Rigid beam with two masses on springs and a damper

Two masses sit on a light rigid beam supported by two springs and a damper. A harmonic force $F_0\cos\omega t$ acts on one of the masses. The paper derives the equations of motion correctly and runs a thorough parameter study in Simulink, and our simulation reproduces all of its plots. The weak point is the interpretation. Most of the transient motion is caused by gravity, not by the force. The damper value used is about ten times too large to damp well. The amplitude reductions are attributed to the wrong causes. The interactive simulator adds the frequency response, the damping as a function of $B$, and the separation of the gravity and force effects.

Problem statement

Two masses $m_1$ and $m_2$ are attached to a rigid beam of negligible mass. The beam is supported by springs of stiffness $k_1$ and $k_2$ and a damper $B$, as shown in the figure. Assuming the beam moves only in the plane of the drawing, determine the equations of motion.

The beam has length $3L$. Spring $k_2$ supports its left end.
Mass $m_2$, with the damper $B$ below it, sits at distance $L$ from the left end.
Mass $m_1$ and spring $k_1$ are at the right end, at $3L$.
The force $F(t) = F_0\cos\omega t$ acts vertically on $m_1$.

The beam can translate and rotate, so it has two degrees of freedom. The paper chooses the vertical displacements $x$ of $m_1$ and $y$ of $m_2$, both positive upward (consistent with its gravity terms $+mgx$, $+mgy$).

Figure 1. Massless rigid beam of length 3L on springs k₂ (left end) and k₁ (right end) and damper B (under m₂); force F on m₁; x and y positive upward

Mathematical model

Kinematics

For small rotations, the beam's angle and the displacement $z$ of its left end follow from $x$ and $y$:

(1)–(3)$$\tan\theta = \frac{x-y}{2L},\qquad z = y - L\tan\theta = \tfrac32y - \tfrac12x$$

For vertical displacements of a rigid beam this relation is exact. The approximation is that the springs and the damper stay vertical.

Energies

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

(4)–(5)$$\mathcal{A} = \tfrac12k_2\big(\tfrac32y-\tfrac12x\big)^2 + \tfrac12k_1x^2 + m_2gy + m_1gx,\qquad \mathcal{R}^* = \tfrac12m_1\dot x^2 + \tfrac12m_2\dot y^2$$

Generalized forces

The force $F$ acts on $m_1$ along $x$. The damper acts on $m_2$ along $y$:

$$\mathcal{F}_x = F_0\cos\omega t,\qquad \mathcal{F}_y = -B\dot y$$

Equations of motion

(8), (10)$$\begin{aligned} m_1\ddot x + \big(k_1 + \tfrac14k_2\big)x - \tfrac34k_2\,y + m_1g &= F_0\cos\omega t\\ m_2\ddot y + B\dot y + \tfrac94k_2\,y - \tfrac34k_2\,x + m_2g &= 0\end{aligned}$$

These are correct. In matrix form, $\mathbf M\ddot{\mathbf q} + \mathbf C\dot{\mathbf q} + \mathbf K\mathbf q = \mathbf f$:

$$\mathbf M = \begin{bmatrix} m_1 & 0\\ 0 & m_2\end{bmatrix},\qquad \mathbf C = \begin{bmatrix} 0 & 0\\ 0 & B\end{bmatrix},\qquad \mathbf K = \begin{bmatrix} k_1 + \tfrac14k_2 & -\tfrac34k_2\\ -\tfrac34k_2 & \tfrac94k_2\end{bmatrix}$$

The masses are coupled only through the stiffness: the spring $k_2$ at the left end turns the beam, which moves both masses. The damper acts directly on one coordinate only.

Static equilibrium

The springs are unstretched at $x=y=0$, so gravity moves the system to $\mathbf q^* = -\mathbf K^{-1}g\,[m_1,\ m_2]^T$. For the base parameters ($k_1=k_2=100$ N/m, $m_1=m_2=0.5$ kg) this gives $x^* = -0.065$ m and $y^* = -0.044$ m. Every simulation in the paper starts from $x=y=0$, so each one also contains this drop under gravity.

