Step 4: bob se mean position tak ka arc highlight hai aur uska label hai s equals L theta, aur purple component ka label hai restoring torque about the support.

String ka pull aur $m g \cos θ$ dono string ke along, seedhe support se hokar act karte hain, isliye inmein se koi bhi bob ko support ke around ghuma nahi sakta. Inka difference, $F_{T} - m g \cos θ,$ ka kaam alag hai: yeh centripetal force $m v^{2} / L$ hai jo bob ko uske circular arc par rakhta hai. Bachta hai $m g \sin θ$ , jo hamesha arc ke along sabse neeche wale point ki taraf point karta hai aur saara speed badhana aur ghatana yahi karta hai. Kyunki bob support ke around ghoomta hai, us point ke around torques sabse natural tool hain: string ke along wale dono forces turant hat jaate hain.

DerivationSimple pendulum ka period

Tangential component $m g \sin θ$ support se doori $L$ par act karta hai. Iska torque $θ$ ko kam karne ki koshish karta hai, isiliye minus sign hai:

τ = - L (m g \sin θ)

(13.19)

Support ke around rotation ke liye Newton ka law torque ko angular acceleration $α$ se support ke around moment of inertia $I$ ke zariye jodta hai:

τ = I α

(13.20)

Torque ke dono expressions ko barabar rakhne par,

I α = - m g L \sin θ

(13.21)

α = - \frac{m g L}{I} \sin θ

(13.22)

Yeh abhi SHM nahi hai, kyunki $α$ proportional hai $\sin θ$ ke, $θ$ ke nahi. Lekin radians mein angle ke liye,

\begin{aligned} \sin θ & = θ - \frac{θ^{3}}{3!} \\ + \frac{θ^{5}}{5!} - \dots \end{aligned}

(13.23)

aur jab $θ$ chhota ho, to doosra aur baad ke terms pehle ke comparison mein bahut chhote hote hain. Sirf $θ$ rakhne par,

α = - \frac{m g L}{I} θ

(13.24)

Ise $a = - ω^{2} x$ (Eq. (13.11)) ke saath rakho: yeh wahi equation hai, jismein angle $θ$ displacement hai aur $ω^{2}$ ki jagah $m g L / I$ hai. Isliye pendulum ke chhote swings simple harmonic hain, jahan

\begin{aligned} ω & = \sqrt{\frac{m g L}{I}} \\ T & = 2 π \sqrt{\frac{I}{m g L}} \end{aligned}

(13.25)

String ka koi mass nahi hai, isliye moment of inertia sirf bob ka hai, $I = m L^{2}$ . Mass cancel ho jaata hai:

Simple pendulum ka period

T = 2 π \sqrt{\frac{L}{g}}

(13.26)

Chhota kitna chhota hai? Radians mein $θ$ ko $\sin θ$ se compare karo (Fig. 13.15 aur Table 13.2). Lagbhag $15^{\circ}$ tak dono mein ek percent se thoda sa hi zyada farq hai, aur $25^{\circ}$ par lagbhag teen percent; $60^{\circ}$ tak gap bada ho jaata hai.

3/3Lagbhag $20^{\circ}$ se neeche dono lagbhag barabar hain; $60^{\circ}$ par unmein $0.18$ ka farq hai.

Fig. 13.15Chhote angles ke liye

\sin θ

aur

θ

(radians mein) lagbhag barabar hain, aur isi wajah se pendulum ke chhote swings simple harmonic hote hain.NCERT Table 13.1 · p. 271

0 se 90 degrees tak angle ke against value ke axes, 20, 40, 60 aur 80 degrees par aur 0.5, 1.0 aur 1.5 par grid lines. Step 1: radians mein theta ke liye ek dashed blue seedhi line, jo 90 degrees par lagbhag 1.57 tak pahunchti hai. Step 2: sine theta ke liye ek vermillion curve, jo pehle line se chipka rehta hai aur phir uske neeche mud jaata hai, 90 degrees par 1 tak pahunchta hai. Step 3: 20 degrees se neeche ka area shaded hai aur uska label nearly equal hai, aur 60 degrees par ek double arrow line aur curve ke beech 0.18 ka gap mark karta hai.

