Mechanical Vibrations

a. Free-body diagram and D’Alembert balance

Isolate the horizontal beam and its machine. The pin at $O$ transmits two forces and no couple. Measure the beam angle $\theta$ from static equilibrium; a positive angle lifts the shaft $P$. The known rotor angle $\omega t$ sets the direction of the excitation; it is not a second unknown coordinate.

Applied forces

Include weight

D’Alembert inertia loads

Both drawings use the same stations: damper $D$ at $L/4$, beam center and spring $G$ at $L/2$, and machine shaft $P$ at $L$. Arrows show reference directions; a negative value reverses an arrow. Their lengths here are schematic.

Balance moments about the pin

Add the moments in both drawings, taking counterclockwise as positive. The hinge reactions pass through $O$. The rotor contributes only its vertical component to the first-order moment about the horizontal equilibrium beam.

$$\begin{aligned} 0={}&L\,[mr_1\omega^2\sin(\omega t)]\\ &-\frac{L}{4}\!\left[c\frac{L}{4}\dot\theta\right] -\frac{L}{2}\!\left[k\frac{L}{2}\theta\right]\\ &-\frac{L}{2}\!\left[m_1\frac{L}{2}\ddot\theta\right] -J_G\ddot\theta\\ &-L\,[ML\ddot\theta]. \end{aligned}$$

The beam contributes both translation of its center and rotation about its center. Together these give $J_G+m_1(L/2)^2=m_1L^2/3$. The machine adds $ML^2$, using the full $M=500\,\mathrm{kg}$. The housing alone has mass $M-m$; the eccentric mass is already part of $M$.

Forces, moment arms, and the static spring load

Force on the system	Distance from O	How it acts
Hinge reactions $H_x,H_y$	$0$	The hinge holds the left end in place and lets it rotate freely.
Beam weight $m_1g$	$L/2$	Downward, at the center of the uniform beam.
Machine weight $Mg$	$L$	Downward, at the machine attachment.
Spring force $F_{k,s}-k(L/2)\theta$	$L/2$	$F_{k,s}$ is the upward spring force at equilibrium. An upward displacement reduces it.
Damper force $-c(L/4)\dot\theta$	$L/4$	Opposes the velocity of the damper attachment.
Rotor force $F_u\sin\omega t$, with $F_u=mr_1\omega^2$	$L$	Vertical component of the force from the eccentric mass.

The horizontal rotor force, $F_u\cos\omega t$, changes $H_x$. For small vibrations, use the moment arms at equilibrium. This drops the term $-L\theta F_u\cos\omega t$, a product of the small angular displacement and the disturbing force. The model also ignores the size of the machine housing. Under these assumptions the horizontal force contributes no moment to the linear equation.

Before the rotor starts, moments about the hinge balance:

$$F_{k,s}\frac L2=MgL+m_1g\frac L2, \qquad F_{k,s}=(2M+m_1)g.$$

Subtract this balance to get the dynamic diagram. It contains the changes in the spring force and hinge reactions, the damper force, and the rotor force. The weights cancel, which explains why $g$ does not appear in the equation of motion.

b. Find the equation of motion

For a small rotation, vertical displacement equals the distance from the hinge times the angle:

$$y_k=\frac L2\theta,\qquad y_c=\frac L4\theta,\qquad y_M=L\theta.$$

Use the parallel-axis theorem to shift the beam’s moment of inertia from its center to the hinge:

$$J_{\text{beam},O} =\frac{m_1L^2}{12}+m_1\left(\frac L2\right)^2 =\frac{m_1L^2}{3}.$$

Treat the machine as a point mass at the beam tip. Its total mass $M$ already includes the eccentric mass $m$, so adding $m$ again would count it twice.

$$J_\theta=ML^2+\frac{m_1L^2}{3} =\left(M+\frac{m_1}{3}\right)L^2.$$

Sum the dynamic moments about the hinge:

$$mr_1\omega^2L\sin\omega t -k\left(\frac L2\theta\right)\frac L2 -c\left(\frac L4\dot\theta\right)\frac L4 =J_\theta\ddot\theta.$$ $$\boxed{\begin{aligned} \left(M+\frac{m_1}{3}\right)L^2\ddot\theta &+\frac{cL^2}{16}\dot\theta +\frac{kL^2}{4}\theta\\ &=mr_1\omega^2L\sin\omega t. \end{aligned}}$$

Each distance enters twice: once to find displacement or velocity, then again to turn force into moment. For example, the damping coefficient contains $(L/4)^2$.

