Natural frequency with series-parallel springs

Data: $m = 12$ kg, $k_1 = 15000$ N/m, $k_2 = 25000$ N/m (parallel), $k_3 = 20000$ N/m (in series with the pair).

Equivalent stiffness

$$k_p = k_1 + k_2 = 15000 + 25000 = 40000\ \text{N/m}$$ $$k_{eq} = \frac{k_p k_3}{k_p + k_3} = \frac{40000 \times 20000}{40000 + 20000} = 13333.3\ \text{N/m}$$

Static deflection

$$\delta_{st} = \frac{mg}{k_{eq}} = \frac{12 \times 9.81}{13333.3} = 0.00883\ \text{m} = 8.83\ \text{mm}$$

Natural frequency and period

$$\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{13333.3}{12}} = 33.33\ \text{rad/s}$$ $$f_n = \frac{\omega_n}{2\pi} = 5.305\ \text{Hz}, \qquad T = \frac{1}{f_n} = 0.1885\ \text{s}$$

Response to the initial conditions

$x(t) = x_0\cos\omega_n t + \dfrac{v_0}{\omega_n}\sin\omega_n t = X\sin(\omega_n t + \phi)$

$$X = \sqrt{x_0^2 + \left(\frac{v_0}{\omega_n}\right)^2} = \sqrt{0.02^2 + \left(\frac{0.5}{33.33}\right)^2} = 0.0250\ \text{m} = 25.0\ \text{mm}$$ $$\phi = \tan^{-1}\left(\frac{\omega_n x_0}{v_0}\right) = \tan^{-1}\left(\frac{33.33 \times 0.02}{0.5}\right) = 53.1^\circ$$ $$v_{max} = \omega_n X = 33.33 \times 0.0250 = 0.833\ \text{m/s}, \qquad a_{max} = \omega_n^2 X = 27.78\ \text{m/s}^2$$

Answer: $k_{eq} = 13333.3$ N/m; $\delta_{st} = 8.83$ mm; $f_n = 5.305$ Hz; $T = 0.1885$ s; $X = 25.0$ mm; $v_{max} = 0.833$ m/s; $a_{max} = 27.78$ m/s².