Simple pendulum — small angle approximation, T = 2π√(L - g) derivation: Visual walkthrough

1.6.6 · D2 · Physics › Oscillations & Waves › Simple pendulum — small angle approximation, T = 2π√(L - g)

Step 1 — Pendulum draw karo aur har part ka naam rakho

KYA. Pendulum ek chhota bhaari ball (the bob) hota hai jo ek fixed point (the pivot) se string ke zariye latka hota hai. Maano:

$L$ = string ki lambai, pivot se bob ke middle tak.
$θ$ (Greek letter "theta") = wo angle jo string seedhi-neeche vertical line se banati hai. Jab bob rest mein latka ho, $θ = 0$ .

Pehle naam kyun rakhein. Neeche har equation inhi symbols se bani ek sentence hai. Agar aap inhe picture nahi karte, equations sirf noise hain. $θ$ hamara single "ghar se kitni dur" number hai — poori kahani yeh hai ki $θ$ time ke saath kaise badalta hai.

PICTURE. Laal bob ek kaali string par latka hai. Dashed vertical line "ghar" hai ( $θ = 0$ ); uske aur string ke beech ka angle $θ$ hai.

Figure — Simple pendulum — small angle approximation, T = 2π√(L - g) derivation

Step 2 — Bob par kheenchne wali do forces

KYA. Sirf do cheezein bob ko touch/kheenchti hain:

Gravity, seedha neeche kheenchti hai. Iska strength $m g$ hai, jahan $m$ bob ka mass hai aur $g \approx 9.8 m/s^{2}$ Earth ki gravitational pull per kilogram hai.
Tension $T_{string}$ , string ke saath pivot ki taraf kheenchti hai.

Sirf yeh do kyun matter karte hain. Ek "simple" pendulum mein koi friction nahi aur koi air drag nahi hota (hum ne inhe jaanboojhkar hataaya). Toh yeh do arrows hi poora cast hain.

PICTURE. Laal arrow gravity $m g$ hai (neeche). String ke saath kaala arrow tension hai. Dhyan karo yeh alag-alag directions mein point karte hain — yahi farq bob ko seedha girne ya latke rehne ki bajay swing karata hai.

Step 3 — Gravity ko "string ke saath" aur "swing ke saath" mein split karo

KYA. Gravity seedha neeche point karti hai, lekin bob ko ek curved arc par move karna hota hai (string usse pivot se fixed distance $L$ par rakhti hai). Toh hum gravity arrow ko do useful directions mein split karte hain:

Radial (string ke saath): $m g cos θ$ . Yeh sirf tension se ladhta hai.
Tangential (arc ke saath, wo direction jismein bob actually move kar sakta hai): $m g sin θ$ .

Aise split kyun karein. Tension string ke saath hoti hai, isliye tension bob ko uske arc ke saath kabhi push nahi kar sakti. Sirf gravity ka tangential slice bob ko speed up ya slow down karta hai. Woh slice swing ka engine hai — isliye hum sirf usi ko chase karte hain.

$sin θ$ kyun aata hai, $cos θ$ kyun nahi? Figure dekho: tangential direction string ke perpendicular hai. Jab aap gravity ke arrow ko us perpendicular direction par drop karte ho, geometry (ek right triangle jiska chhota angle $θ$ ke barabar hai) opposite side deti hai, aur opposite/hypotenuse $= sin θ$ . Radial slice adjacent side par padti hai $= cos θ$ .

PICTURE. Laal arrow tangential slice $m g sin θ$ hai — sirf yahi piece matter karta hai. Faint kaale arrows poori gravity aur uski radial slice hain.

Step 4 — Check karo ki bob kisi bhi direction mein ho (saare cases)

KYA. Hume ensure karna hai ki minus sign kisi bhi position par sahi hai — left, right, aur bilkul centre.

KYU. Ek formula tabhi trustworthy hota hai jab woh har case mein survive kare — left, right, aur dead-centre.

