 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/7/b/i/solution

For two bars, after subtracting the gravitational equilibrium,

K = (2 - 1 - 1 2), M = (10 0 m_{2}) .

η = v e^{iω t}

(K - ω^{2} M) v = 0

. With

z = ω^{2}

0 = (2 - z) (2 - m_{2} z) - 1 = m_{2} z^{2} - 2 (m_{2} + 1) z + 3.

Thus, for

m_{2} > 0

ω_{1, 2} = \frac{m _{2} + 1 \mp m _{2}^{2} - m _{2} + 1}{m _{2}}, ω_{1} < ω_{2} .

Both squared angular frequencies are positive. A corresponding eigenvector is

(1, 2 - ω_{j}^{2})^{T}

, from the first row of the matrix equation.

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