Noesis
1.6.6Oscillations & Waves

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

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🇮🇳 Yeh note Hinglish mein padho →

WHAT is a simple pendulum?

WHY these idealizations? They strip away complications so the physics is pure: no string mass to wobble, no stretching to store energy, no friction to damp the swing.

HOW to derive T=2πL/gT = 2\pi\sqrt{L/g} from scratch

Step 1 — Identify the forces and the restoring direction

When the string makes angle θ\theta with the vertical, two forces act on the bob: gravity mgmg (down) and tension TstringT_{\text{string}} (along the string).

Why split gravity into components? Tension lies along the string, so it can't speed the bob up or down along the arc. Only the component of gravity along the arc (tangential) drives the motion.

Ftangential=−mgsin⁡θF_{\text{tangential}} = -mg\sin\theta

Why the minus sign? Because the force points back toward θ=0\theta=0. If θ>0\theta>0, the force is negative (restoring). That negative sign is the heart of all oscillation.

Step 2 — Write Newton's law along the arc

The bob moves along a circular arc of radius LL. Arc length (displacement) is s=Lθs = L\theta, so the tangential acceleration is:

a=d2sdt2=Ld2θdt2a = \frac{d^2 s}{dt^2} = L\frac{d^2\theta}{dt^2}

Newton's second law tangentially (ma=Ftangentialma = F_{\text{tangential}}):

mLd2θdt2=−mgsin⁡θmL\frac{d^2\theta}{dt^2} = -mg\sin\theta

Why this step? We turned forces into an equation of motion. Cancel mm (mass drops out — pendulums don't care about the bob's weight!):

d2θdt2=−gLsin⁡θ\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin\theta

Step 3 — The small-angle approximation

This equation has sin⁡θ\sin\theta — it is not SHM yet (SHM needs the restoring term linear in θ\theta).

The key trick: Taylor-expand sin⁡θ\sin\theta about θ=0\theta=0:

sin⁡θ=θ−θ36+θ5120−⋯\sin\theta = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots

For small θ\theta (in radians), the θ3\theta^3 term is tiny, so:

sin⁡θ≈θ\boxed{\sin\theta \approx \theta}

Why is this allowed? At θ=0.1 \theta = 0.1\,rad (≈5.7°\approx 5.7°), sin⁡θ=0.09983\sin\theta = 0.09983, error ≈0.17%\approx 0.17\%. Negligible!

d2θdt2=−gLθ\frac{d^2\theta}{dt^2} = -\frac{g}{L}\theta

Step 4 — Recognize SHM and read off the period

Compare with the standard SHM equation:

d2xdt2=−ω2x⟹ω2=gL\frac{d^2 x}{dt^2} = -\omega^2 x \quad\Longrightarrow\quad \omega^2 = \frac{g}{L}

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

Reading the formula (the 80/20 insight)

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Try before revealing
  • Where does the minus sign in F=−mgsin⁡θF=-mg\sin\theta come from?
  • At what step does mass cancel, and why does that matter?
  • Why must θ\theta be in radians?
  • What does T∝LT \propto \sqrt{L} mean physically?
Recall Feynman: explain to a 12-year-old

Imagine a swing in a park. Gravity always tries to pull the swing back to the lowest point — the higher you pull it, the harder gravity tugs it back. That "always pulling back, more when farther" rule is what makes it go back-and-forth in a steady rhythm. The neat surprise: it doesn't matter if a heavy kid or a light kid is on the swing — both take the same time for one full swing, as long as the chain length is the same! Long chains = slow lazy swings; short chains = quick swings. A clockmaker uses exactly this steady rhythm to keep time.

