A team of technicians is analysing the dynamic behavior of a two-degree-of-freedom springmass system used for structural vibration control. The system consists of two identical
masses, 𝑚1 = 𝑚2 = 4 𝑘𝑔, connected in series by three identical springs, each with a
stiffness of k=120 N/m.
Calculate the mode shapes corresponding to each natural frequency.
We had the equations of motion in matrix form (after substituting harmonic motion):
We solved the characteristic equation and found the natural frequencies:
Step-by-Step: Finding the Mode Shapes
Case 1:
Substitute into the matrix equation:
Now write the system of equations:
Let’s normalize by setting 𝑋1 = 1, then
This means both masses move together in phase with equal amplitudes. The system
vibrates like a rigid body, both masses move in the same direction at the same time.
Case 2:
Substitute into the matrix equation:
Now the equation becomes:
Let’s normalize by setting 𝑋1 = 1, then
Summary
This means the masses move in opposite directions with equal magnitude. It’s an out-ofphase mode: when one mass goes up, the other goes down.
Mode Number
Natural Frequency (rad/s)
1
𝜔1 = √30 = 5.48
2
𝜔2 = √90 = 9.49
Mode Shape ϕ
Physical Interpretation
1
[ ]
1
1
[ ]
−1
Masses move in phase
Masses move out of phase