Write $C_\theta=cL^2/16$ and $K_\theta=kL^2/4$. This gives the single-degree-of-freedom form on the exam sheet, with angular displacement $\theta$ and rotational inertia $J_\theta$.

c. Find the steady response

The steady response repeats at the rotor frequency $\omega$. Its amplitude is $\Theta$; its phase lag $\phi$ tells us how far the beam motion trails the vertical rotor force.

$$\boxed{\theta_{\mathrm{ss}}(t)=\Theta\sin(\omega t-\phi)}$$ $$\boxed{\Theta= \frac{mr_1\omega^2L} {\sqrt{(K_\theta-J_\theta\omega^2)^2+(C_\theta\omega)^2}}}$$ $$\phi=\operatorname{atan2}\!\left(C_\theta\omega, K_\theta-J_\theta\omega^2\right).$$

For positive damping, $\phi$ lies between $0$ and $\pi$. Using both arguments in $\operatorname{atan2}$ selects the correct quadrant. The ordinary $\arctan$ in the exam formula needs a quadrant correction when $\omega>\omega_n$.

$$\omega_n=\sqrt{\frac{K_\theta}{J_\theta}} =\sqrt{\frac{k}{4(M+m_1/3)}},\qquad \zeta=\frac{C_\theta}{2\sqrt{K_\theta J_\theta}} =\frac{c}{16\sqrt{k(M+m_1/3)}}.$$

With the frequency ratio $r=\omega/\omega_n$, the amplitude becomes

$$\boxed{\Theta=\frac{mr_1}{(M+m_1/3)L} \frac{r^2}{\sqrt{(1-r^2)^2+(2\zeta r)^2}}}.$$

The numerator contains $r^2$ because the force from the eccentric mass grows with $\omega^2$. At rest, the rotor exerts no dynamic force. With nonzero imbalance, zero damping and $\omega=\omega_n$, there is no bounded steady response. With zero damping away from resonance, the formula gives the forced component; any free vibration set by the initial conditions persists.

d. Find the vertical force amplitude at the floor

Add the forces that the spring and damper exert on the floor. The moment arms have already set the motion of the attachment points. Let $q=\omega t-\phi$. Taking upward as positive gives

$$\begin{aligned} F_{\mathrm{tr}}(t) &=k\frac L2\theta+c\frac L4\dot\theta\\ &=\frac{kL\Theta}{2}\sin q +\frac{cL\omega\Theta}{4}\cos q. \end{aligned}$$

The spring force follows displacement. The damper force follows velocity, so it reaches its peak a quarter cycle earlier. Add their squared amplitudes and take the square root:

$$\boxed{F_{\mathrm{tr},0} =L\Theta\sqrt{\left(\frac k2\right)^2 +\left(\frac{c\omega}{4}\right)^2}}.$$

In the original problem variables, this gives

$$\boxed{F_{\mathrm{tr},0} =\frac{mr_1\omega^2 \sqrt{(k/2)^2+(c\omega/4)^2}} {\sqrt{[k/4-(M+m_1/3)\omega^2]^2+(c\omega/16)^2}}}.$$

The question asks for the path through the spring and damper only, so the answer excludes the hinge reaction. The static spring load on the floor is $(2M+m_1)g$ downward. It shifts the mean force but does not change the vibration amplitude $F_{\mathrm{tr},0}$.

Model assumptions: a rigid beam and housing, a point mass at the tip, an ideal hinge, a linear spring and damper, constant rotor speed, and small vibrations about static equilibrium. The exam data do not determine numerical values for $\Theta$ or $F_{\mathrm{tr},0}$.

