Question

The mass and stiffness matrices and the mode shapes of a two-degree-of-freedom system in the absence of damping elements are given by =ቂ1 0

0 4ቃ, =൤ 12 −ଵଶ
−ଵଶ ଶଶ
൨, ଵ = ቂ 1
9.11ቃ, ଶ = ቂ−36.44
1. ቃ
If the first natural frequency is given by ଵ = 1.7, determine the stiffness
coefficients ଵଶ and ଶଶ and the natural frequency of vibration, ଶ.

Answer 1

The stiffness coefficients for the two-degree-of-freedom system are 2.89 and 11.56, and the natural frequency of vibration is 1.7 rad/s.

Step-by-step explanation:

The stiffness matrix of a two-degree-of-freedom system is given as:

[1 0]

[0 4]

The mode shapes matrix is:

[1]

[9.11]

The first natural frequency is given as 1.7. To determine the stiffness coefficients and the natural frequency of vibration, we can use the equation:

ω = √(k/m)

where ω is the natural frequency, k is the stiffness coefficient, and m is the mass.

Substituting the given value of the natural frequency, we have:

1.7 = √(k/1)

Squaring both sides of the equation:

2.89 = k

Therefore, the stiffness coefficients are 2.89 and 11.56 (multiplied by 4 in the matrix). The natural frequency of vibration (ω) is 1.7 rad/s.