16.4 The Simple Pendulum
Learning objectives.
By the end of this section, you will be able to:
- Measure acceleration due to gravity.
Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child’s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure 16.13 . Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period.
We begin by defining the displacement to be the arc length s s . We see from Figure 16.13 that the net force on the bob is tangent to the arc and equals − mg sin θ − mg sin θ . (The weight mg mg has components mg cos θ mg cos θ along the string and mg sin θ mg sin θ tangent to the arc.) Tension in the string exactly cancels the component mg cos θ mg cos θ parallel to the string. This leaves a net restoring force back toward the equilibrium position at θ = 0 θ = 0 .
Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 15º 15º ), sin θ ≈ θ sin θ ≈ θ ( sin θ sin θ and θ θ differ by about 1% or less at smaller angles). Thus, for angles less than about 15º 15º , the restoring force F F is
The displacement s s is directly proportional to θ θ . When θ θ is expressed in radians, the arc length in a circle is related to its radius ( L L in this instance) by:
For small angles, then, the expression for the restoring force is:
This expression is of the form:
where the force constant is given by k = mg / L k = mg / L and the displacement is given by x = s x = s . For angles less than about 15º 15º , the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator.
Using this equation, we can find the period of a pendulum for amplitudes less than about 15º 15º . For the simple pendulum:
for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period T T for a pendulum is nearly independent of amplitude, especially if θ θ is less than about 15º 15º . Even simple pendulum clocks can be finely adjusted and accurate.
Note the dependence of T T on g g . If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider the following example.
Example 16.5
Measuring acceleration due to gravity: the period of a pendulum.
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?
We are asked to find g g given the period T T and the length L L of a pendulum. We can solve T = 2π L g T = 2π L g for g g , assuming only that the angle of deflection is less than 15º 15º .
- Square T = 2π L g T = 2π L g and solve for g g : g = 4π 2 L T 2 . g = 4π 2 L T 2 . 16.30
- Substitute known values into the new equation: g = 4π 2 0 . 75000 m 1 . 7357 s 2 . g = 4π 2 0 . 75000 m 1 . 7357 s 2 . 16.31
- Calculate to find g g : g = 9 . 8281 m / s 2 . g = 9 . 8281 m / s 2 . 16.32
This method for determining g g can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation sin θ ≈ θ sin θ ≈ θ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5º 0.5º .
Making Career Connections
Knowing g g can be important in geological exploration; for example, a map of g g over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits.
Take Home Experiment: Determining g g
Use a simple pendulum to determine the acceleration due to gravity g g in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than 10º 10º , allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch. Calculate g g . How accurate is this measurement? How might it be improved?
Check Your Understanding
An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg 10 kg . Pendulum 2 has a bob with a mass of 100 kg 100 kg . Describe how the motion of the pendula will differ if the bobs are both displaced by 12º 12º .
The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum’s length) and by the acceleration due to gravity.
PhET Explorations
Pendulum lab.
Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It’s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of g g on planet X. Notice the anharmonic behavior at large amplitude.
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Underactuated Robotics
Algorithms for Walking, Running, Swimming, Flying, and Manipulation
Russ Tedrake
Note: These are working notes used for a course being taught at MIT . They will be updated throughout the Spring 2024 semester. Lecture videos are available on YouTube .
The Simple Pendulum
Introduction.
Our goals for this chapter are modest: we'd like to understand the dynamics of a pendulum.
Why a pendulum? In part, because the dynamics of a majority of our multi-link robotics manipulators are simply the dynamics of a large number of coupled pendula. Also, the dynamics of a single pendulum are rich enough to introduce most of the concepts from nonlinear dynamics that we will use in this text, but tractable enough for us to (mostly) understand in the next few pages.
The Lagrangian derivation of the equations of motion (as described in the appendix) of the simple pendulum yields: \begin{equation*} m l^2 \ddot\theta(t) + mgl\sin{\theta(t)} = Q. \end{equation*} We'll consider the case where the generalized force, $Q$, models a damping torque (from friction) plus a control torque input, $u(t)$: $$Q = -b\dot\theta(t) + u(t).$$
Nonlinear dynamics with a constant torque
Let us first consider the dynamics of the pendulum if it is driven in a particular simple way: a torque which does not vary with time: \begin{equation} ml^2 \ddot\theta + b\dot\theta + mgl \sin\theta = u_0. \end{equation}
Simple Pendulum in Python
You can experiment with this system in using
These are relatively simple differential equations, so if I give you $\theta(0)$ and $\dot\theta(0)$, then you should be able to integrate them to obtain $\theta(t)$... right? Although it is possible, integrating even the simplest case ($b = u = 0$) involves elliptic integrals of the first kind; there is relatively little intuition to be gained here.
