Problems

Two particles \(X\) and \(Y\), of equal mass \(m\), lie on a smooth horizontal table and are connected by a light elastic spring of natural length \(a\) and modulus of elasticity \(\lambda\). Two more springs, identical to the first, connect \(X\) to a point \(P\) on the table and \(Y\) to a point \(Q\) on the table. The distance between \(P\) and \(Q\) is \(3a\). Initially, the particles are held so that \(XP=a\), \(YQ= \frac12 a\,\), and \(PXYQ\) is a straight line. The particles are then released. At time \(t\), the particle \(X\) is a distance \(a+x\) from \(P\) and the particle \(Y\) is a distance \(a+y\) from \(Q\). Show that \[ m \frac{\.d ^2 x}{\.d t^2} = -\frac\lambda a (2x+y) \] and find a similar expression involving \(\dfrac{\.d^2 y}{\.d t^2}\). Deduce that \[ x-y = A\cos \omega t +B \sin\omega t \] where \(A\) and \(B\) are constants to be determined and \(ma\omega^2=\lambda\). Find a similar expression for \(x+y\). Show that \(Y\) will never return to its initial position.