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

Throughout, the master formula is the one built in the parent note the parent note:

Take $g = 9.8 m/s^{2}$ on Earth unless a problem says otherwise. Use $π = 3.1416$ .

Level 1 — Recognition

Can you spot the right formula and read it correctly?

L1·Q1

A simple pendulum has length $L = 2.45 m$ . Write down the expression for its period and compute $T$ .

Recall Solution

WHAT: Direct use of $T = 2 π L / g$ . WHY: Recognition — no rearranging needed. $T = 2 π \frac{2.45}{9.8} = 2 π 0.25 = 2 π (0.5) = 3.14 s$ So $T \approx 3.14 s$ .

L1·Q2

Two pendulums are identical except one has a heavy iron bob and one a light cork bob (same size). Which has the longer period?

Recall Solution

WHAT: Compare periods. WHY: Test whether you remember mass cancels. Neither — they have the same period. Mass $m$ does not appear in $T = 2 π L / g$ . This is isochronism. See Restoring Force and Equilibrium.

L1·Q3

Which symbol sits on top inside the square root, $L$ or $g$ ? What physical consequence follows?

Recall Solution

$L$ is on top: $T = 2 π L / g$ . Consequence: a longer string gives a longer period (Long pendulums are sLow). Bigger $g$ (bottom) gives a shorter period.

Level 2 — Application

Plug in, rearrange, convert units.

L2·Q1

Find the length of a "seconds pendulum" — one whose period is exactly $T = 2.0 s$ .

Recall Solution

WHAT: Solve for $L$ . WHY: We rearrange algebraically first, then substitute (fewer errors). $T = 2 π \frac{L}{g} \Rightarrow \frac{T}{2 π} = \frac{L}{g} \Rightarrow L = g (\frac{T}{2 π})^{2}$ $L = 9.8 (\frac{2.0}{6.2832})^{2} = 9.8 (0.3183)^{2} = 9.8 (0.10132) = 0.993 m$ So $L \approx 0.99 m$ — about one metre.

L2·Q2

A pendulum on the Moon ( $g_{Moon} = 1.6 m/s^{2}$ ) has $L = 1.0 m$ . Find $T$ and compare to Earth.

Recall Solution

WHAT: Same formula, smaller $g$ . $T_{Moon} = 2 π \frac{1.0}{1.6} = 2 π 0.625 = 2 π (0.7906) = 4.97 s$ On Earth the same pendulum gives $T = 2 π 1.0/9.8 = 2.01 s$ . WHY it makes sense: Weaker gravity restores the bob more gently, so it swings slower — about $2.5 \times$ slower here. See Measuring g with a Pendulum.

L2·Q3

Convert a swing amplitude of $θ_{0} = 8. 0^{\circ}$ into radians, and check whether $sin θ \approx θ$ holds to within $1%$ .

Recall Solution

WHAT: Convert and test the approximation. WHY radians? The series $sin θ = θ - θ^{3} /6 + \dots$ is only true when $θ$ is measured in radians. $θ_{0} = 8. 0^{\circ} \times \frac{π}{180} = 0.13963 rad$ $sin (0.13963) = 0.13918$ Relative error $= \frac{0.13963 - 0.13918}{0.13918} = 0.00325 = 0.33%$ . Yes, well within $1%$ .

Level 3 — Analysis

How do the quantities depend on each other?

L3·Q1

A pendulum's length is increased by a factor of $9$ . By what factor does its period change? Its frequency?

Recall Solution

WHAT: Use the proportionality $T \propto L$ . WHY: Constants ( $2 π$ , $g$ ) don't change, so only the $L$ dependence matters. $\frac{T _{new}}{T _{old}} = \frac{9 L}{L} = 9 = 3$ Period triples. Frequency $f = 1/ T$ , so frequency becomes $1/3$ of before.

L3·Q2

On a planet where $g$ is $4$ times Earth's value, what happens to the period of a fixed-length pendulum?

