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

1.6.6 · D4 · Physics › Oscillations & Waves › Simple pendulum — small angle approximation, T = 2π√(L - g)

Poore note mein, master formula wahi hai jo parent note mein build ki gayi hai the parent note:

Earth par $g = 9.8 m/s^{2}$ lo jab tak problem kuch aur na kahe. $π = 3.1416$ use karo.

Level 1 — Recognition

Kya tum sahi formula spot kar sakte ho aur use sahi se padh sakte ho?

L1·Q1

Ek simple pendulum ki length $L = 2.45 m$ hai. Iske period ka expression likho aur $T$ calculate karo.

Recall Solution

WHAT: $T = 2 π L / g$ ka direct use. WHY: Recognition — koi rearranging nahi chahiye. $T = 2 π \frac{2.45}{9.8} = 2 π 0.25 = 2 π (0.5) = 3.14 s$ Toh $T \approx 3.14 s$ .

L1·Q2

Do pendulums bilkul identical hain, bas ek mein heavy iron bob hai aur ek mein light cork bob (same size). Kis ka period zyada lamba hoga?

Recall Solution

WHAT: Periods compare karo. WHY: Test karo ki tum yaad rakhte ho ki mass cancel hota hai. Koi nahi — dono ka same period hoga. Mass $m$ $T = 2 π L / g$ mein appear hi nahi karta. Ise isochronism kehte hain. Dekho Restoring Force and Equilibrium.

L1·Q3

Square root ke andar upar kaun sa symbol hai, $L$ ya $g$ ? Isse kya physical consequence nikalti hai?

Recall Solution

$L$ upar hai: $T = 2 π L / g$ . Consequence: lamba string lamba period deta hai (Long pendulums are sLow). Bada $g$ (neeche) chhota period deta hai.

Level 2 — Application

Plug in karo, rearrange karo, units convert karo.

L2·Q1

Ek "seconds pendulum" ki length dhundho — woh jiska period exactly $T = 2.0 s$ ho.

Recall Solution

WHAT: $L$ ke liye solve karo. WHY: Pehle algebraically rearrange karo, phir substitute karo (errors kam hote hain). $T = 2 π \frac{L}{g} \Rightarrow \frac{T}{2 π} = \frac{L}{g} \Rightarrow L = g (\frac{T}{2 π})^{2}$ $L = 9.8 (\frac{2.0}{6.2832})^{2} = 9.8 (0.3183)^{2} = 9.8 (0.10132) = 0.993 m$ Toh $L \approx 0.99 m$ — almost ek metre.

L2·Q2

Moon par ( $g_{Moon} = 1.6 m/s^{2}$ ) ek pendulum ki $L = 1.0 m$ hai. $T$ nikalo aur Earth se compare karo.

Recall Solution

WHAT: Same formula, chhota $g$ . $T_{Moon} = 2 π \frac{1.0}{1.6} = 2 π 0.625 = 2 π (0.7906) = 4.97 s$ Earth par same pendulum $T = 2 π 1.0/9.8 = 2.01 s$ deta hai. WHY it makes sense: Kamzor gravity bob ko zyada gently restore karti hai, isliye woh slower jhulta hai — yahan lagbhag $2.5 \times$ slower. Dekho Measuring g with a Pendulum.

L2·Q3

Swing amplitude $θ_{0} = 8. 0^{\circ}$ ko radians mein convert karo, aur check karo ki $sin θ \approx θ$ $1%$ ke andar hold karta hai ya nahi.

Recall Solution

WHAT: Convert karo aur approximation test karo. WHY radians? Series $sin θ = θ - θ^{3} /6 + \dots$ tabhi sach hai jab $θ$ radians mein measured ho. $θ_{0} = 8. 0^{\circ} \times \frac{π}{180} = 0.13963 rad$ $sin (0.13963) = 0.13918$ Relative error $= \frac{0.13963 - 0.13918}{0.13918} = 0.00325 = 0.33%$ . Haan, $1%$ ke andar hai.

Level 3 — Analysis

Quantities ek doosre par kaise depend karti hain?

L3·Q1

Ek pendulum ki length $9$ ke factor se badha di jaati hai. Iske period mein kitna factor change aata hai? Frequency mein?

Recall Solution

WHAT: Proportionality $T \propto L$ use karo. WHY: Constants ( $2 π$ , $g$ ) change nahi hote, toh sirf $L$ dependence matter karti hai. $\frac{T _{new}}{T _{old}} = \frac{9 L}{L} = 9 = 3$ Period teen guna ho jaata hai. Frequency $f = 1/ T$ , toh frequency pehle ki $1/3$ ho jaati hai.

L3·Q2

Ek aisi planet par jahan $g$ Earth ki value ka $4$ guna hai, fixed-length pendulum ke period ka kya hoga?

