Linear Vibrations Lab

Worked examples › Example 8

Example 8Pulley system driven by a moving support

pulleysbase excitationresonance

Problem

A support B moves horizontally as $x_B=b\cos\omega t$. It is connected by a spring $k$ to a cable that passes over a fixed pulley, down around a light pulley attached to mass $m$, and back up to the fixed pulley's support. A second spring $k$ connects the mass to the ground. $x$ is the mass's displacement, positive downward, measured from static equilibrium.

Find: (a) the differential equation of motion; (b) the resonant 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)Steady state

Steady-state amplitude vs forcing frequency; the dot is your current ω.

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 mass's displacement, positive downward, measured from static equilibrium. Let $x_C$ be the displacement of the cable end where it meets the top spring (positive toward the pulley), and $x_B=b\cos\omega t$ the prescribed support motion.

Step 2 · Draw the free-body diagrams
Mass + light pulleymTTkx+xmg is balanced by the static forces (x from equilibrium)Junction of cable and top springTk(xC − xB)massless point: T = k(xC − xB)
Left: the mass and its pulley (dynamic forces only). Right: the massless point where the cable meets the top spring.
Step 3 · Kinetics (Newton–Euler equations)

Mass + pulley (down +, dynamic forces only): $\;m\ddot x=-2T-kx$

Massless junction: the cable tension equals the top spring force, $\;T=k(x_C-x_B)$

Step 4 · Kinematics (constraints)

Cable length is constant. When the mass moves down $x$, both strands holding it lengthen by $x$, so the cable end is pulled toward the pulley by twice as much:

$$x_C=2x$$
Step 5 · Solve

$T=k(2x-x_B)$. Substitute into the mass equation:

$$m\ddot x=-2k(2x-x_B)-kx\qquad\Longrightarrow\qquad m\ddot x+5kx=2kx_B=2kb\cos\omega t$$ $$\text{Resonant frequency: }\ \omega_n=\sqrt{\frac{5k}m}$$

Energy check: $V=\tfrac12k(2x-x_B)^2+\tfrac12kx^2$ gives effective stiffness $4k+k=5k$. ✓ The pulley doubles both the motion and the force, so the top spring counts four times.

Answer

(a) $\displaystyle m\ddot x+5kx=2kb\cos\omega t$

(b) $\displaystyle \omega_n=\sqrt{\frac{5k}m}$

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