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

1.6.6 · D3 · Physics › Oscillations & Waves › Simple pendulum — small angle approximation, T = 2π√(L - g)

Shuru karne se pehle, teen symbols ko simple shabdon mein fix kar lete hain jo hum baar baar use karte hain:

Scenario matrix

Pendulum ka har woh problem jo tum kabhi bhi dekhoge, inhi cells mein se ek hogi. Neeche diye worked examples mein har ek ke saath (Cell A) jaisa tag hai taaki tum dekh sako ki poora map cover ho gaya hai.

Cell	Scenario class	Kya unknown hai	Trick / danger
A	Forward: diya $L, g$ → dhundo $T$	$T$	seedha substitution
B	Backward: diya $T, g$ → dhundo $L$	$L$	invert karo aur square karo
C	Gravity measure karo: diya $L, T$ → dhundo $g$	$g$	$g = 4 π^{2} L / T^{2}$
D	Ratio / scaling: $L$ ya $g$ change ho to $T$ kaise change hota hai?	ratio $T_{2} / T_{1}$	numbers ki zaroorat nahi
E	Alag planet/Moon (chhota $g$ )	$T$ ya comparison	sirf $g$ change hota hai
F	Degenerate / limiting inputs ( $L \to 0$ , $g \to 0$ , $L \to \infty$ )	behaviour	conceptual limits
G	Many-swings timing (real lab)	$T$ phir $g$	total time ko count se divide karo
H	Large-amplitude correction (formula toot jaati hai)	sach wala $T$	$T \approx 2 π L / g (1 + θ_{0}^{2} /16)$
I	Exam twist: units, radians vs degrees, "half a swing"	trap-aware answer	wording dhyan se padho

Ab hum har cell ko cover karte hain.

Cell A — Forward: period dhundo

Worked example Example 1 — Earth par 2.5 m ka pendulum

(Cell A) Ek bob $L = 2.5 m$ ki string par latka hua hai, $g = 9.8 m/s^{2}$ . $T$ dhundo.

Forecast: 1 m wala pendulum $\approx 2$ s tha. Yeh $2.5 \times$ lamba hai — lekin $T \propto L$ , toh guess karo thoda $3$ s se zyada, $5$ s nahi.

Formula likho: $T = 2 π L / g$ . Yeh step kyun? $L$ aur $g$ dono known hain, $T$ hi ek akela unknown hai — pure substitution.
Root ke andar divide karo: $L / g = 2.5/9.8 = 0.2551$ . Yeh step kyun? Division pehle square root se karo taaki number tidy rahe.
Square root: $0.2551 = 0.5051$ . Yeh step kyun? Formula ne ratio ko square root mein wrap kiya hai, toh hame woh root undo karni hai taaki woh number mile jise $2 π$ multiply karta hai.
$2 π$ se multiply karo: $T = 6.2832 \times 0.5051 = 3.17 s$ . Yeh step kyun? Aage ka $2 π$ "ratio ki root" ko ek poore period mein convert karta hai — yeh phase ke ek complete cycle se bacha hua factor hai (dekho Angular Frequency and Period).

Verify: Root ke andar units: $m \div (m/s^{2}) = s^{2}$ , aur $s^{2} = s$ ✓. Hamara forecast "thoda $3$ s se zyada" bilkul sahi nikla. $✓$

Cell B — Backward: length dhundo

Worked example Example 2 — "seconds pendulum" ki length

(Cell B) Ek seconds pendulum $T = 2.0 s$ mein ek poora period complete karta hai. Earth par yeh kitni length se milega?

Forecast: Parent ka Example 1 mein dikhaya tha ki $L = 1.0 m$ se $T \approx 2.0 s$ milta hai, toh $L$ kaafi $1$ m ke paas hoga.