Recall Solution

WHAT: $T \propto 1/ g$ . $\frac{T _{new}}{T _{old}} = \frac{g}{4 g} = \frac{1}{4} = \frac{1}{2}$ Period halves — the pendulum swings twice as fast. Bigger $g$ = stronger restoring pull = quicker oscillation.

L3·Q3

A clock pendulum is designed for $T = 2.00 s$ but is found to run with $T = 2.02 s$ (slightly slow). Should you make the pendulum shorter or longer, and by roughly what fraction?

Recall Solution

WHAT: Relate a small change in $L$ to a small change in $T$ . WHY differentiate? For tiny adjustments we want the rate $d T / d L$ , not a full recompute — this is where a derivative earns its place. Since $T = 2 π L / g = (2 π / g) L^{1/2}$ , $\frac{d T}{T} = \frac{1}{2} \frac{d L}{L} .$ This says a fractional length change produces half that fractional period change. We are $\frac{2.02 - 2.00}{2.00} = 1.0%$ too slow (too long a period), so we need $T$ smaller ⇒ $L$ shorter. $\frac{d L}{L} = 2 \frac{d T}{T} = 2 (- 0.01) = - 0.02 = - 2%.$ Shorten the pendulum by about $2%$ .

Level 4 — Synthesis

Combine energy, corrections, and other systems.

L4·Q1

A bob of mass $m = 0.20 kg$ on a string of length $L = 1.5 m$ is released from a small angle $θ_{0} = 0.15 rad$ . Using energy, find its maximum speed at the bottom. (Use small-angle height $h \approx \frac{1}{2} L θ_{0}^{2}$ .)

Recall Solution

WHAT: Convert potential energy at the top into kinetic energy at the bottom. See Energy in Oscillations. WHY the height formula? The bob rises by $h = L (1 - cos θ_{0})$ . For small $θ_{0}$ , $cos θ_{0} \approx 1 - θ_{0}^{2} /2$ , so $h \approx \frac{1}{2} L θ_{0}^{2}$ — the same small-angle spirit as the parent note. $h \approx \frac{1}{2} (1.5) (0.15)^{2} = \frac{1}{2} (1.5) (0.0225) = 0.016875 m$ Energy conservation $m g h = \frac{1}{2} m v^{2}$ (mass cancels): $v = 2 g h = 2 (9.8) (0.016875) = 0.33075 = 0.575 m/s .$ Maximum speed $\approx 0.58 m/s$ .

L4·Q2

A mass-spring system and this pendulum ( $L = 1.5 m$ ) are to have the same period. What spring constant $k$ is needed if the spring carries $m = 0.20 kg$ ?

Recall Solution

WHAT: Match periods of two oscillators. See Mass-Spring System. WHY: Both obey $\overset{x}{¨} = - ω^{2} x$ ; equal period means equal $ω^{2}$ . Pendulum: $ω^{2} = g / L = 9.8/1.5 = 6.533 s^{- 2}$ . Spring: $ω^{2} = k / m$ , so $k = m ω^{2} = 0.20 \times 6.533 = 1.307 N/m .$ So $k \approx 1.31 N/m$ .

L4·Q3

The same pendulum ( $L = 1.5 m$ ) is swung with a large amplitude $θ_{0} = 3 0^{\circ}$ . Using the first correction $T \approx 2 π L / g (1 + \frac{θ _{0}^{2}}{16})$ , find the period and the percentage it exceeds the small-angle value.

Recall Solution

WHAT: Apply the amplitude correction. WHY: The pure formula assumed $sin θ \approx θ$ ; at $3 0^{\circ}$ the dropped $θ^{3}$ term matters, so we add the next term. Small-angle base: $T_{0} = 2 π 1.5/9.8 = 2 π 0.15306 = 2 π (0.39123) = 2.4581 s$ . Convert $θ_{0} = 3 0^{\circ} = 0.5236 rad$ , so $θ_{0}^{2} /16 = 0.27416/16 = 0.017135$ . $T = 2.4581 (1 + 0.017135) = 2.4581 (1.017135) = 2.5002 s .$ Excess $= 1.71%$ above the small-angle value.