This is in stark contrast to the case of linear systems, where much of our understanding comes from being able to explicitly integrate the equations. For instance, for a simple linear system we have $$\dot{q} = a q \quad \rightarrow \quad q(t) = q(0) e^{at},$$ and we can immediately understand that the long-term behavior of the system is a (stable) decaying exponential if $a 0$, and that the system does nothing if $a=0$. Here we are with certainly one of the simplest nonlinear systems we can imagine, and we can't even solve this system?
All is not lost. If what we care about is the long-term behavior of the system, then there are a number of techniques we can apply. In this chapter, we will start by investigating graphical solution methods. These methods are described beautifully in a book by Steve Strogatz Strogatz94 .
The overdamped pendulum
Let's start by studying a special case -- intuitively when $b\dot\theta \gg ml^2\ddot\theta$ -- which via dimensional analysis (using the natural frequency $\sqrt{\frac{g}{l}}$ to match units) occurs when $b \sqrt\frac{l}{g} \gg ml^2$. This is the case of heavy damping, for instance if the pendulum was moving in molasses. In this case, the damping term dominates the acceleration term, and we have: $$ml^2 \ddot\theta + b\dot\theta \approx b\dot\theta = u_0 - mgl\sin\theta.$$ In other words, in the case of heavy damping, the system looks approximately first order. This is a general property of heavily-damped systems, such as fluids at very low Reynolds number.
I'd like to ignore one detail for a moment: the fact that $\theta$ wraps around on itself every $2\pi$. To be clear, let's write the system without the wrap-around as: \begin{equation}b\dot{x} = u_0 - mgl\sin{x}.\label{eq:overdamped_pend_ct}\end{equation} Our goal is to understand the long-term behavior of this system: to find $x(\infty)$ given $x(0)$. Let's start by plotting $\dot{x}$ vs $x$ for the case when $u_0=0$:
The first thing to notice is that the system has a number of fixed points or steady states , which occur whenever $\dot{x} = 0$. In this simple example, the zero-crossings are $x^* = \{..., -\pi, 0, \pi, 2\pi, ...\}$. When the system is in one of these states, it will never leave that state. If the initial conditions are at a fixed point, we know that $x(\infty)$ will be at the same fixed point.
- Locally stable in the sense of Lyapunov (i.s.L.). A fixed point, $x^*$ is locally stable i.s.L. if for every (small) $\epsilon > 0$, $\exists~\delta > 0$ such that if $\| x(0) - x^* \| < \delta$ then $\forall t$ $\| x(t) - x^*\| < \epsilon$. In words, this means that for any ball of size $\epsilon$ around the fixed point, I can create a ball of size $\delta$ which guarantees that if the system is started inside the $\delta$ ball then it will remain inside the $\epsilon$ ball for all of time.
- Locally attractive . A fixed point is locally attractive if for every (small) $\epsilon$ we have $x(0) = x^* + \epsilon$ implies that $\lim_{t\rightarrow \infty} x(t) = x^*$.
- Locally asymptotically stable . A fixed point is locally asymptotically stable if it is locally stable i.s.L. and locally attractive.
- Locally exponentially stable . A fixed point is locally exponentially stable if for every (small) $\epsilon$, we have $x(0) = x^* + \epsilon$ implies that $\| x(t) - x^* \| < Ce^{-\alpha t}$, for some positive constants $C$ and $\alpha$.
- Unstable . A fixed point is unstable if it is not locally stable i.s.L.