$T = 2 π L / g$ se shuru karo aur root ko isolate karo: $\frac{T}{2 π} = \frac{L}{g}$ . Yeh step kyun? Hum pehle algebraically solve karte hain — numbers jaldi daaloge toh rounding ki galtiyaan zyada hoti hain.
Dono sides ko square karo: $(\frac{T}{2 π})^{2} = \frac{L}{g}$ . Yeh step kyun? Square root woh wrapper hai jiske andar $L$ chhupi hai; squaring uska inverse operation hai jo use nikaalta hai.
$g$ se multiply karo: $L = g (\frac{T}{2 π})^{2}$ . Yeh step kyun? $L$ abhi $g$ se divide ho raha hai; dono sides ko $g$ se multiply karne par $L$ akela ek taraf reh jaata hai — yahi goal hai.
Plug in karo: $L = 9.8 (\frac{2.0}{6.2832})^{2} = 9.8 \times (0.3183)^{2} = 9.8 \times 0.1013 = 0.993 m$ . Yeh step kyun? Sirf ab, jab $L$ poori tarah isolate ho chuki hai, numbers daalte hain — ek clean single evaluation jismein koi algebra bacha hi nahi slip hone ko.

Verify: $L = 0.993$ wapas Cell A mein daalte hain: $T = 2 π 0.993/9.8 = 2.00 s$ ✓. Hamara forecast "just under 1 m" confirm hua. $✓$

Cell C — $g$ measure karna

Worked example Example 3 — ek clean period se

$g$ dhundo (Cell C) $L = 0.80 m$ wale pendulum ka period $T = 1.80 s$ paya gaya. Yahan $g$ kya hai?

Forecast: Earth ka $g$ $\approx 9.8$ hai. Agar yeh bahut zyada alag aaye, toh humse koi galti hui hai.

Formula ko square karo: $T^{2} = 4 π^{2} \frac{L}{g}$ . Yeh step kyun? $g$ ek square root ke andar phansa hua hai; squaring se woh ek seedhe fraction mein aa jaata hai.
$g$ ke liye rearrange karo: $g = \frac{4 π ^{2} L}{T ^{2}}$ . Yeh step kyun? Dono sides ko $g$ se multiply karo, $T^{2}$ se divide karo — standard algebra se target isolate hota hai.
Numbers daalte hain: $g = \frac{4 π ^{2} ( 0.80 )}{( 1.80 ) ^{2}} = \frac{31.58}{3.24} = 9.75 m/s^{2}$ . Yeh step kyun? $g$ left side par akela hone par, substitution ek safe single evaluation hai; upar ( $4 π^{2} \times 0.80$ ) aur neeche ( $1.8 0^{2}$ ) alag alag compute karte hain taaki galtiyaan na ho.

Verify: $9.75$ , $9.8$ se $0.5%$ ke andar hai — ek believable Earth value ✓. Units: $\frac{m}{s ^{2}} = m/s^{2}$ ✓. Dekho Measuring g with a Pendulum. $✓$

Cell D — Ratios aur scaling (calculator ki zaroorat nahi)

Yahan poori baat yeh hai ki tum kabhi $g$ plug in nahi karte — woh cancel ho jaata hai. Ek baar samajh lo toh yeh sabse fast type ka problem hai.

Neeche ka figure dekho. Horizontal axis length $L$ hai; vertical axis period $T$ hai. Yellow dashed markers $L = 1$ m par aur red markers $L = 4$ m par notice karo: length $4 \times$ jump karti hai lekin height (period) sirf double hoti hai. Poora curve jhuk jaata hai — woh jhukna hi "square root" ka picture hai.

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

Cell E — Doosri duniya: Moon

Ab hum same pendulum ko do jagahon par compare karte hain. Dono ko alag rakhne ke liye hum ek chhota subscript lagate hain: ek chhota letter neeche likhte hain jo batata hai quantity kis world ki hai. Yahan $g_{E}$ aur $T_{E}$ matlab gravity aur period Earth par (subscript $E$ ), jabki $g_{M}$ aur $T_{M}$ matlab same quantities Moon par (subscript $M$ ).