Coordinates and spring extension

Measure $x_1$ uphill and $x_2$ downward from static equilibrium. Take clockwise rotation $\varphi$ as positive. The two positions are independent coordinates: specifying the cylinder position alone does not locate the hanging mass, because the upper spring can stretch. Once we release the system, the spring forces couple their motions.

No slip at the incline gives $x_1=R\varphi$. At the top of the cylinder, the translational and rotational velocities point in the same direction. With uphill positive, the velocity along the cord is

Thus the cord moves the cylinder-side end of $k_2$ uphill by $2x_1$. As the mass moves down by $x_2$, the pulley-side end of $k_2$ moves uphill by $x_2$. Subtract the motion of the first end from that of the second. The changes in length from equilibrium are

For example, the mixed release at +100% starts with $x_1=28\,\mathrm{mm}$ and $x_2=21\,\mathrm{mm}$. The first end moves 56 mm uphill and the second moves 21 mm uphill, so the spring shortens by 35 mm. The cord already carried tension at equilibrium, so a shorter spring can still pull.

Use small vibrations, linear springs, a taut cord that does not stretch, and a pulley with no mass or friction. The exam assumes rolling without slip. That requires $\lvert f\rvert\leq\mu_s N$; without a coefficient of static friction, we cannot check that condition for a particular pair of surfaces.

b. Equations of motion

Return to the cylinder’s two zero balances in part a. Divide the moment equation by $R$, use $\ddot\varphi=\ddot{x}_1/R$, and add the result to the uphill force equation. The two $\Delta f$ terms cancel. The upper spring appears twice: once as a force and once through its torque.

For this solid cylinder, $J_G=\tfrac12m_1R^2$, so $m_1+J_G/R^2=\tfrac32m_1$. This effective mass already counts the center’s translation and the cylinder’s rotation. Do not add a separate rolling inertia term.

Mass and stiffness matrices

Use $\mathbf x=[x_1,\,x_2]^{\mathsf T}$, in that order. The two rows become

Gravity sets the static equilibrium but adds no linear stiffness for motion along this straight incline. Static friction enforces rolling. The point of contact has zero velocity, so friction does no work in this model and gives no damping term.

Check the result using energy

Add the energy of the moving center, the turning cylinder, and the moving mass. The vibration potential energy counts both springs after the equilibrium terms cancel:

$$\begin{aligned} \mathcal T&=\tfrac12m_1\dot{x}_1^2+\tfrac12J_G\left(\frac{\dot{x}_1}{R}\right)^2+\tfrac12m_2\dot{x}_2^2,\\ V&=\tfrac12k_1x_1^2+\tfrac12k_2(x_2-2x_1)^2. \end{aligned}$$

Lagrange’s equation $\tfrac{d}{dt}(\partial\mathcal T/\partial\dot x_i)+\partial V/\partial x_i=0$ gives the same matrices. In particular, $\partial V/\partial x_1=k_1x_1-2k_2(x_2-2x_1)$ gives the terms $4k_2$ and $-2k_2$. The stiffness matrix is symmetric and $\det\mathbf K=k_1k_2>0$ when both springs have positive stiffness.

If $x_2=2x_1$ at one instant, the ends of the upper spring have moved by the same amount. Then $e_2=0$ and $\Delta T_2=0$ at that instant. It does not follow that the bodies keep this ratio after release.

How the motion patterns follow from the matrices

Look for motion of the form $\mathbf x=\mathbf u\cos(\omega t)$. Substitution gives $(\mathbf K-\omega^2\mathbf M)\mathbf u=\mathbf0$. For the demo values the natural angular frequencies are 3.808 and 12.865 rad/s. The low mode has $x_2/x_1\approx+2.637$; the high mode has $x_2/x_1\approx-1.137$. These ratios refer to the signed axes in the drawing. Each single mode preserves its ratio throughout the motion, including sign reversals.