Table 13.2

\sin θ

kitne kareeb se follow karta hai

θ

(radians mein) koNCERT Table 13.1 · p. 271

$θ$ (degrees)	$θ$ (radians)	$\sin θ$	farq
4	0.0698	0.0698	0.08 %
8	0.1396	0.1392	0.33 %
12	0.2094	0.2079	0.73 %
16	0.2793	0.2756	1.3 %
25	0.4363	0.4226	3.2 %
40	0.6981	0.6428	8.6 %

Aisa period jo amplitude par depend na kare, bilkul wahi hai jo clock ko chahiye. Asli pendulum ke liye yeh sirf lagbhag sach hai, aur neeche wali lab tumhe ise test karne deti hai. Dono taraf $20^{\circ}$ ka swing Eq. (13.26) ki prediction se lagbhag $0.8$ percent zyada time leta hai, aur $60^{\circ}$ ka swing lagbhag $7$ percent zyada. Galileo ka lamp, jo kuch hi degrees mein jhool raha tha, safe range ke kaafi andar tha.

Fig. 13.16Length, release angle aur planet chuno, phir Play dabao. Naape gaye period ko is se compare karo,

2 π \sqrt{L / g},

aur dekho ki bade angles par grey small-angle pendulum kaise aage nikal jaata hai.

Eq. (13.25) simple pendulum se zyada general hai. Fixed pivot ke around jhoolti koi bhi rigid body, jaise ek end par keel se latka ruler, chhote swings ke liye ise follow karti hai, jahan $I$ pivot ke around uska moment of inertia hai aur $L$ pivot se uske centre of mass tak ki doori. Aisi body ko physical pendulum kehte hain.

Example 13.8NCERT Example 13.8 · p. 271

Aadhe second wala pendulum

Ek clockmaker ko theek $1.00 s$ period wala pendulum chahiye, taaki woh har aadhe second par tick kare. (a) Iski length kitni honi chahiye? Lo $g = 9.80 m / s^{2} .$ (b) Period double karne ke liye length kitne factor se badalni chahiye, aur woh pendulum kitna lamba hoga?

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

Recap

Summary aur key points

Jo motion barabar intervals ke baad khud ko bilkul waisa repeat kare, woh periodic hai; oscillation equilibrium position ke aas-paas ek periodic aage-peeche wala motion hai (Section 13.2).
Period $T$ ek repeat ka sabse chhota time hai; frequency $ν = 1 / T$ hai, jo hertz mein naapi jaati hai ( $1 H z = 1 s^{- 1}$ ) (Section 13.2.1).
Displacement ka matlab hai apni equilibrium value se naapi gayi koi bhi oscillate karti quantity; $A \sin ω t + B \cos ω t$ ek single sinusoid hai, $D \sin (ω t + ϕ),$ aur koi bhi periodic function sinusoids ka sum hai (Fourier) (Section 13.2.2).
SHM mein, $x (t) = A \cos (ω t + ϕ) :$ amplitude $A$ , phase $ω t + ϕ,$ phase constant $ϕ$ , angular frequency $ω = 2 π / T = 2 π ν$ (Section 13.3).
Circle par steady ghoomta point, ek diameter ke along dekhne par, SHM mein move karta hai: reference circle dono motions ko jodta hai (Section 13.4).
$v = - ω A \sin (ω t + ϕ)$ aur $a = - ω^{2} x;$ maximum values $ω A$ aur $ω^{2} A$ hain, aur $v = \pm ω \sqrt{A^{2} - x^{2}}$ (Section 13.5).
Velocity displacement se $π / 2$ aage hai; acceleration displacement ke ulta hai (phase difference $π$ ) (Section 13.5).
SHM theek tab hota hai jab $F = - k x$ : tab $ω = \sqrt{k / m}$ aur $T = 2 π \sqrt{m / k},$ jo amplitude par depend nahi karta (Section 13.6).
Gravity vertical spring ka sirf equilibrium khiskati hai; springs parallel mein $k_{1} + k_{2}$ aur series mein $k_{1} k_{2} / (k_{1} + k_{2})$ ki tarah judte hain (Section 13.6).
$K = \frac{1}{2} k (A^{2} - x^{2})$ aur $U = \frac{1}{2} k x^{2}$ dono ka period $T / 2$ hai; unka total $E = \frac{1}{2} k A^{2}$ constant rehta hai (Section 13.7).
Chhote angles ke liye simple pendulum SHM mein move karta hai, $T = 2 π \sqrt{L / g},$ jo bob ke mass par depend nahi karta (Section 13.8).
Approximation $\sin θ \approx θ$ lagbhag $1 %$ tak sahi hai, $15^{\circ}$ tak; bade swings formula ki prediction se zyada time lete hain (Section 13.8).

