 Solution

At equilibrium, the two horizontal equations imply ${\hat{X}}_2={\hat{X}}_1$ and then $k_1{\hat{X}}_1=0$, so both vanish. Adding the two vertical equations gives $k_1{\hat{Y}}_1=-(m_1+m_2)g$; the second separately gives $k_2({\hat{Y}}_2-{\hat{Y}}_1)=-m_{2}g$. HenceRetain the imposed constraint $X_2=0$ and set $X_1=x_1$, $Y_1={\hat{Y}}_1+y_1$, $Y_2={\hat{Y}}_2+y_2$. Subtracting the equilibrium equations cancels all constant gravitational terms. For $q=(x_1,y_1,y_2{)}^T$, the small-oscillation mass and stiffness matrices areThe horizontal reaction enforcing the constraint replaces the removed $X_2$ equation; it is not necessary to demand that the original unconstrained horizontal force on mass 2 vanish during motion. Because the zero-natural-length spring potentials are exactly quadratic, these displacement equations are actually exact within the ideal spring model, not merely first-order approximations.