Worked example Example 5 — same pendulum, Earth vs Moon

(Cell E) Ek pendulum ka $T_{E} = 2.00 s$ Earth par hai ( $g_{E} = 9.8 m/s^{2}$ ). Moon par $g_{M} = 1.62 m/s^{2}$ hai. Wahan uska period $T_{M}$ dhundo.

Forecast: Kamzor gravity = aalsee, dhimi swing = bada $T$ . Kuch guna lamba expect karo.

Sirf $g$ change hota hai, $L$ fixed hai, toh ratio lo: $\frac{T _{M}}{T _{E}} = \frac{g _{E}}{g _{M}}$ . Yeh step kyun? $g$ root ke andar neeche hai, toh chhota $g$ bada $T$ banata hai — aur ratio lene se unknown $L$ khatam ho jaata hai.
Compute karo: $9.8/1.62 = 6.049 = 2.460$ . Yeh step kyun? Yeh bare number woh factor hai jisse Moon period ko stretch karta hai; dhyaan do $g_{E}$ upar hai toh Moon ki kamzor gravity ( $g_{M}$ chhota) ratio ko 1 se bada banati hai, confirming dhimi swing.
Toh $T_{M} = 2.460 \times 2.00 = 4.92 s$ . Yeh step kyun? Step 2 ka ratio batata hai ki Moon ka period Earth ke period se kitne guna lamba hai; known Earth period $T_{E} = 2.00$ s se multiply karne par woh pure factor ek actual time in seconds mein convert ho jaata hai — wahi quantity jo poochi gayi hai.

Verify: Moon par lamba period — "dhimi aalsee swing" ki prediction se match karta hai ✓. Sanity: $6 \approx 2.45$ , toh $\approx 4.9$ s ✓. $✓$

Cell F — Degenerate aur limiting inputs

Kabhi bhi kisi extreme se surprised nahi hona chahiye. Yeh conceptual hain, lekin formula har ek ka saaf jawab deta hai.

Worked example Example 6 — limits

$L \to 0$ , $g \to 0$ , $L \to \infty$ (Cell F) $T = 2 π L / g$ se teen extreme cases mein period ka behaviour predict karo.

Forecast: Padhne se pehle socho ki root ke andar har ratio ka kya hoga.

$L \to 0$ (string sar tak simat jaaye): $T = 2 π 0/ g = 0$ . Yeh kyun sense karta hai: zero-length pendulum ke paas swing karne ki jagah hi nahi — period zero ho jaata hai. Physically woh infinitely fast "buzz" karta hai.
$L \to \infty$ (endlessly lamba string): $L / g \to \infty$ , toh $T \to \infty$ . Yeh kyun sense karta hai: bahut lamba pendulum poora slow swing karta hai; limit mein woh kabhi wapas nahi aata.
$g \to 0$ (deep space, koi gravity nahi): $T = 2 π L /0 \to \infty$ . Yeh kyun sense karta hai: koi restoring pull nahi toh bob ko wapas laane wala kuch nahi — woh kabhi oscillate nahi karta. Compare karo Restoring Force and Equilibrium: no restoring force ⇒ no period.

Verify: Try karo $L = 1 0^{- 6}$ m, $g = 9.8$ : $T = 2 π 1 0^{- 6} /9.8 = 2.01 \times 1 0^{- 3} s$ , tiny ✓. Try karo $g = 1 0^{- 6}$ , $L = 1$ : $T = 6.28 \times 1 0^{3} s$ , huge ✓. Dono apne predicted extremes ki taraf trend kar rahe hain. $✓$

Cell G — Real-lab mein many swings ka timing

Worked example Example 7 — 20 swings, phir

$g$ dhundo (Cell G) Ek student $L = 1.20 m$ wale pendulum ke $20$ complete swings time karta hai: total $43.9 s$ . $T$ dhundo, phir $g$ .

