Answer to Question #193730 in Mechanical Engineering for Lemony

Question

Find the natural frequencies of the system shown on the right, with m1=m, m2=2m, k1=k, and k2=2k. Determine the response of the system when k=1000 N/m, m=20 kg, and the initial values of the displacements of the masses m1 and m2 are l and -1 respectively.

Answer 1

The equation of motion for m₁

$m_1\ddot{x} _1+(k_1+k_2)x_1-k_2x_2=0$

Substituting $- \omega^2x_1$ for $\ddot{x} _1 \implies -\omega^2m_1x_1+(k_1+k_2)x_1-k_2x_2=0$

The equation of motion for m₂

$m_2\ddot{x} _2+(k_1+k_2)x_2-k_2x_2=0$

Equation 1 and 2 in form of a matrix

$\begin{bmatrix} -\omega^2m_1+k_1+k_2 & -k_2 \\ -k_2 & k_2-m_2\omega^2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$\omega^4-(\frac{k_1+k_2}{m_1}+\frac{k_2}{m_2})\omega^2+\frac{k_1k_2}{m_1m_2}=0$

$\omega^2_1,\omega^2_2 = \frac{k_1+k_2}{2m_1}+\frac{k_2}{2m_2} \mp \sqrt{\frac{1}{4}(\frac{k_1+k_2}{m_1}+\frac{k_2}{2m_2})^2-\frac{k_1k_2}{m_1m_2}}$

Substituting in the values

$\omega^2_1,\omega^2_2 = \frac{1000+2000}{2*20}+\frac{2000}{2*40} \mp \sqrt{\frac{1}{4}(\frac{1000+2000}{20}+\frac{2000}{2*20})^2-\frac{1000*2000}{20*40}}$

$\omega_1=3.6606rad/s,\omega_2 =13.6603 rad/s$

$r_1=\frac{X_2}{X_1}=\frac{-m_1\omega_2^2+k_1+k_2}{k_2}=\frac{-20*13.6603^2+1000+2000}{2000}=-0.366$

$r_2=\frac{X_2}{X_1}=\frac{-m_1\omega_2^2+k_1+k_2}{k_2}=\frac{-20*13.6603^2+1000+2000}{2000}=-0.366$

$x_1=-0.366 cos 3.6603t-1.366cos13.6603t$

$x_2=-0.5 cos 3.6603t+0.5cos13.6603t$