Balanced beam with three masses and springs, driven by a torque

A light beam turns about its centre and carries a mass at each end. The left end hangs from a spring, and the right end carries a second spring, a third mass and a third spring to the floor. A sinusoidal torque drives the beam. The paper derives the equations of motion three times: with Lagrange's equations, Hamilton's principle and the direct method. All three derivations are correct. The interpretation of the results is not. The large oscillations in the plots come from gravity, not from the torque. The "instability" in the parameter study is just a larger sag. The legend of that study does not match the curves. The simulator separates the gravity, free and forced parts of the motion.

Problem statement

A rigid beam of negligible weight and length $2l$ turns about an axle at its centre in the vertical plane, as in the figure. Determine the equations of motion of the system.

Three balls of equal mass $m$: one at each end of the beam and one between the two right-hand springs.
Three springs of equal stiffness $k$. The left end of the beam hangs from the ceiling. The right end is connected to the third ball, which is connected to the floor.
An external torque $\tau = \tau_0\sin\omega t$ acts on the beam.
The springs are unstretched when the beam is horizontal and the third ball is at its initial height.

The paper chooses the spring elongations as generalized coordinates: $x_2$ at the right end of the beam and $x_1$ for the third ball. By similar triangles, the left end moves by the same amount as the right end, in the opposite direction: $\theta = x_2/l = x_3/l$. The system therefore has two degrees of freedom. Below, $x_1$ and $x_2$ are measured upward, consistent with the paper's gravity term $+mgx_1$.

Figure 1. Massless beam of length 2l on a central axle; three masses m and three springs k; torque τ on the beam. Coordinates x₂ (beam end) and x₁ (third mass), positive upward.

Lagrange's equations

TerminologyAs in the course: attractive energy $\mathcal{A}$ = potential energy, repulsive coenergy $\mathcal{R}^*$ = kinetic coenergy, $\mathcal{L}=\mathcal{R}^*-\mathcal{A}$.

Energies

The two end balls both move with speed $\dot x_2$, and the third ball moves with $\dot x_1$:

$$\mathcal{R}^* = \tfrac12m\dot x_1^2 + \tfrac12m\dot x_2^2 + \tfrac12m\dot x_2^2 = \tfrac12m\dot x_1^2 + m\dot x_2^2$$

The springs stretch by $x_2$ (top left), $x_2-x_1$ (middle) and $x_1$ (bottom). The gravity terms of the two end balls cancel, because one rises while the other falls (the beam is balanced):

$$\mathcal{A} = \tfrac12kx_2^2 + \tfrac12k(x_2-x_1)^2 + \tfrac12kx_1^2 + mgx_2 - mgx_2 + mgx_1 = kx_1^2 + kx_2^2 - kx_1x_2 + mgx_1$$

Generalized force

The torque does the virtual work $\delta W = \tau\,\delta\theta$, and $\theta = x_2/l$, so

$$\mathcal{F}_1 = 0,\qquad \mathcal{F}_2 = \tau\frac{\partial\theta}{\partial x_2} = \frac{\tau}{l}$$

Equations of motion

(1)–(2)$$\begin{aligned} m\ddot x_1 + 2kx_1 - kx_2 + mg &= 0\\ 2m\ddot x_2 + 2kx_2 - kx_1 &= \frac{\tau_0}{l}\sin\omega t\end{aligned}$$

These are correct. In matrix form, $\mathbf M\ddot{\mathbf x} + \mathbf K\mathbf x = \mathbf f(t)$:

$$\mathbf M = \begin{bmatrix} m & 0\\ 0 & 2m\end{bmatrix},\qquad \mathbf K = \begin{bmatrix} 2k & -k\\ -k & 2k\end{bmatrix},\qquad \mathbf f = \begin{bmatrix} -mg\\ \tau_0\sin(\omega t)/l\end{bmatrix}$$

Both matrices are symmetric, as they must be for a conservative system.