PICTURE + cases:

Centre ke daayein ( $θ > 0$ ): $sin θ > 0$ , toh $F = - m g sin θ < 0$ → force baayein point karta hai (ghar). ✓
Centre ke baayein ( $θ < 0$ ): $sin θ < 0$ , toh $F = - m g sin θ > 0$ → force daayein point karta hai (ghar). ✓
Bilkul centre par ( $θ = 0$ ): $sin 0 = 0$ , toh $F = 0$ → koi push nahi. Yeh equilibrium hai: bob yahan hamesha latka rahega agar rest mein rakha jaaye.

Step 5 — Force ko equation of motion mein badlo

KYA. Bob radius $L$ ke circular arc par ride karta hai. Arc ke saath uski travel ki gayi distance $s = L θ$ hai (arc length = radius × angle — yahi wajah hai ki hum ne Step 1 mein radians par zor diya). Kyunki $L$ fixed hai, arc ke saath acceleration hai

$a = L \frac{d ^{2} θ}{d t ^{2}} .$

Ab Newton's second law " $mass \times acceleration = force$ " arc ke saath apply karo:

$mass m tangential accel. L \frac{d ^{2} θ}{d t ^{2}} = tangential force - m g sin θ$

KYU. Newton's law "koi force act kar rahi hai" se "woh kaise move hogi" ka bridge hai. Symbol $\frac{d ^{2} θ}{d t ^{2}}$ (padho "d-two-theta-d-t-squared") ka matlab sirf angle ki speed kitni tezi se badal rahi hai — angular acceleration.

PICTURE. Bob apne arc ke beech mein drawn hai; arc length $s = L θ$ laal mein marked hai, acceleration arrow arc ke tangent par.

Ab mass cancel karo. $m$ dono sides par hai:

$\frac{d ^{2} θ}{d t ^{2}} = - \frac{g}{L} sin θ .$

Step 6 — Yeh "nice" equation kyun NAHI hai abhi

KYA. Equation $\ddot{θ} = - \frac{g}{L} sin θ$ mein $sin θ$ hai. Simple Harmonic Motion — clean, solvable, sine-wave motion — ke liye restoring term ko $θ$ ke saath straight-line proportional hona chahiye, yaani $\ddot{θ} = - (const) \cdot θ$ . Hamara $sin θ$ curve karta hai, toh hum abhi wahan nahi hain.

KYU care karein. Sirf tabhi jab equation $θ$ mein linear ho, hum keh sakte hain "period har swing ke liye same hai" aur ek clean formula padh sakte hain. Dekho Simple Harmonic Motion.

PICTURE. Plot karo $y = sin θ$ (laal curve) against $y = θ$ (kaali seedhi line). Origin ke paas yeh ek doosre se chipke rehte hain; door jaane par alag ho jaate hain. Origin ke paas woh chipkna hi woh loophole hai jo hum aage exploit karte hain.

Step 7 — Small-angle approximation, drawn

KYA. Chhote $θ$ ke liye, curve aur line almost identical hain, toh hum ek ko doosre se replace kar dete hain:

$sin θ \approx θ (θ small, in radians) .$

Yeh Taylor series se aata hai:

$sin θ = keep this θ - tiny when θ small \frac{θ ^{3}}{6} + \frac{θ ^{5}}{120} - \dots$

Kyun allowed hai. $θ = 0.1$ rad ( $\approx 5. 7^{\circ}$ ) par, drop kiya hua $θ^{3} /6$ term sirf $0.000167$ hai — ek $0.17%$ correction. Picture dikhati hai ki gap wahan invisibly chhota hai.

Radians kyun (degrees nahi). Upar wali series sirf radians mein sahi hai. Degrees mein, $sin (1 0^{\circ}) = 0.174$ lekin " $θ$ " $10$ hoga — 57 ke factor se alag. Radians arc length $s = L θ$ ko honest banate hain, aur wahi honesty $sin θ \approx θ$ ko kaam karwati hai.

PICTURE. $θ = 0$ ke paas ke region mein zoom karo: laal $sin θ$ aur kaala $θ$ overlap karte hain; inke beech shaded gap error hai, aur woh zero ki taraf simat jaata hai.