Level 5 — Mastery

Full multi-step reasoning; watch every trap.

L5·Q1 — Lab measurement of $g$

In a lab a student times $50$ complete oscillations of an $L = 0.993 m$ pendulum and gets $99.4 s$ . (a) Find the period. (b) Compute $g$ . (c) Explain why timing $50$ swings rather than one is smarter.

Recall Solution

(a) WHAT: Period is total time ÷ number of swings. $T = \frac{99.4}{50} = 1.988 s .$ (b) WHAT: Solve $T = 2 π L / g$ for $g$ by squaring both sides. WHY square? $g$ is trapped under a square root; squaring frees it. $g = \frac{4 π ^{2} L}{T ^{2}} = \frac{4 π ^{2} ( 0.993 )}{( 1.988 ) ^{2}} = \frac{39.196}{3.9521} = 9.918 m/s^{2} .$ So $g \approx 9.92 m/s^{2}$ . (c) WHY 50 swings: A human stopwatch has a fixed reaction error (say $\pm 0.2$ s). Spread over $50$ periods, that error is divided by $50$ per period, shrinking the uncertainty in $T$ by $50 \times$ . See Measuring g with a Pendulum.

L5·Q2 — Deriving the length change for a temperature-lengthened clock

A pendulum clock keeps perfect time in winter. In summer the metal rod expands so $L$ grows by $0.05%$ . (a) Does the clock run fast or slow? (b) By how many seconds per day does it drift?

Recall Solution

(a) WHAT/WHY: Longer $L$ ⇒ longer $T$ ⇒ each "second" the clock counts is actually longer than a true second ⇒ it ticks too few times ⇒ it runs slow. (b) WHAT: Use the sensitivity relation from L3: $\frac{d T}{T} = \frac{1}{2} \frac{d L}{L}$ . $\frac{d T}{T} = \frac{1}{2} (0.0005) = 0.00025 = 0.025%.$ A day has $86400 s$ . The clock loses this fraction of a day: $Δ t = 0.00025 \times 86400 = 21.6 s/day .$ The clock runs about $21.6$ s slow each day.

L5·Q3 — Two pendulums beating together

Pendulum A has $L_{A} = 1.00 m$ ; pendulum B has $L_{B} = 1.02 m$ . Both start swinging in phase. After how long do they next return to swinging exactly in phase together? (Take $g = 9.8$ .)

Recall Solution

WHAT: They realign after a whole number of periods each — specifically after a time equal to the "beat period", when B has done exactly one fewer full swing than A. WHY: Their periods differ slightly, so a phase gap opens each cycle; they re-synchronise when that gap has grown to one full period. $T_{A} = 2 π \frac{1.00}{9.8} = 2 π (0.31944) = 2.0071 s,$ $T_{B} = 2 π \frac{1.02}{9.8} = 2 π (0.32261) = 2.0270 s .$ Let $t$ be the resync time. In it A completes $n$ periods and B completes $n - 1$ : $t = n T_{A} = (n - 1) T_{B} .$ Solve for $n$ : $n T_{A} = (n - 1) T_{B} \Rightarrow n (T_{B} - T_{A}) = T_{B} \Rightarrow n = \frac{T _{B}}{T _{B} - T _{A}} .$ $n = \frac{2.0270}{2.0270 - 2.0071} = \frac{2.0270}{0.01994} = 101.66.$ Then $t = n T_{A} = 101.66 \times 2.0071 = 204.05 s \approx 204 s .$ So roughly every $3.4$ minutes they swing together again. (See the beat picture below.)

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

Active recall

Recall Quick self-check

What factor does $T$ change by if $L$ is multiplied by $9$ ? ::: $3$ (since $T \propto L$ ).
A $1%$ period error needs what $%$ length correction? ::: $2%$ (factor-of-two from the square root).
Why time many swings to measure $g$ ? ::: Reaction-time error is divided by the number of swings.
What must the amplitude be in, inside the $θ_{0}^{2} /16$ correction? ::: Radians.

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