The model is linearThe relation $\theta = x_2/l$ and the assumption that the springs stay vertical hold only for small rotations. Later the paper writes $\sin\theta = x_2/l$. For small angles the two agree, but the model as a whole is valid only for $|x_2| \ll l$.

Hamilton's principle and the direct method

Hamilton's principle

The paper writes the variational indicator

$$\mathrm{V.I.} = \int_{t_1}^{t_2}\Big[\delta(\mathcal{R}^*-\mathcal{A}) + \sum_j\mathcal{F}_j\,\delta x_j\Big]dt$$

It integrates the velocity terms by parts (the variations vanish at $t_1$ and $t_2$) and collects the coefficients of $\delta x_1$ and $\delta x_2$. Requiring them to vanish for arbitrary variations gives again (1)–(2). This check is correct and well presented.

Direct method

The paper writes Newton's law for each ball separately:

$$\text{third ball: } m\ddot x_1 + kx_1 - k(x_2-x_1) + mg = 0,\qquad \text{right ball: } m\ddot x_2 + k(x_2-x_1) + mg = \frac\tau l,\qquad \text{left ball: } m\ddot x_2 + kx_2 - mg = 0$$

Adding the last two gives (2). The result is correct, but the two end-ball equations are not valid on their own. The balls are fixed to the beam, which pushes on them with forces that are not shown, and the torque is not a force on the right ball. The proper direct method is the moment balance of the beam with its two balls about the axle:

$$2ml^2\,\ddot\theta = \tau - l\,k x_2 - l\,k(x_2-x_1) + l\,mg - l\,mg$$

With $\theta = x_2/l$, dividing by $l$ gives exactly (2). The unknown beam forces never appear because they are internal to the rigid beam. Adding the two ball equations removes them for the same reason.

Simulink model

Solving (1)–(2) for the accelerations:

(11)$$\ddot x_1 = -\frac{2k}{m}x_1 + \frac km x_2 - g,\qquad \ddot x_2 = \frac{k}{2m}x_1 - \frac km x_2 + \frac{\tau_0}{2ml}\sin\omega t$$

The model has an input level (a constant 9.81 and a sine source for the torque), a subsystem with two integrator chains and gains $2k/m$, $k/m$, $k/m$, $k/(2m)$, and two scopes. The gains match (11).

Parameters: $m = 0.5$ kg, $k = 6$ N/m, $l = 1$ m, $\tau_0 = 1$ N·m, $T = 1$ s ($\omega = 2\pi$ rad/s). Numerically:

$$\ddot x_1 = -24x_1 + 12x_2 - 9.81,\qquad \ddot x_2 = 6x_1 - 12x_2 + \sin2\pi t$$

Sign slip in the matrix formIn the matrices (8), (9), (12) and (13) the element $A_{12}$ is printed as $-k/m$ (or $-1$), but (11) and the expanded (12) correctly have $+k/m$. The Simulink model follows (11), so the results are not affected. Had the printed matrix been simulated, the stiffness matrix would be non-symmetric and the system unstable. The simulator includes this variant ("printed matrix").

Results from the paper

Plot	Content
Fig. 6, 7, 9	$x_1(t)$ and $x_2(t)$, 10 s and 20 s, base parameters, released from $x_1=x_2=0$ at rest
Fig. 8	Beam angle $\theta = \arcsin(x_2/l)$, 20 s
Fig. 10	Parameter study: $e = k/m$ = 0.5, 2, 15, 25 (according to the legend), 30 s

The paper's interpretation is as follows:

The elongations start negative "because both springs are compressed", in line with the direction of the torque at $t=0$.
They then vary periodically, because the excitation is periodic. The system is stable and the oscillations are bounded.
If the torque stopped, the oscillations would be damped out.
The ratio $e = k/m$ has the greatest influence ($e\propto f^2$). For small $e$ the elongations become very large (up to 30 m), so "the system tends to instability".

Our simulation of (11) reproduces the base case exactly ($x_1$ between −1.11 and 0.04 m, $x_2$ between −0.79 and 0.22 m).

