Linear Vibrations Lab

Worked examples › Example 1

Example 1Block on an incline driving a pulley

EOMpulleysnatural frequency

Problem

Block A (mass $m$) rests on a smooth incline of angle $\theta$ and is attached to the wall by a spring of stiffness $k$. A cable from A runs over a fixed pulley, down around a light pulley that carries block B (mass $m$), and back up to the fixed support. A force $f(t)$ acts downward on B. The coordinate $x$ is A's displacement up the incline.

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 A's displacement up the incline (toward the pulley), measured from the position where the spring is unstretched. Let $x_B$ be B's displacement, positive downward.

Step 2 · Draw the free-body diagrams
Block A (on the incline)ATkxNmgθBlock B + light pulleyBTTmgf(t)
Free-body diagrams of A and of B with its pulley. The cable tension T is the same everywhere (massless, frictionless pulleys).
Step 3 · Kinetics (Newton–Euler equations)

Block A, along the incline: $\;m\ddot x = T - kx - mg\sin\theta$

Block B, vertical (down +): $\;m\ddot x_B = mg + f(t) - 2T$

Pulleys are massless and frictionless, so the tension is the same in every strand. The normal force N on A has no component along the incline.

Step 4 · Kinematics (constraints)

The cable length is constant. If A moves $x$ toward the pulley, that length is shared by the two strands holding B, so B drops half as far:

$$x_B=\tfrac12 x\qquad\Longrightarrow\qquad \ddot x_B=\tfrac12\ddot x$$
Step 5 · Solve

From B: $T=\tfrac12\left(mg+f-\tfrac12 m\ddot x\right)$. Substitute into A:

$$\tfrac54 m\ddot x+kx=\tfrac12 f(t)+mg\left(\tfrac12-\sin\theta\right)$$

The constant term is balanced by a static spring stretch $\delta_{st}=mg(\tfrac12-\sin\theta)/k$. Measuring $x$ from that equilibrium removes it:

$$\tfrac54 m\ddot x+kx=\tfrac12 f(t),\qquad \omega_n=\sqrt{\frac{k}{5m/4}}=2\sqrt{\frac{k}{5m}}$$

Energy check: $T=\tfrac12 m\dot x^2+\tfrac12 m(\dot x/2)^2=\tfrac12(\tfrac54m)\dot x^2$ and $V=\tfrac12kx^2$. ✓

Answer

(a) $\displaystyle \frac54 m\ddot x+kx=\frac12 f(t)$   (with $x$ from static equilibrium)

(b) $\displaystyle \omega_n=\sqrt{\frac{k}{5m/4}}=2\sqrt{\frac{k}{5m}}$

The incline angle $\theta$ only changes the static stretch, not $\omega_n$.

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