Physical pendulum — compound pendulum — Hinglish notes

1.6.7 · Physics › Oscillations & Waves

Ek physical (compound) pendulum koi bhi aisa rigid body hota hai jo gravity ke under ek fixed horizontal axis ke baare mein swing karta hai — yeh koi massless string par point mass nahi hota. Socho ek swinging meter stick, ek darwaza, ya tumhara forearm.

1. Setup — WHAT hum describe kar rahe hain

Ek rigid body jiska mass $m$ hai, ek fixed horizontal axis $O$ (pivot) ke baare mein rotate karne ke liye free hai. Uska center of mass (CM) pivot se $d$ distance par hai. Jab angle $θ$ se displace hota hai, gravity CM ko equilibrium ki taraf wapas kheenchti hai.

Figure — Physical pendulum — compound pendulum

==Pivot $O$ ==: fixed axis jiske baare mein body rotate karti hai.
== $d$ ==: pivot se center of mass tak ki distance.
== $I$ ==: moment of inertia pivot axis $O$ ke baare mein (CM ke baare mein nahi!).
== $θ$ ==: vertical (equilibrium) se angular displacement.

2. Scratch se Derivation — HOW hum period nikalte hain

Step 1 — Restoring torque locate karo. Gravity $m g$ CM par neecha act karti hai, $O$ se $d$ distance par. Jab body $θ$ se tilted hoti hai, $O$ ke baare mein gravity ka perpendicular lever arm $d sin θ$ hota hai.

$τ = - m g d sin θ$

Yeh step kyun? Minus sign kehta hai ki torque displacement ka virodh karta hai (restoring). Lever arm $d sin θ$ hai kyunki torque = force × line of action se axis tak ki perpendicular distance.

Step 2 — Rotational Newton's law apply karo.

$I \frac{d ^{2} θ}{d t ^{2}} = τ = - m g d sin θ$

Yeh step kyun? $τ = I α$ aur $α = \ddot{θ}$ . $I$ pivot $O$ ke baare mein liya jata hai kyunki wahi rotation ki axis hai.

Step 3 — Small-angle approximation. Chhote $θ$ ke liye, $sin θ \approx θ$ (radians mein):

$I \ddot{θ} = - m g d θ \Rightarrow \ddot{θ} = - \frac{m g d}{I} θ$

Yeh step kyun? Ab iska SHM form $\ddot{θ} = - ω^{2} θ$ clearly dikh raha hai.

Step 4 — $ω$ aur $T$ padho.

$ω = \frac{m g d}{I}, T = 2 π \frac{I}{m g d}$

3. Equivalent simple-pendulum length — 80/20 insight

Simple pendulum $T = 2 π L / g$ se compare karo. Dono ki same period hogi jab

$\frac{L}{g} = \frac{I}{m g d} \Rightarrow L_{eq} = \frac{I}{m d}$

Parallel-axis theorem $I = I_{c m} + m d^{2} = m k^{2} + m d^{2}$ use karke (jahan $k$ = CM ke baare mein radius of gyration):

$L_{eq} = \frac{m k ^{2} + m d ^{2}}{m d} = \frac{k ^{2}}{d} + d, T = 2 π \frac{k ^{2} + d ^{2}}{g d}$

4. Minimum period aur center of oscillation

$T^{2} \propto \frac{k ^{2} + d ^{2}}{d} = \frac{k ^{2}}{d} + d$ ko minimize karo. Derivative zero set karo:

$\frac{d}{dd} (\frac{k ^{2}}{d} + d) = - \frac{k ^{2}}{d ^{2}} + 1 = 0 \Rightarrow d = k$

$d = k$ par: $L_{eq} = k^{2} / k + k = 2 k$ , aur

$T_{m i n} = 2 π \frac{2 k}{g} .$

Pivot se (CM ke through line par) $L_{eq}$ distance par wala point center of oscillation hai. Ek famous reversibility: center of oscillation par pivot karne se same period milti hai (yahi Kater's pendulum ka basis hai $g$ measure karne ke liye).

5. Worked examples

6. Common mistakes (Steel-man + fix)

7. Forecast-then-Verify

Recall Padhne se pehle predict karo

Q: Agar tum ek rod ka pivot end se center ki taraf move karo, toh kya period pehle lambi hogi ya chhoti? Pehle forecast karo, phir check karo: Yeh decrease hoti hai jab tak $d = k = ℓ / 12$ na ho jaye, $T_{m i n}$ tak pahunchti hai, phir increase hoti hai jab $d \to 0$ . Log galti se expect karte hain ki yeh CM ki taraf bas decrease hoti rahegi — lekin CM ke paas restoring torque khatam ho jaata hai aur $T \to \infty$ .

8. Active-recall flashcards

Compound pendulum period formula

T = 2 π I / (m g d)

I

pivot ke baare mein,

d

= pivot-to-CM.

Formula mein kaun sa moment of inertia jaata hai?

I

pivot axis ke baare mein, =

I_{c m} + m d^{2}

Equivalent simple-pendulum length

L_{e q} = I / (m d) = k^{2} / d + d

Ek end par pivot ki gayi uniform rod ki period

T = 2 π 2 ℓ / (3 g)

, toh

L_{e q} = 2 ℓ /3

Kis pivot distance par period minimum hoti hai?

d = k

(CM ke baare mein radius of gyration), jisse

L_{e q} = 2 k

milta hai.

Jab pivot CM ke paas aata hai toh

T \to \infty

kyun?

Restoring torque

m g d sin θ \to 0

jabki

I

finite rehta hai.

Center of oscillation kya hai?

Pivot se

L_{e q}

distance par wala point; wahan pivot karne se same

T

milta hai (Kater's pendulum).

Kya period mass par depend karti hai?

Nahi —

m

cancel ho jaata hai (kisi bhi gravity pendulum mein chhote angles ke liye).

SHM (small-angle) form hold karne ki condition

sin θ \approx θ

, yani chhote amplitudes.

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek nail ke through ek end se ek ruler swing karna. Yeh ek jhule ki tarah aage-peechhe swing karti hai. Yeh kitni fast swing karti hai yeh do cheezoon par depend karta hai: nail se door wale hisse kitne heavy hain (yahi "mass ka phailav" hai, moment of inertia), aur ruler ka beech nail se kitna door hai (wahan gravity kheenchti hai). Agar tum ise bilkul beech mein nail karo, toh yeh bilkul nahi swing karegi — kheenchne ke liye kuch nahi hai. Clever trick yeh hai: har swinging object bilkul theek length ke simple string-aur-ball pendulum jaisi act karti hai, aur woh length $I / (m d)$ hai.

Connections

Simple pendulum — special case jahan saari mass $L$ distance par hai, $I = m L^{2}$ , $d = L$ ⟹ $T = 2 π L / g$ .
Moment of inertia aur Parallel axis theorem — $I = I_{c m} + m d^{2}$ dete hain.
Radius of gyration — $k$ ko $I_{c m} = m k^{2}$ se define karta hai.
Simple Harmonic Motion — $\ddot{θ} = - ω^{2} θ$ ka engine iske peeche hai.
Kater's pendulum — center of oscillation ke baare mein reversibility use karke $g$ measure karta hai.
Torsional pendulum — rotational oscillation lekin gravity nahi, wire ke torsion se restore hota hai.

Physical pendulum — compound pendulum