Critical review

The start is not an equilibrium, and gravity drives the response

The paper calls $x_1=x_2=0$ the equilibrium. Gravity, however, pulls the third ball down, so the static equilibrium of (1)–(2) is

$$x_1^* = -\frac{2mg}{3k} = -0.545\ \text{m},\qquad x_2^* = -\frac{mg}{3k} = -0.273\ \text{m}$$

Released from $x=0$, the system swings about this equilibrium with an amplitude of the same size. The negative values and the large slow oscillation in the plots are this free vibration under gravity, not the effect of the torque. Without any torque, $x_1$ still reaches −1.09 m. The steady forced response to $\tau_0 = 1$ N·m at $\omega = 2\pi$ is only a few centimetres (section 7). Preset "No torque" and preset "From equilibrium" separate the two effects. The gravity swing also makes the rotation large: $x_2$ reaches −0.79 m with $l = 1$ m, i.e. the beam turns by about 52°, far beyond the small-angle range the model assumes.

No damping, so nothing dies out

The model contains no dissipative element. If the torque stopped, the system would keep oscillating at its natural frequencies forever, with constant energy. The free vibration started by the initial condition never decays either. That is why the response is not a clean periodic function of period 1 s.

The parameter study: wrong legend and no instability

The curves in Fig. 10 do not match the values in the legend. Simulating (11) with $\tau_0/(2ml)=1$ reproduces them for $e \approx 0.3$ (red), $e \approx 0.6$ (blue) and $e = 2$ (purple). For example, the red $x_2$ reaches −29.8 m at 7.3 s. With the legend's $e = 0.5$, the minimum would be −17.8 m. A plausible explanation is that the plots were made with $k = 6$ and $m = 20, 10, 3$.

Large elongations do not mean instability. For every $e>0$ the eigenvalues of $\mathbf M^{-1}\mathbf K$ are positive, so the system is (marginally) stable. The excursions grow because the static sag $2g/(3e)$ and the free vibration around it grow as $1/e$. Moreover, $|x_2|$ of 15–30 m on a beam with $l = 1$ m is physically impossible ($|x_2|\le l$). The linear model has long left its range of validity.

Smaller points

The Lagrange equation for $x_2$ is written with $\partial\mathcal{L}/\partial\dot x_1$ (a typo). Equation (13) uses $\tau$ where it should be $\tau_0\sin\omega t$. The relation $e = cf^2$ is stated without saying which frequency $f$ is meant. The natural frequencies, the quantities that actually decide the behaviour, are never computed. Laplace analysis is mentioned but not carried out, even though it would directly give the forced response.

What is goodThe choice of coordinates is justified carefully, the generalized force $\tau/l$ is derived correctly, and the equations are confirmed by three independent methods. The Simulink model matches the equations.

Extension: normal modes and forced response

Natural frequencies

From $\det(\mathbf K - \omega^2\mathbf M) = 0$:

$$2m^2\omega^4 - 6km\,\omega^2 + 3k^2 = 0\qquad\Rightarrow\qquad \omega_{1,2}^2 = \frac km\cdot\frac{3\mp\sqrt3}{2},\qquad \frac{X_2}{X_1} = \frac{1\pm\sqrt3}{2}$$

The frequencies are $\omega_1 = 0.796\sqrt{k/m}$ and $\omega_2 = 1.538\sqrt{k/m}$. This confirms the paper's remark that only $e = k/m$ matters, since $\omega^2\propto e$.

In the slow mode the third ball and the beam end move together ($X_2/X_1 = 1.37$).
In the fast mode they move against each other ($X_2/X_1 = -0.37$).

For the base parameters ($e = 12$): $\omega_1 = 2.76$ rad/s ($T_1 = 2.28$ s) and $\omega_2 = 5.33$ rad/s ($T_2 = 1.18$ s). The 1 s period of the torque lies above both.