Recall Solution

WHAT: $T \propto 1/ g$ . $\frac{T _{new}}{T _{old}} = \frac{g}{4 g} = \frac{1}{4} = \frac{1}{2}$ Period aadha ho jaata hai — pendulum do guna tez jhulta hai. Bada $g$ = zyada strong restoring pull = jaldi oscillation.

L3·Q3

Ek clock pendulum $T = 2.00 s$ ke liye design kiya gaya hai lekin $T = 2.02 s$ se chalta hua paaya jaata hai (thoda slow). Kya pendulum ko chhota karna chahiye ya lamba, aur lagbhag kitne fraction se?

Recall Solution

WHAT: $L$ mein ek chote change ko $T$ mein ek chote change se relate karo. WHY differentiate? Choti adjustments ke liye hume rate $d T / d L$ chahiye, poora recompute nahi — yahi jagah hai jahan derivative kaam aati hai. Kyunki $T = 2 π L / g = (2 π / g) L^{1/2}$ , $\frac{d T}{T} = \frac{1}{2} \frac{d L}{L} .$ Ye kehta hai ki ek fractional length change aadha fractional period change produce karta hai. Hum $\frac{2.02 - 2.00}{2.00} = 1.0%$ slow hain (period zyada lamba hai), toh hume $T$ chhota chahiye ⇒ $L$ chhota karo. $\frac{d L}{L} = 2 \frac{d T}{T} = 2 (- 0.01) = - 0.02 = - 2%.$ Pendulum ko lagbhag $2%$ chhota karo.

Level 4 — Synthesis

Energy, corrections, aur doosre systems ko combine karo.

L4·Q1

$m = 0.20 kg$ mass ka bob $L = 1.5 m$ ki string par chota angle $θ_{0} = 0.15 rad$ se release kiya jaata hai. Energy use karke, bottom par maximum speed nikalo. (Small-angle height $h \approx \frac{1}{2} L θ_{0}^{2}$ use karo.)

Recall Solution

WHAT: Upar ka potential energy neeche kinetic energy mein convert karo. Dekho Energy in Oscillations. WHY the height formula? Bob $h = L (1 - cos θ_{0})$ se utha. Chote $θ_{0}$ ke liye, $cos θ_{0} \approx 1 - θ_{0}^{2} /2$ , toh $h \approx \frac{1}{2} L θ_{0}^{2}$ — wohi small-angle spirit jaise parent note mein. $h \approx \frac{1}{2} (1.5) (0.15)^{2} = \frac{1}{2} (1.5) (0.0225) = 0.016875 m$ Energy conservation $m g h = \frac{1}{2} m v^{2}$ (mass cancel ho jaata hai): $v = 2 g h = 2 (9.8) (0.016875) = 0.33075 = 0.575 m/s .$ Maximum speed $\approx 0.58 m/s$ .

L4·Q2

Ek mass-spring system aur is pendulum ( $L = 1.5 m$ ) ka same period hona chahiye. Agar spring $m = 0.20 kg$ carry karta hai toh kitna spring constant $k$ chahiye?

Recall Solution

WHAT: Do oscillators ke periods match karo. Dekho Mass-Spring System. WHY: Dono $\overset{x}{¨} = - ω^{2} x$ follow karte hain; equal period matlab equal $ω^{2}$ . Pendulum: $ω^{2} = g / L = 9.8/1.5 = 6.533 s^{- 2}$ . Spring: $ω^{2} = k / m$ , toh $k = m ω^{2} = 0.20 \times 6.533 = 1.307 N/m .$ Toh $k \approx 1.31 N/m$ .

L4·Q3

Same pendulum ( $L = 1.5 m$ ) ko large amplitude $θ_{0} = 3 0^{\circ}$ se jhulaya jaata hai. First correction $T \approx 2 π L / g (1 + \frac{θ _{0}^{2}}{16})$ use karke, period nikalo aur percentage batao ki ye small-angle value se kitna zyada hai.

Recall Solution

WHAT: Amplitude correction apply karo. WHY: Pure formula ne assume kiya tha $sin θ \approx θ$ ; $3 0^{\circ}$ par dropped $θ^{3}$ term matter karta hai, isliye hum next term add karte hain. Small-angle base: $T_{0} = 2 π 1.5/9.8 = 2 π 0.15306 = 2 π (0.39123) = 2.4581 s$ . Convert $θ_{0} = 3 0^{\circ} = 0.5236 rad$ , toh $θ_{0}^{2} /16 = 0.27416/16 = 0.017135$ . $T = 2.4581 (1 + 0.017135) = 2.4581 (1.017135) = 2.5002 s .$ Excess $= 1.71%$ small-angle value se zyada.

Level 5 — Mastery

Poori multi-step reasoning; har trap se bachna.