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 mein	Yahan practice question
NCERT Exercise 13.1	Practice Q1
NCERT Exercise 13.2	Practice Q2
NCERT Exercise 13.3	Practice Q3
NCERT Exercise 13.4	Practice Q4
NCERT Exercise 13.5	Practice Q5
NCERT Exercise 13.6	Practice Q6
NCERT Exercise 13.7	Practice Q7
NCERT Exercise 13.8	Practice Q8
NCERT Exercise 13.9	Practice Q9
NCERT Exercise 13.10	Practice Q10
NCERT Exercise 13.11	Practice Q11
NCERT Exercise 13.12	Practice Q12
NCERT Exercise 13.13	Practice Q13
NCERT Exercise 13.14	Practice Q14
NCERT Exercise 13.15	Practice Q15
NCERT Exercise 13.16	Practice Q16
NCERT Exercise 13.17	Practice Q17
NCERT Exercise 13.18	Practice Q18

Quick check: MCQs

13.1 Introduction2 questions
13.2.2 Displacement3 questions
13.3 Simple harmonic motion3 questions
13.4 Simple harmonic motion aur uniform circular motion3 questions
13.5 Simple harmonic motion mein velocity aur acceleration3 questions
13.6 Simple harmonic motion ka force law3 questions
13.7 Simple harmonic motion mein energy3 questions
13.8 Simple pendulum4 questions

Glossary · 14

Key terms (English + Hindi)

amplitudeआयामaayaam: Oscillation mein equilibrium se displacement ka sabse bada magnitude, $A .$ Chapter mein pehli baar yahan aaya →
angular frequencyकोणीय आवृत्तिkoniya aavritti: $ω = 2 π / T = 2 π ν,$ radians per second mein.Chapter mein pehli baar yahan aaya →

frequencyआवृत्तिaavritti: Har second mein repeats ki ginti, $ν = 1 / T,$ jo hertz (Hz) mein naapi jaati hai.Chapter mein pehli baar yahan aaya →

linear harmonic oscillatorरैखिक आवर्ती दोलित्रraikhik aavartee dolitra: Aisi body jis par uske displacement ke bilkul proportional restoring force act karta hai.Chapter mein pehli baar yahan aaya →

oscillatory motionदोलन गतिdolan gati: Equilibrium position ke aas-paas periodic aage-peeche wala motion, jo usi position ki taraf wapas point karne wale force se hota hai.Chapter mein pehli baar yahan aaya →

periodआवर्तकालaavartkaal: Woh sabse chhota time jiske baad periodic motion repeat hota hai, $T;$ SI unit second.Chapter mein pehli baar yahan aaya →
periodic motionआवर्ती गतिaavartee gati: Aisa motion jo barabar time intervals ke baad khud ko bilkul waisa hi repeat karta hai.Chapter mein pehli baar yahan aaya →
phaseकलाkala: $x = A \cos (ω t + ϕ),$ mein angle $ω t + ϕ$ , jo position aur motion ki direction tay karta hai, kisi bhi time $t .$ Chapter mein pehli baar yahan aaya →
phase constantकला स्थिरांकkala sthirank: Phase ki value $ϕ$ , jab $t = 0.$ Chapter mein pehli baar yahan aaya →

reference circleनिर्देश वृत्तnirdesh vritt: Aisa circle jis par particle uniformly ghoomta hai; diameter par uska projection SHM karta hai.Chapter mein pehli baar yahan aaya →
restoring forceप्रत्यानयन बलpratyanayan bal: Aisa force jo hamesha equilibrium position ki taraf wapas point karta hai; SHM mein yeh hai $- k x .$ Chapter mein pehli baar yahan aaya →

simple harmonic motionसरल आवर्त गतिsaral aavart gati: Aisa oscillation jismein displacement time ka sinusoidal function hota hai, jo is force se banta hai: $F = - k x .$
simple pendulumसरल लोलकsaral lolak: Halki, na khinchne wali string par point jaisa bob; chhote swings ke liye $T = 2 π \sqrt{L / g} .$ Chapter mein pehli baar yahan aaya →
spring constantस्प्रिंग नियतांकspring niyatank: Spring ka per unit extension force, $k,$ N/m mein; SHM mein $k = m ω^{2} .$ Chapter mein pehli baar yahan aaya →

Oscillations: Class 11 Physics Chapter 13 Hinglish mein

Is chapter mein aap kya seekhenge

13.1 Introduction

13.2 Periodic aur oscillatory motions

13.2.1 Period aur frequency

13.2.2 Displacement

13.3 Simple harmonic motion

13.4 Simple harmonic motion aur uniform circular motion

13.5 Simple harmonic motion mein velocity aur acceleration

13.6 Simple harmonic motion ka force law

13.7 Simple harmonic motion mein energy

13.8 Simple pendulum

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye