Linear Vibrations Lab

Worked examples › Example 3

Example 3Two carts linked by a pivoting bar

EOMsmall anglesmassless links

Problem

Two masses, each $m$, are connected by links to a massless bar of length $L$ that pivots about a pin at its lower end. The lower mass attaches at $L/2$, the upper at $L$. Each mass is connected to ground by a spring $k$ and moves on a smooth surface. $x$ is the upper mass's position, measured from where the bar is vertical (springs unstretched). Assume small rotations.

Find: (a) the equation of motion in terms of $x$; (b) the natural frequency.

InteractiveThe system in motion

Displacements in the drawing are scaled up so the motion is visible; the plot shows the true values.

x(t)

Solution

Every solution follows the same five steps: coordinates, free-body diagrams, kinetics, kinematics, solve. Try the problem first, then open one step at a time to check your work.

Step 1 · Choose a coordinate system

Let $x$ be the upper cart's position and $x_1$ the lower cart's position, both positive to the right and zero when the bar is vertical (springs unstretched). Let $\theta$ be the bar's small rotation.

Step 2 · Draw the free-body diagrams
Upper cartmkxPAmg and N are verticalMassless barPAPBOxOyLL/2Lower cartmkx₁PBmg and N are vertical
Free-body diagrams of both carts and the massless bar. PA and PB are the link forces, drawn in +x on the carts and equal and opposite on the bar. Both springs push back in −x.
Step 3 · Kinetics (Newton–Euler equations)

Upper cart: $\;m\ddot x=-kx+P_A$

Lower cart: $\;m\ddot x_1=-kx_1+P_B$

Massless bar, moments about the pin O (zero because the bar has no inertia): $\;P_A L+P_B\tfrac L2=0\;\Rightarrow\;P_B=-2P_A$

Step 4 · Kinematics (constraints)

For a small rotation $\theta$ of the bar, the top moves $L\theta$ and the midpoint moves $\tfrac L2\theta$:

$$x=L\theta,\qquad x_1=\tfrac L2\theta=\tfrac12x$$
Step 5 · Solve

From the upper cart, $P_A=m\ddot x+kx$. Then $P_B=-2(m\ddot x+kx)$, and the lower cart with $x_1=x/2$ gives

$$\tfrac12m\ddot x+\tfrac12kx=-2(m\ddot x+kx)\;\;\Longrightarrow\;\;\tfrac52\left(m\ddot x+kx\right)=0$$ $$\tfrac54m\ddot x+\tfrac54kx=0,\qquad \omega_n=\sqrt{\frac km}$$

L cancels: both the inertia and the stiffness pick up the same lever factor. Energy check: $T=\tfrac12(\tfrac54m)\dot x^2$, $V=\tfrac12(\tfrac54k)x^2$. ✓

Answer

(a) $\displaystyle \tfrac54m\ddot x+\tfrac54kx=0\;\;\Longleftrightarrow\;\;\ddot x+\frac km x=0$

(b) $\displaystyle \omega_n=\sqrt{\frac km}$, independent of $L$.

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