Steady-state forced response

For $\mathbf f = [0,\ F]^T\sin\omega t$ with $F = \tau_0/l$, the steady amplitudes follow from $(\mathbf K - \omega^2\mathbf M)\mathbf X = \mathbf f$:

$$X_1 = \frac{kF}{\Delta(\omega)},\qquad X_2 = \frac{(2k - m\omega^2)F}{\Delta(\omega)},\qquad \Delta(\omega) = (2k-m\omega^2)(2k-2m\omega^2) - k^2$$

For the base case $X_1 = +0.034$ m (in phase with the torque) and $X_2 = -0.044$ m (opposite phase). That is more than ten times smaller than the gravity-induced swing. At $\omega = \sqrt{2k/m}$ the numerator of $X_2$ vanishes: the beam does not move at all in steady state, while the third ball oscillates with $X_1 = -F/k$. This is an antiresonance. The third ball and its springs act as a vibration absorber for the beam. The frequency-response chart in the simulator shows the two resonances and this antiresonance.

Interactive simulator

The simulator solves (1)–(2) numerically (4th-order Runge–Kutta, 1 ms step). The "From equilibrium" option starts at the static equilibrium instead of $x=0$, which removes the free vibration caused by gravity.

Cases

Model

Start from static equilibrium (adds to x(0)) Start on the steady-state forced motion Compare with the same run without torque

System parameters

m [kg]

k [N/m]

l [m]

τ₀ [N·m]

ω [rad/s]

g [m/s²]

x₁(0) [m]

x₂(0) [m]

Duration [s]

x is measured upward from the unstretched position. ω accepts expressions such as 2*pi.

Third ball x₁(t)

Beam end x₂(t)

Beam angle θ(t) = arcsin(x₂/l)

Shown only while |x₂| ≤ l. Beyond that the model has no physical meaning.

Frequency response (steady-state amplitude per unit torque)

|X₁| and |X₂| for F = τ₀/l, clipped at 5× the static value. The vertical lines are the natural frequencies; the marker is the current ω.

Animation

Results table (every 0.05 s)

Review questions

Why do the gravity terms of the two end balls cancel?

The beam is balanced about its centre. When the right end rises by $x_2$, the left end falls by $x_2$. The changes of potential energy, $+mgx_2$ and $-mgx_2$, cancel. Only the third ball's weight remains.

Why is $\mathcal{F}_2 = \tau/l$?

The torque's virtual work is $\tau\,\delta\theta$. With $\theta = x_2/l$ this is $(\tau/l)\,\delta x_2$, and the coefficient of $\delta x_2$ is the generalized force.

Where does the system oscillate around, and why does that matter here?

Around the static equilibrium $x_1^* = -2mg/(3k)$, $x_2^* = -mg/(3k)$. Starting from $x = 0$ excites a free vibration of the same size, which is larger than the forced response in the paper's case.

Does a small $e = k/m$ make the system unstable?

No. Both natural frequencies $\omega^2 = e(3\mp\sqrt3)/2$ stay real and positive. The motion only becomes slower and the sag larger. Instability would need a negative or complex eigenvalue, e.g. from a non-symmetric stiffness matrix such as the misprinted one.

What is an antiresonance, and where is it in this system?

A frequency at which a driven coordinate does not move in steady state. Here $X_2 = 0$ at $\omega = \sqrt{2k/m}$. The third ball and its two springs then form a tuned absorber that cancels the motion of the beam.

When is the linear model ($\theta = x_2/l$, vertical springs) acceptable?

For small beam rotations, say $|x_2| < 0.1\,l$ to $0.2\,l$. For larger rotations the ends move on circles, the springs tilt, and the geometry becomes nonlinear.

Plot	Content
Fig. 6, 7, 9	\(x_1(t)\) and \(x_2(t)\), 10 s and 20 s, base parameters, released from \(x_1=x_2=0\) at rest
Fig. 8	Beam angle \(\theta = \arcsin(x_2/l)\), 20 s
Fig. 10	Parameter study: \(e = k/m\) = 0.5, 2, 15, 25 (according to the legend), 30 s