Two degrees of freedom: modes and the frame

Two masses, two frequencies

• The pump sits on mounts, and the mounts sit on a steel frame on the floor. Two masses and two springs give two natural frequencies, and either can be in the wrong place.
• Newton's law for each mass gives $\mathbf M\ddot{\mathbf x} + \mathbf C\dot{\mathbf x} + \mathbf K\mathbf x = \mathbf F$.

The eigenvalue problem

• Try $\mathbf x = \mathbf u\sin\omega t$ without damping: $(\mathbf K - \omega^2\mathbf M)\mathbf u = 0$. This eigenvalue problem gives a natural frequency for each root and a mode shape for each vector.
• Mode shapes are orthogonal through $\mathbf M$ and $\mathbf K$, so mode superposition splits the coupled equations into one single-degree equation per mode.

The pump on its frame

The response at 24 Hz

• Solving $(\mathbf K - \omega^2\mathbf M)\mathbf X = (27.3, 0)$ at 150.8 rad/s: the pump moves $\mathbf{11.4}$ µm, the frame $\mathbf{1.5}$ µm, and the floor feels $k_2X_2 = \mathbf{3.0}$ N.
• 24 Hz is above both modes, so the result is close to the single-mass estimate. A stiffer, lighter frame with mode 2 near 24 Hz would raise the floor force sharply.

Check the modes are orthogonal

• Mode shapes must satisfy $\mathbf u_1^T\mathbf M\mathbf u_2 = 0$. Here $120 \times 1 \times 1 + 200 \times 0.160 \times (-3.76) = 120 - 120.3 \approx 0$.
• A non-zero result means an arithmetic slip in the modes. This one-line check catches it before any response is computed from them.

Worked example: a tuned absorber

• Hang a small mass on a spring from the pump, tuned to the 24 Hz drive: $k_a/m_a = 150.8^2$. With $m_a = 6$ kg, $k_a = \mathbf{136.4}$ kN/m.
• At exactly 24 Hz the absorber's spring pushes back on the pump with the whole 27.3 N, so the pump stands still and the absorber swings $27.3/136\,400 = \mathbf{0.20}$ mm. It works only at its tuned speed.

Three matrix slips

Recap