The mixed release adds both modes. Its reference pair gives each mode a 14 mm contribution to the initial center displacement. With no damping or external force, the sum $E=\mathcal T+V$ stays constant. Since $\tfrac12 k_2e_2^2\le E$, the total cord tension satisfies $T_2\ge m_2g-\sqrt{2k_2E}$. This bound is positive for every release offered here.

a. Natural frequencies and mode shapes

The given equation is $\mathbf J\ddot{\boldsymbol\theta}+\mathbf K\boldsymbol\theta=\mathbf0$. Here $\boldsymbol\theta$ holds the two disc angles. For a vibrating mode, try $\boldsymbol\theta(t)=\boldsymbol\phi\cos(\omega_n t)$. Differentiating twice gives $-\omega_n^2\boldsymbol\phi\cos(\omega_n t)$, so:

For a nonzero mode shape $\boldsymbol\phi$, the matrix must have zero determinant. Let $\lambda=\omega_n^2$ and insert the given inertias and stiffness:

The roots are $\lambda_1=0$ and $\lambda_2=6$. Take the square root to get angular frequency, then use $f_n=\omega_n/(2\pi)$ to get cycles per second:

Mode 1: the discs turn together

With $\lambda_1=0$, the first row gives $4\phi_1-4\phi_2=0$. The angles match, so the shaft does not twist.

When the whole shaft turns together, it exerts no restoring torque. Both discs can stay at a fixed angle or turn at a constant speed: $\theta_1=\theta_2=a+bt$. This explains the zero natural frequency: there is no back-and-forth motion in this mode.

Mode 2: the discs turn in opposite directions

With $\lambda_2=6$, the first row gives $-2\phi_1-4\phi_2=0$, so $\phi_1=-2\phi_2$.

The minus sign means the discs turn in opposite directions. Disc 1 turns through twice the angle of disc 2. Multiplying both entries by the same nonzero factor gives the same mode shape.

b. Node location in each mode

A node is a cross section whose angle stays zero throughout the motion. Measure $x$ from disc 1. The model puts all the inertia in the discs. The uniform shaft therefore carries the same torque along its length, and its angle changes linearly:

In mode 1, $\phi(x)=1$ everywhere. No cross section has zero amplitude.

In mode 2, set $\phi_1=1$ and $\phi_2=-1/2$. At the node, $\phi(x_s)=0$:

Each cross section passes through zero angle when the discs pass their neutral positions. Only the node stays at zero for the entire cycle. The centers of the discs also stay in place, but the node question asks where a whole cross section does not rotate.

a. Damped natural frequency

From one positive peak to the next, one damped period $T_d$ elapses. The bracket spans two periods:

The damped angular frequency is $\omega_d=2\pi f_d\approx187.56\,\mathrm{rad/s}$. Using $67\,\mathrm{ms}$ as one period would give $14.93\,\mathrm{Hz}$, half the correct frequency.

b. Critical motor speed

For very small damping, $\omega_d=\omega_n\sqrt{1-\zeta^2}$ gives $f_n\approx f_d$. Motor imbalance applies one force cycle per revolution. Near the critical speed, the motor turns about $29.85$ times per second. Multiply by 60 to get revolutions per minute:

This estimate assumes very small damping and one force cycle per revolution. With damping, the peak response can occur slightly away from the natural frequency. The exact shift depends on the forcing and damping; the problem gives no measured damping ratio for that correction.

c. Phase lag above the natural frequency

Compare displacement with the applied force. Write the force as $F(t)=F_0\sin(\omega t)$ and the steady response as $x(t)=X\sin(\omega t-\phi)$. A positive $\phi$ means the displacement reaches its peak after the force does.

Above the natural frequency, $r>1$, so $1-r^2<0$. With positive damping, $2\zeta r>0$. These signs place the phase lag in the second quadrant:

For the very small damping in the problem, and sufficiently far above resonance, this gives the usual estimate:

As the force approaches its peak in one direction, displacement approaches its peak in the other. With zero damping, the lag is exactly $180^\circ$ for any $r>1$. At $r=1$ with positive damping, displacement lags by $90^\circ$. These angles describe displacement relative to force.

The formula sheet uses $\tan^{-1}[2\zeta r/(1-r^2)]$. The function atan2 uses the signs of both arguments to choose the quadrant. If an ordinary inverse tangent gives a negative angle for $r>1$, add $180^\circ$.