Simulink model

The model solves (8) and (10) for the accelerations, with two chains of two integrators. The gains are $(4k_1+k_2)/(4m_1)$, $3k_2/(4m_1)$, $9k_2/(4m_2)$, $3k_2/(4m_2)$ and $B/m_2$, and constant blocks supply $g$. The input is a Clock block followed by a gain $\omega$, a cos block and a gain $F_0/m_1$. The outputs go to the workspace and to two scopes. The model matches the equations. Writing the clock-based input explicitly is a good idea, because the input stays readable and is easy to change.

Results from the paper

Case	Change from the base set	Paper's interpretation
Base	k₁=k₂=100, B=100, m₁=m₂=0.5, F₀=10, ω=10	Periodic after the transient, since energy is supplied continuously
Stiffness	k₁=k₂=10 (50 s)	Softer springs allow larger elongation, so larger amplitudes
Masses	m₁=m₂=5 (20 s)	Smaller amplitude: the force "cannot supply enough energy" because much of it is dissipated in the damper
Damping	B=1000	y changes very little, because most of its energy is used up in the damper; x is almost unchanged
"Impulse"	m₂=1, force 1 N for 0.5 s, B=100	Quasi-oscillatory response; energy delivered once and then dissipated
"Impulse"	m₂=1, force 1 N for 0.5 s, B=1	y decays more slowly, like a resistor in an electric circuit
Force	F₀=1	Less energy supplied, so smaller amplitude
Frequency	ω=1 (30 s)	After the transient, the coordinates follow the force at the same frequency

Simulating the equations with zero initial conditions reproduces every plot. For example, in the base case $x$ ranges from −0.26 to 0.16 m, and with $k=10$ $x$ reaches −1.41 m. The results are numerically sound.

Critical review

Gravity drives the transients

Each run starts with unstretched springs, i.e. $|y^*| = 4.4$ cm away from the static equilibrium for the base case and $10\times$ further for $k=10$ or $m=5$. The large initial swings and the offsets of the curves are this drop under gravity. They show up as "$x$ and $y$ mostly negative".

In the "impulse" runs the force (1 N for 0.5 s) changes $x$ by at most 1.6 cm, while gravity moves it by 10 cm. What those plots show is the system settling under gravity, not its response to the impulse. Starting from the static equilibrium (option in the simulator) isolates the effect of the force.

The damper is far too strong to damp well

With $B = 100$ N·s/m the damper almost locks $m_2$. The beam then pivots about $m_2$, and $m_1$ oscillates on the springs almost undamped, at $\sqrt{(k_1+k_2/4)/m_1} = 15.8$ rad/s. The slowest mode decays with a rate of only 0.23 s⁻¹, about 4 s per factor $e$. That is why the transient lasts so long in the base case.

The best damping, a decay rate of about 1.85 s⁻¹, is reached at $B \approx 10$ N·s/m (section 6). With $B = 1000$, $y$ hardly moves because the damper acts almost like a rigid support, and it dissipates very little energy (power $B\dot y^2$ with $\dot y\approx0$). The paper's statement that most of the energy is used up in the damper is the opposite of what happens.

Larger masses: inertia, not dissipation

With $m_1 = m_2 = 5$ kg the natural frequencies drop to 4.1 and 7.3 rad/s. The forcing at 10 rad/s lies above both, where the response is controlled by inertia: the amplitude is roughly $F_0/(m_1\omega^2)$. The steady amplitude of $x$ falls from 13.5 cm to 2.7 cm because the masses are harder to accelerate, not because the damper absorbs more energy.

Softer springs: static sag and resonance

With $k = 10$ the curves move down mainly because the static sag grows tenfold ($x^* = -0.65$ m). The natural frequencies also drop from 13.0/23.0 to 4.1/7.3 rad/s. The steady amplitude of $x$ doubles (13.5 → 26.7 cm), because the forcing frequency is now closer to a natural frequency.

Smaller points

The "impulse" plots have a horizontal axis from 0 to 300 and from 0 to 120, which cannot be seconds: the oscillation periods (0.34 and 0.55 s) would then be over a factor ten too long. They are most likely plotted against the sample index. The input there is a rectangular pulse, not an impulse. The natural frequencies, which explain all the observed changes, are never computed.