Substitute karne par clean equation milti hai:

$\frac{d ^{2} θ}{d t ^{2}} = - \frac{g}{L} θ .$

Step 8 — SHM pehchano aur period padho

KYA. Hamari clean equation ko master SHM equation se compare karo:

$\frac{d ^{2} x}{d t ^{2}} = - ω^{2} x ⟺ \frac{d ^{2} θ}{d t ^{2}} = - \frac{g}{L} θ .$

Term-by-term match karne par, constant $ω^{2}$ (squared angular frequency) $g / L$ ke barabar hona chahiye:

$ω = \frac{g}{L} .$

Phir ek poora swing $2 π$ phase ka hota hai, aur $ω$ phase-per-second hai, isliye ek cycle ka time hai

$T = \frac{2 π}{ω} = 2 π \frac{L}{g} .$

$2 π / ω$ kyun. $ω$ batata hai kitne radians of phase per second milte hain; ek poora cycle $2 π$ radians of phase ka hota hai; divide karne par seconds per cycle milte hain. Dekho Angular Frequency and Period.

PICTURE. Angle $θ (t)$ time ke against trace kiya hua ek pure cosine wave hai. Period $T$ do crests ke beech ke time ke roop mein marked hai; $ω$ set karta hai wave kitni tight hai.

Ek-picture summary

PICTURE. Ek figure, paanch panels left-to-right: (1) swinging bob → (2) gravity split karo, laal tangential slice $- m g sin θ$ rakho → (3) Newton's law $m L \ddot{θ} = - m g sin θ$ , mass cross out → (4) $sin θ \to θ$ curve ko line mein seedha karo → (5) cosine wave period $T = 2 π L / g$ ke saath.

Recall Feynman retelling — poora walkthrough simple words mein

Ek ball ko string par latkaao. Gravity use seedha neeche kheenchti hai, lekin string use curved path par force karti hai, isliye gravity ka sirf sideways slice actually use us path ke saath push karta hai — aur woh slice hamesha neeche ki taraf point karti hai. Newton ka "force makes motion" rule us sideways push ke liye likho, aur ek chhota sa miracle hota hai: ball ka weight dono sides par aata hai aur cancel ho jaata hai, isliye ek bhaari ball aur ek halka ball bilkul same time mein swing karte hain. Push $sin θ$ par depend karti hai, jo ek curve hai — lekin chhote swings ke liye woh curve practically ek seedhi line hai, isliye hum use seedha kar dete hain. Jab seedhi ho jaati hai, to motion woh clean back-and-forth sine wave ban jaati hai jise SHM kehte hain, aur uska swing-time $T = 2 π L / g$ hai: lambi string, dheemi swing; strong gravity, tezi swing; aur ball ka weight kabhi enter hi nahi karta.

Recall Checkpoints

Gravity ka kaunsa slice swing drive karta hai? ::: Tangential slice $m g sin θ$ , string ke perpendicular. Kis step mein mass cancel hota hai? ::: Step 5, Newton's law $m L \ddot{θ} = - m g sin θ$ — $m$ dono sides par. $θ$ radians mein kyun hona chahiye? ::: Taaki $s = L θ$ aur $sin θ \approx θ$ hold karein; Taylor series radians-only hai. $2 π$ kahan se aata hai? ::: Ek full cycle $2 π$ radians of phase ka hota hai, aur $T = 2 π / ω$ .

Connections

Simple Harmonic Motion — woh clean equation jo hum ne Step 8 mein match ki
Restoring Force and Equilibrium — Step 3/4 mein minus sign
Taylor Series and Small-Angle Approximations — Step 7 mein seedha karna
Angular Frequency and Period — $T = 2 π / ω$
Energy in Oscillations
Mass-Spring System — wahi $ω = k / m$ pattern
Measuring g with a Pendulum

Visual walkthrough — Simple pendulum — small angle approximation, T = 2π√(L - g) derivation