Forecast: Ek swing roughly $2.2$ s hona chahiye (1 m ≈ 2 s wale pendulum se thoda zyada), aur $g$ kaafi $9.8$ ke paas aayega.

Count se period: $T = \frac{total time}{number of swings} = \frac{43.9}{20} = 2.195 s$ . Yeh step kyun? Zyada swings time karna aur divide karna tumhari reaction-time error ko 20 periods par spread kar deta hai, toh har $T$ kaafi accurate hoti hai — yeh key lab trick hai.
Cell C se $g = \frac{4 π ^{2} L}{T ^{2}}$ use karo. Yeh step kyun? Same rearranged formula; $g$ hi woh hai jo experiment actually chaahta hai.
Compute karo: $g = \frac{4 π ^{2} ( 1.20 )}{( 2.195 ) ^{2}} = \frac{47.37}{4.818} = 9.83 m/s^{2}$ . Yeh step kyun? Upar ( $4 π^{2} \times 1.20$ ) aur neeche ( $2.19 5^{2}$ ) alag alag evaluate karte hain, phir divide karte hain — dono operations alag rakhne se calculator-order mistakes se bachte hain.

Verify: $9.83 \approx 9.8$ , ek excellent Earth measurement ✓. Forecast " $T \approx 2.2$ s" confirm hua. $✓$

Cell H — Jab swing bahut badi ho (formula toot jaati hai)

Formula $T = 2 π L / g$ ek approximation hai jo chhote angles ke liye valid hai (yeh parent mein $sin θ \approx θ$ se aayi thi — dekho Taylor Series and Small-Angle Approximations). Wide swing ke liye pehla correction chahiye:

Worked example Example 8 —

$θ_{0} = 30°$ par simple formula kitna galat hai? (Cell H) Ek $L = 1.00 m$ wala pendulum $θ_{0} = 30°$ amplitude ke saath jhulta hai. Small-angle period aur corrected period dhundo, aur percent difference bhi.

Forecast: Correction chhoti hogi — true period thoda lamba hoga, around $1$ – $2%$ .

Pehle small-angle period — ise $T_{0}$ kaho, matlab woh period jo simple formula predict karta hai kisi bhi large-swing correction se pehle: $T_{0} = 2 π 1.00/9.8 = 2.007 s$ . Yeh step kyun? Yeh $T_{0}$ woh baseline hai jise correction factor multiply karta hai; use nudge karne se pehle hamein chahiye.
Amplitude ko radians mein convert karo: $θ_{0} = 30° = 30 \times \frac{π}{180} = 0.5236 rad$ . Yeh step kyun? Correction term $θ_{0}^{2} /16$ ek Taylor coefficient hai — yeh sirf radians mein kaam karta hai, degrees mein kabhi nahi.
Correction factor: $1 + \frac{( 0.5236 ) ^{2}}{16} = 1 + \frac{0.2742}{16} = 1 + 0.01714 = 1.01714$ . Yeh step kyun? Amplitude ko square karke 16 se divide karne par fractional extra period milta hai; 1 add karne se yeh ek multiplier ban jaata hai jise hum $T_{0}$ par apply kar sakte hain.
Corrected period: $T = T_{0} \times 1.01714 = 2.007 \times 1.01714 = 2.041 s$ . Yeh step kyun? Baseline $T_{0}$ ko 1 se bade factor se multiply karne par woh stretch ho jaata hai — wide swing sach mein zyada waqt leta hai.
Percent difference: $\frac{2.041 - 2.007}{2.007} \times 100% = 1.71%$ . Yeh step kyun? Corrected value ko $T_{0}$ se percentage ke roop mein compare karne par pata chalta hai ki simple formula $30°$ par kitni error carry karta hai — practical takeaway yahi hai.