What is goodThe kinematics are explained clearly with an exaggerated sketch. The generalized forces of the force and of the damper are introduced correctly. The derivation and the Simulink model are correct, and the parameter study is broad and systematic.

Extension: optimal damping and frequency response

Damping as a function of $B$

The free motion is governed by the eigenvalues of the state matrix $\begin{bmatrix}\mathbf 0 & \mathbf I\\ -\mathbf M^{-1}\mathbf K & -\mathbf M^{-1}\mathbf C\end{bmatrix}$. The slowest-decaying mode decides how long a transient lasts. For the base masses and springs:

B [N·s/m]	0.1	1	3	10	30	100	1000
Slowest decay rate [s⁻¹]	0.02	0.22	0.67	1.85	0.76	0.23	0.02

A weak damper barely acts.
A very strong damper locks $m_2$, and the remaining motion of $m_1$ is undamped.
The optimum lies in between, at $B\approx10$. The damping and decay chart in the simulator shows this curve and the current $B$.

The same principle governs the tuning of vibration dampers in practice.

Steady-state forced response

Once the transient has died out, $\mathbf q(t) = \mathrm{Re}\{\mathbf Q e^{i\omega t}\}$ with

$$\big(\mathbf K - \omega^2\mathbf M + i\omega\mathbf C\big)\,\mathbf Q = \begin{bmatrix}F_0\\ 0\end{bmatrix}$$

For the base case $|X| = 13.5$ cm and $|Y| = 1.0$ cm. For $\omega = 1$ rad/s, far below resonance, the response is almost static ($|X| = 9.6$ cm, in phase with the force), as the paper observes. Fast decay of transients and a small forced amplitude are not the same goal. At $\omega = 10$ rad/s, $B = 10$ gives $|X| = 19$ cm against 13.5 cm for $B = 100$. The strong damper locks $m_2$ and moves the resonance of $m_1$ up to 15.8 rad/s, away from the forcing frequency. The frequency-response chart makes this visible.

Interactive simulator

The simulator integrates the equations with 4th-order Runge–Kutta (1 ms step). In the animation the vertical displacements are scaled up so they stay visible; the scale factor is shown.

Cases

Force

Start from static equilibrium Compare with the same run without force

System parameters

k₁ [N/m]

k₂ [N/m]

B [N·s/m]

m₁ [kg]

m₂ [kg]

g [m/s²]

F₀ [N]

ω [rad/s]

t_p [s] (pulse)

Duration [s]

Mass m₁: x(t)

Mass m₂: y(t)

Frequency response

Steady amplitudes |X|, |Y| for F₀; the marker is the current ω.

Damping of the slowest mode vs B

Decay rate −max Re(λ); the marker is the current B (log scale of B).

Animation

Results table (every 0.05 s)

Review questions

Why does the left end move by $z = \tfrac32y - \tfrac12x$?

For a rigid beam, the vertical displacement varies linearly along the beam. Extrapolating from the points at $L$ (displacement $y$) and $3L$ (displacement $x$) back to 0 gives $z = y - \tfrac12(x-y)$.

Why does a very large $B$ give poor damping?

A damper dissipates power $B\dot y^2$. A huge $B$ forces $\dot y\approx0$, so the dissipated power tends to zero. The damper then acts like a support, and the remaining mode, $m_1$ on the springs with the beam pivoting about $m_2$, is undamped.

Why does the amplitude drop when the masses grow, at fixed $\omega = 10$ rad/s?

The natural frequencies fall below the forcing frequency, into the inertia-controlled range, where the amplitude is about $F_0/(m\omega^2)$.

How do you separate the effect of the force from the effect of gravity?

Start from the static equilibrium $\mathbf q^* = -\mathbf K^{-1}g[m_1,m_2]^T$, or subtract a run without the force. Because the system is linear, the two responses simply add.

What would the response to a true impulse look like?

A sudden velocity of $m_1$, $\Delta\dot x = I/m_1$, followed by free damped oscillation of both modes. A 0.5 s pulse is not short compared with the periods of 0.34 and 0.55 s, so it is not an impulse in the strict sense.