L5·Q1 — Lab measurement of $g$

Lab mein ek student $L = 0.993 m$ pendulum ke $50$ complete oscillations time karta hai aur $99.4 s$ paata hai. (a) Period nikalo. (b) $g$ calculate karo. (c) Explain karo ki ek swing ki jagah $50$ swings time karna zyada samajhdari kyun hai.

Recall Solution

(a) WHAT: Period = total time ÷ number of swings. $T = \frac{99.4}{50} = 1.988 s .$ (b) WHAT: $T = 2 π L / g$ ko $g$ ke liye solve karo, dono sides square karke. WHY square? $g$ square root ke neeche trapped hai; square karne se woh free ho jaata hai. $g = \frac{4 π ^{2} L}{T ^{2}} = \frac{4 π ^{2} ( 0.993 )}{( 1.988 ) ^{2}} = \frac{39.196}{3.9521} = 9.918 m/s^{2} .$ Toh $g \approx 9.92 m/s^{2}$ . (c) WHY 50 swings: Human stopwatch mein ek fixed reaction error hota hai (maano $\pm 0.2$ s). $50$ periods par spread hone se, woh error per period $50$ se divide ho jaata hai, $T$ mein uncertainty $50 \times$ kam ho jaati hai. Dekho Measuring g with a Pendulum.

L5·Q2 — Temperature se lamba hue clock ke liye length change derive karna

Ek pendulum clock sardi mein perfect time rakhta hai. Garmi mein metal rod expand ho jaati hai isliye $L$ $0.05%$ badh jaata hai. (a) Kya clock fast chalega ya slow? (b) Har din kitne seconds drift karega?

Recall Solution

(a) WHAT/WHY: Lamba $L$ ⇒ lamba $T$ ⇒ har "second" jo clock count karta hai woh actually ek true second se lamba hota hai ⇒ woh bahut kam times tick karta hai ⇒ woh slow chalta hai. (b) WHAT: L3 waali sensitivity relation use karo: $\frac{d T}{T} = \frac{1}{2} \frac{d L}{L}$ . $\frac{d T}{T} = \frac{1}{2} (0.0005) = 0.00025 = 0.025%.$ Ek din mein $86400 s$ hote hain. Clock din ka itna fraction lose karta hai: $Δ t = 0.00025 \times 86400 = 21.6 s/day .$ Clock har din lagbhag $21.6$ s slow chalta hai.

L5·Q3 — Do pendulums saath beat karte hain

Pendulum A ki $L_{A} = 1.00 m$ hai; pendulum B ki $L_{B} = 1.02 m$ hai. Dono in phase swing karna shuru karte hain. Kitne time baad woh agli baar exactly in phase mein saath jhulenge? ( $g = 9.8$ lo.)

Recall Solution

WHAT: Woh ek doosre ke saath tab realign hote hain jab dono poori periods complete karte hain — specifically ek aisi time ke baad jo "beat period" ke barabar ho, jab B ne A se exactly ek kam full swing ki ho. WHY: Unke periods thode alag hain, isliye har cycle mein ek phase gap khulta hai; woh tab re-synchronise karte hain jab woh gap ek poori period ke barabar ho jaaye. $T_{A} = 2 π \frac{1.00}{9.8} = 2 π (0.31944) = 2.0071 s,$ $T_{B} = 2 π \frac{1.02}{9.8} = 2 π (0.32261) = 2.0270 s .$ Maano $t$ resync time hai. Isme A $n$ periods complete karta hai aur B $n - 1$ : $t = n T_{A} = (n - 1) T_{B} .$ $n$ ke liye solve karo: $n T_{A} = (n - 1) T_{B} \Rightarrow n (T_{B} - T_{A}) = T_{B} \Rightarrow n = \frac{T _{B}}{T _{B} - T _{A}} .$ $n = \frac{2.0270}{2.0270 - 2.0071} = \frac{2.0270}{0.01994} = 101.66.$ Phir $t = n T_{A} = 101.66 \times 2.0071 = 204.05 s \approx 204 s .$ Toh lagbhag har $3.4$ minutes mein woh saath jhulte hain. (Neeche beat picture dekho.)

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

Active recall

Recall Quick self-check

Agar $L$ ko $9$ se multiply kiya jaaye toh $T$ kitne factor se change hoga? ::: $3$ (kyunki $T \propto L$ ).
$1%$ period error ke liye kitne $%$ length correction chahiye? ::: $2%$ (square root ki wajah se factor-of-two).
$g$ measure karne ke liye kaafi saari swings time kyun karni chahiye? ::: Reaction-time error swings ki number se divide ho jaata hai.
$θ_{0}^{2} /16$ correction ke andar amplitude kis unit mein honi chahiye? ::: Radians mein.

Connections

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