Verify: Parent note ne kaha tha " $θ_{0} = 30°$ par error $\sim 1.7%$ hai" — humne $1.71%$ paya ✓. Corrected period sach mein small-angle $T_{0}$ se thoda bada hai ✓, aur dono 1 m ke pendulum ke liye expected $2$ s ke paas hain. $✓$

Neeche ka figure study karo. Horizontal axis amplitude $θ_{0}$ degrees mein hai; vertical axis batata hai ki true period simple formula se kitne percent lamba hai. Chhote angles par (yellow arrow) curve zero se chipka rehta hai — simple formula excellent hai. $30°$ par (red marker) yeh approximately $1.7%$ tak chad chuka hai, Example 8 se match karta hai. Message: error amplitude squared ki tarah badhti hai, pehle dheere phir tez.

Cell I — Exam twists (wording dhyan se padho!)

Worked example Example 9 — "half a swing" aur degrees ka trap

(Cell I) Ek pendulum ek extreme se doosre extreme tak jaane mein $0.90 s$ leta hai (yaani, half a full cycle). Uska amplitude $8°$ bataya gaya hai. Earth par $L$ dhundo.

Forecast: "Extreme to extreme" sirf half period hai, toh full period $1.80$ s hai, $0.90$ s nahi. Aur $8°$ chhota hai — simple formula theek hai, koi correction nahi chahiye.

Real period identify karo: ek full cycle hai extreme → doosra extreme → wapas = do half-swings, toh $T = 2 \times 0.90 = 1.80 s$ . Yeh step kyun? Ek taraf se doosri taraf ka "swing" half period hota hai; yeh bhulaoge toh baad mein har answer half ya double ho jaayega. Yeh sabse common exam trap hai.
Angle check karo: $8° = 0.1396 rad$ , tiny — small-angle formula valid hai, correction ignore karo. Yeh step kyun? $T = 2 π L / g$ use karne se pehle justify karna zaroori hai (Cell H ne dikhaya tha ki bade swings ke liye yeh fail karta hai).
$L$ ke liye solve karo (jaise Cell B mein): $L = g (\frac{T}{2 π})^{2} = 9.8 (\frac{1.80}{6.2832})^{2}$ . Yeh step kyun? $L$ unknown hai; hum Cell B ka wahi rearrangement reuse karte hain jo use isolate karta hai.
Compute karo: $L = 9.8 \times (0.28648)^{2} = 9.8 \times 0.08207 = 0.804 m$ . Yeh step kyun? Sahi full period $T = 1.80$ s daalte hain toh yeh single evaluation sach wali length deta hai; agar trap wali value $0.90$ s yahan use karte toh answer kharaab ho jaata.

Verify: Wapas feed karo: $T = 2 π 0.804/9.8 = 1.800 s$ , ek full cycle ✓, aur uska aadha $0.90$ s hai jaisa bataya gaya tha ✓. Agar galti se $T = 0.90$ s use karte toh $L = 0.201$ m milta — chaar guna chhota. $✓$

Active recall

Recall Har cell ko uske move se match karo

Diya $L, g$ , chahiye $T$ ::: seedha $T = 2 π L / g$ mein substitute karo (Cell A).
Diya $T, g$ , chahiye $L$ ::: $L = g (T /2 π)^{2}$ (Cell B).
Diya $L, T$ , chahiye $g$ ::: $g = 4 π^{2} L / T^{2}$ (Cell C).
Length $9 \times$ badi ho, period kya hoga? ::: $9 = 3$ se multiply hoga (Cell D).
Kamzor gravity (Moon), period kya karega? ::: bada hoga, kyunki $g$ root ke denominator mein hai (Cell E).
"Ek swing side to side" period ka kitna hissa hai? ::: ek aadha (Cell I).

Connections

Parent: the derivation
Simple Harmonic Motion
Restoring Force and Equilibrium
Taylor Series and Small-Angle Approximations
Angular Frequency and Period
Energy in Oscillations
Mass-Spring System
Measuring g with a Pendulum

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