Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 7 / ii / Solution

The displacement directions specified at release give $x(0)=(-x_0,x_0{)}^T$. Let $\Delta r=r_{-}-r_{+}=2D/m_2>0$. By initial-data decomposition into two normal modes, write $x(0)=c_{-}{v}_{-}+c_{+}{v}_{+}$. Solving the two coordinate equations givesZero initial velocities remove the sine terms, soIn components,Substitution at zero gives $x_1=-x_0$, $x_2=x_0$, and both derivatives vanish. For unequal masses both normal modes are generally present. When $m_1=m_2=m$, $r_{-}=1$, $r_{+}=-1$, so only the higher mode survives: $x_1=-x_0\cos (\sqrt{3k/m}t)$ and $x_2=x_0\cos (\sqrt{3k/m}t)$.