PhysicsGrade 10–12

Simple pendulum

Pull a pendulum aside, let it go and measure how long one swing takes. What does it depend on? Also with large angles and damping.

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What is it about?

The restoring force Fr=mgsin⁡φF_r = m g \sin\varphi pulls the bob back to equilibrium. For small amplitudes (sin⁡φ≈φ\sin\varphi \approx \varphi) the motion is harmonic with

T0=2πlg.T_0 = 2\pi \sqrt{\frac{l}{g}}.

The period depends only on the length and on gg – not on the mass. Since T2=4π2g lT^2 = \frac{4\pi^2}{g}\, l, a graph of T2T^2 against ll is a straight line through the origin with slope 4π2/g4\pi^2/g. For large amplitudes the period gets longer: up to about 60∘60^\circ, T≈T0 (1+φ02/16)T \approx T_0 \,(1 + \varphi_0^2/16) with φ0\varphi_0 in radians; the simulation itself solves the equation of motion numerically and computes the period exactly as T=T0/AGM⁡(1,cos⁡φ02)T = T_0 / \operatorname{AGM}(1, \cos\tfrac{\varphi_0}{2}), more than 1 %1\,\% above 22.8∘22.8^\circ and 18 %18\,\% at 90∘90^\circ. Above 90∘90^\circ a string would go slack, so the simulation uses a rigid rod.

Try it

  1. Drag the bob aside and let go. Watch the energy bars while the total stays the same.
  2. Turn on the force arrows. Forces from 90°
  3. Compare with the small-angle approximation. Graphs
  4. Time 10 oscillations at several lengths and determine gg. Measurements
  5. Damping turns the phase-space path into a spiral. Damped

Tasks

Task 1: Seconds pendulum

How long must a pendulum be for T=2 sT = 2\,\text{s}?

Show solution

l=g(T2π)2≈0.994 ml = g \left(\frac{T}{2\pi}\right)^2 \approx 0.994\,\text{m}. Show

Task 2: On the Moon

Find T0T_0 of a 1 m1\,\text{m} pendulum on the Moon (g=1.62 ms2g = 1.62\,\tfrac{\text{m}}{\text{s}^2}).

Show solution

T0≈4.94 sT_0 \approx 4.94\,\text{s}. Show in the simulation

Task 3: Speed and tension

m=0.5 kgm = 0.5\,\text{kg}, l=1 ml = 1\,\text{m}, released from 20∘20^\circ: how fast is the bob at the bottom? What is the tension there when released from 90∘90^\circ?

Show solution

h≈0.0603 mh \approx 0.0603\,\text{m}, v=2gh≈1.09 msv = \sqrt{2gh} \approx 1.09\,\tfrac{\text{m}}{\text{s}}. From 90∘90^\circ: FT=3mg≈14.7 NF_T = 3mg \approx 14.7\,\text{N}.

Ideas for the classroom

  • The measurement series (with random reaction time) prepares a real stopwatch experiment: why time many periods, why use a small amplitude?
  • The exact equation φ¨=−glsin⁡φ−2γφ˙\ddot\varphi = -\tfrac{g}{l}\sin\varphi - 2\gamma\dot\varphi is solved numerically; the small-angle approximation runs alongside as a dashed pendulum.
  • Common stumbling blocks: “A heavier pendulum swings faster.” – “The tension always equals the weight.” (At the bottom it is larger by mv2/lm v^2 / l.)