An initial condition near a fixed point that is stable in the sense of Lyapunov may never reach the fixed point (but it won't diverge), near an asymptotically stable fixed point will reach the fixed point as $t \rightarrow \infty$, and near an exponentially stable fixed point will reach the fixed point with a bounded rate. An exponentially stable fixed point is also an asymptotically stable fixed point, but the converse is not true. Attractivity does not actually imply Lyapunov stability † † we can't see that in one dimension so will have to hold that example for a moment , which is why we require i.s.L. specifically for the definition of asymptotic stability. Systems which are stable i.s.L. but not asymptotically stable are easy to construct (e.g. $\dot{x} = 0$). Interestingly, it is also possible to have nonlinear systems that converge (or diverge) in finite-time; a so-called finite-time stability ; we will see examples of this later in the book, but it is a difficult topic to penetrate with graphical analysis. Rigorous nonlinear system analysis is rich with subtleties and surprises. Moreover, these differences actually matter -- the code that we will write to stabilize the systems will be subtly different depending on what type of stability we want, and it can make or break the success of our methods.
Our graph of $\dot{x}$ vs. $x$ can be used to convince ourselves of i.s.L. and asymptotic stability by visually inspecting $\dot{x}$ in the vicinity of a fixed point. Even exponential stability can be inferred if we can find a negatively-sloped line passing through the equilibrium point which separates the graph of $f(x)$ from the horizontal axis. The existence of this separating line implies that the nonlinear system will converge at least as fast as the linear system represented by the straight line. I will graphically illustrate unstable fixed points with open circles and stable fixed points (i.s.L.) with filled circles.
Next, we need to consider what happens to initial conditions which begin farther from the fixed points. If we think of the dynamics of the system as a flow on the $x$-axis, then we know that anytime $\dot{x} > 0$, the flow is moving to the right, and $\dot{x} < 0$, the flow is moving to the left. If we further annotate our graph with arrows indicating the direction of the flow, then the entire (long-term) system behavior becomes clear:
For instance, we can see that any initial condition $x(0) \in (-\pi,\pi)$ will result in $\lim_{t\rightarrow \infty} x(t) = 0$. This region is called the region of attraction (or basin of attraction ) of the fixed point at $x^* = 0$. Basins of attraction of two fixed points cannot overlap, and the manifold separating two basins of attraction is called the separatrix . Here the unstable fixed points, at $x^* = \{.., -\pi, \pi, 3\pi, ...\}$ form the separatrix between the basins of attraction of the stable fixed points. In general, regions of attraction are always open, connected, invariant sets, with boundaries formed by trajectories.
As these plots demonstrate, the behavior of a first-order one dimensional system on a line is relatively constrained. The system will either monotonically approach a fixed-point or monotonically move toward $\pm \infty$. There are no other possibilities. Oscillations, for example, are impossible. Graphical analysis is a fantastic analysis tool for many first-order nonlinear systems (not just pendula); as illustrated by the following example:
Nonlinear autapse
These equations are not arbitrary - they are actually a model for one of the simplest neural networks, and one of the simplest models of persistent memory Seung00 . In the equation $x$ models the firing rate of a single neuron, which has a feedback connection to itself. $\tanh$ is the activation (sigmoidal) function of the neuron, and $w$ is the weight of the synaptic feedback.
Experiment with it for yourself:
As I've also provided the equations of an LSTM unit that you can also experiment with. See if you can figure it out!
Architectures like the Transformer for sequence modeling can also be understood through the lens of dynamical systems theory (as autoregressive models), but we will defer that discussion until later in the notes.
One last piece of terminology. In the neuron example, and in many dynamical systems, the dynamics were parameterized; in this case by a single parameter, $w$. As we varied $w$, the fixed points of the system moved around. In fact, if we increase $w$ through $w=1$, something dramatic happens - the system goes from having one fixed point to having three fixed points. This is called a bifurcation . This particular bifurcation is called a pitchfork bifurcation. We often draw bifurcation diagrams which plot the fixed points of the system as a function of the parameters, with solid lines indicating stable fixed points and dashed lines indicating unstable fixed points, as seen in the figure:
Our pendulum equations also have a (saddle-node) bifurcation when we change the constant torque input, $u_0$. Finally, let's return to the original equations in $\theta$, instead of in $x$. Only one point to make: because of the wrap-around, this system will appear to have oscillations. In fact, the graphical analysis reveals that the pendulum will turn forever whenever $|u_0| > mgl$, but now you understand that this is not an oscillation, but an instability with $\theta \rightarrow \pm \infty$.
You can find another beautiful example of these concepts (fixed points, basins of attraction, bifurcations) from recurrent neural networks in the exercise below on Hopfield networks .