Flashcards

What approximation makes the pendulum equation linear (SHM)?
sin⁡θ≈θ\sin\theta \approx \theta for small θ\theta in radians.
Why does the bob's mass not affect the period?
mm appears on both sides (mLθ¨=−mgsin⁡θmL\ddot\theta = -mg\sin\theta) and cancels.
State the period of a simple pendulum.
T=2πL/gT = 2\pi\sqrt{L/g}.
What is the tangential restoring force on the bob?
F=−mgsin⁡θF = -mg\sin\theta (minus = points back to equilibrium).
How does TT change if LL is quadrupled?
TT doubles, since T∝LT\propto\sqrt{L}.
Express ω2\omega^2 for a simple pendulum.
ω2=g/L\omega^2 = g/L.
How to find gg from pendulum data?
g=4π2L/T2g = 4\pi^2 L/T^2.
Why must θ\theta be in radians for sin⁡θ≈θ\sin\theta\approx\theta?
The Taylor series sin⁡θ=θ−θ3/6+…\sin\theta=\theta-\theta^3/6+\dots holds only in radians.
First correction term to the period for larger amplitude θ0\theta_0?
T≈2πL/g(1+θ02/16)T\approx 2\pi\sqrt{L/g}(1+\theta_0^2/16).
What is "isochronism"?
Period independent of amplitude (for small swings) — Galileo's observation.

Connections

  • Simple Harmonic Motion — pendulum is a special case
  • Restoring Force and Equilibrium
  • Taylor Series and Small-Angle Approximations
  • Angular Frequency and Period
  • Energy in Oscillations — KE/PE exchange in a swing
  • Mass-Spring System — analogous ω=k/m\omega=\sqrt{k/m}
  • Measuring g with a Pendulum (lab application)

Concept Map

forces acting

tangential component

minus sign means

Newton 2nd law on arc

mass cancels

small angle in radians

linearizes equation

matches -omega^2 x

T = 2pi/omega

independent of mass

Simple pendulum idealized

Gravity mg and tension

F = -mg sin theta

Restoring toward theta=0

mL d2theta/dt2 = -mg sin theta

d2theta/dt2 = -g/L sin theta

sin theta approx theta

d2theta/dt2 = -g/L theta

omega^2 = g/L

T = 2pi sqrt L/g

Steady timekeeping

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, simple pendulum ka matlab hai ek chhota bob (point mass) jo ek massless dhaage (string) se latka hua hai aur gravity ke under aage-peeche jhoolta hai. Jab bob ko thoda side mein le jaate ho, toh gravity ka ek component — exactly mgsin⁡θmg\sin\theta — use wapas equilibrium (neeche) ki taraf kheechta hai. Yahi restoring force hai, aur iske saamne minus sign isliye lagta hai kyunki yeh hamesha center ki taraf point karta hai.

Ab asli trick yeh hai: jab swing chhota hota hai (small angle), toh hum sin⁡θ≈θ\sin\theta \approx \theta maan lete hain — par dhyaan rakho, θ\theta radians mein hona chahiye, degrees mein nahi! Is approximation ke baad equation ban jaati hai θ¨=−(g/L)θ\ddot\theta = -(g/L)\theta, jo bilkul SHM ki shakal hai. Isi se nikalta hai ω=g/L\omega = \sqrt{g/L} aur period T=2πL/gT = 2\pi\sqrt{L/g}.

Sabse mazedaar baat — mass cancel ho jaata hai! Heavy bob ho ya light, period same rahega (agar length same hai). Isko isochronism kehte hain, Galileo ne discover kiya tha. Aur T∝LT \propto \sqrt{L}, matlab dhaaga 4 guna lamba karoge toh period sirf 2 guna badhega. Exam mein common trap: sin⁡θ≈θ\sin\theta\approx\theta ko degrees mein use mat karna, aur gg nikaalne ke liye g=4π2L/T2g = 4\pi^2 L/T^2 yaad rakho. Yeh formula clock banane se lekar gravity measure karne tak — har jagah kaam aata hai.

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Connections

Simple harmonic motion — definition, restoring force F = −kxPhysics · 1.6.1Spring-mass system — horizontal, verticalPhysics · 1.6.8Simple harmonic motion — definition, restoring force F = −kxPhysics · 1.6.1SHM differential equation — solution - x = A cos(ωt + φ)Physics · 1.6.2Physical pendulum — compound pendulumPhysics · 1.6.7Spring-mass system — horizontal, verticalPhysics · 1.6.8