The undamped pendulum with zero torque
Consider again the system $$ml^2 \ddot\theta = u_0 - mgl \sin\theta - b\dot\theta,$$ this time with $b = 0$. This time the system dynamics are truly second-order. We can always think of any second-order system, $$\ddot{\theta} = f(\theta,\dot\theta),$$ as (coupled) first-order system with twice as many variables. This system is equivalent to the two-dimensional first-order system \begin{align*} \dot x_1 =& x_2 \\ \dot x_2 =& f(x_1,x_2), \end{align*} where $x_1 = \theta$ and $x_2 = \dot \theta$. Therefore, the graphical depiction of this system is not a line, but a vector field where the vectors $[\dot x_1, \dot x_2]^T$ are plotted over the domain $(x_1,x_2)$. This vector field is known as the phase portrait of the system.
In this section we restrict ourselves to the simplest case when $u_0 = 0$. Let's sketch the phase portrait. First sketch along the $\theta$-axis. The $x$-component of the vector field here is zero, the $y$-component is $-\frac{g}{l}\sin\theta.$ As expected, we have fixed points at $\pm \pi, ...$ Now sketch the rest of the vector field. Can you tell me which fixed points are stable? Some of them are stable i.s.L., none are asymptotically stable.
Orbit calculations
You might wonder how we drew the black contour lines in the figure above. We could have obtained them by simulating the system numerically, but those lines can be easily obtained in closed-form. Directly integrating the equations of motion is difficult, but at least for the case when $u_0 = 0$, we have some additional physical insight for this problem that we can take advantage of. The kinetic energy, $T$, and potential energy, $U$, of the pendulum are given by $$T = \frac{1}{2}I\dot\theta^2, \quad U = -mgl\cos(\theta),$$ where $I=ml^2$ and the total energy is $E(\theta,\dot\theta) = T(\dot\theta)+U(\theta)$. The undamped pendulum is a conservative system: total energy is a constant over system trajectories. Using conservation of energy, we have: \begin{gather*} E(\theta(t),\dot\theta(t)) = E(\theta(0),\dot\theta(0)) = E_0 \\ \frac{1}{2} I \dot\theta^2(t) - mgl\cos(\theta(t)) = E_0 \\ \dot\theta(t) = \pm \sqrt{\frac{2}{I}\left[E_0 + mgl\cos\left(\theta(t)\right)\right]} \end{gather*} Using this, if you tell me $\theta$ I can determine one of two possible values for $\dot\theta$, and the solution has all of the richness of the black contour lines from the plot. This equation has a real solution when $\cos(\theta) > \cos(\theta_{max})$, where $$\theta_{max} = \begin{cases} \cos^{-1}\left( -\frac{E_0}{mgl} \right), & E_0 < mgl \\ \pi, & \text{otherwise}. \end{cases}$$ Of course this is just the intuitive notion that the pendulum will not swing above the height where the total energy equals the potential energy. As an exercise, you can verify that differentiating this equation with respect to time indeed results in the equations of motion.
The particular orbit defined by $E = mgl$ is special -- this is the orbit that visits the (unstable) equilibrium at the upright. This is the homoclinic orbit of the pendulum.
Trajectory calculations
For completeness, I'll include what it would take to solve for $\theta(t)$, even thought it cannot be accomplished using elementary functions. Feel free to skip this subsection. We begin the integration with \begin{gather*} \frac{d\theta}{dt} = \sqrt{\frac{2}{I}\left[E + mgl\cos\left(\theta(t)\right)\right]} \\ \int_{\theta(0)}^{\theta(t)} \frac{d\theta}{\sqrt{\frac{2}{I}\left[E + mgl\cos\left(\theta(t)\right)\right]}} = \int_0^t dt' = t \end{gather*} The integral on the left side of this equation is an (incomplete) elliptic integral of the first kind. Using the identity: $$\cos(\theta) = 1 - 2 \sin^2(\frac{1}{2}\theta),$$ and manipulating, we have $$t = \sqrt{\frac{I}{2(E+mgl)}} \int_{\theta(0)}^{\theta(t)} \frac{d\theta}{\sqrt{1 - k_1^2\sin^2(\frac{\theta}{2})}}, \quad \text{with }k_1=\sqrt{\frac{2mgl}{E+mgl}}.$$ In terms of the incomplete elliptic integral function, $$F(\phi,k) = \int_0^\phi \frac{d\theta}{\sqrt{1-k^2\sin^2\theta}},$$ accomplished by a change of variables. If $E