3 problems found
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.
Given that \[ \frac{\mathrm{d}x}{\mathrm{d}t}=4(x-y)\qquad\mbox{ and }\qquad\frac{\mathrm{d}y}{\mathrm{d}t}=x-12(\mathrm{e}^{2t}+\mathrm{e}^{-2t}), \] obtain a differential equation for \(x\) which does not contain \(y\). Hence, or otherwise, find \(x\) and \(y\) in terms of \(t\) given that \(x=y=0\) when \(t=0\).
Solution: \begin{align*} && \frac{\d x}{\d t} &= 4(x-y) \\ && \frac{\d y}{\d t} &= x - 12(e^{2t}+e^{-2t}) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} &= 4 \frac{\d x}{\d t}-4\frac{\d y}{\d t} \\ &&&= 4 \frac{\d x}{\d t}-4 \left ( x - 12(e^{2t}+e^{-2t}) \right) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} - 4 \frac{\d x}{\d t}+4x &= 48 (e^{2t}+e^{-2t}) \end{align*} This differential equation has characteristic polynomial \(\lambda^2 - 4\lambda + 4 = (\lambda-2)^2\). Therefore we should expect a general solution of \((At+B)e^{2t}\). For particular integrals we should try \(ke^{-2t}\) and \(Ct^2 e^{2t}\). For the former, we have: \begin{align*} && 48 &= 4k+8k+k \\ \Rightarrow && k &= \frac{48}{13} \end{align*} For the latter we have: \begin{align*} &&4Ct^2e^{2t} -4C(2te^{2t}+2t^2e^{2t})+2C((1+2t)e^{2t}+2t^2e^{2t}) &= 48e^{2t} \\ \Rightarrow && 2C &= 48 \\ \Rightarrow && C &= 24 \end{align*} Therefore the solution should be: \begin{align*} x = (At+B)e^{2t} + \frac{48}{13}e^{-2t} + 24t^2 e^{2t} \\ x(0) = B + \frac{48}{13} \\ x'(0) = 2B+A-\frac{96}{13} \\ x =\frac{48}{13}((4t-1)e^{2t}+e^{-2t})+24t^2e^{2t} \\ y = x - \frac{1}{4} \frac{\d x}{\d t} \end{align*}
The functions \(x(t)\) and \(y(t)\) satisfy the simultaneous differential equations \begin{alignat*}{1} \dfrac{\mathrm{d}x}{\mathrm{d}t}+2x-5y & =0\\ \frac{\mathrm{d}y}{\mathrm{d}t}+ax-2y & =2\cos t, \end{alignat*} subject to \(x=0,\) \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=0\) at \(t=0.\) Solve these equations for \(x\) and \(y\) in the case when \(a=1\). Without solving the equations explicitly, state briefly how the form of the solutions for \(x\) and \(y\) if \(a>1\) would differ from the form when \(a=1.\)
Solution: Letting \(\mathbf{x} =\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}\) and \(\mathbf{A} = \begin{pmatrix} -2 & 5 \\ -a & 2 \end{pmatrix}\) then our differential equation is \(\mathbf{x}' = \mathbf{Ax} + \begin{pmatrix} 0 \\2 \cos t \end{pmatrix}\). Looking at the eigenvalues of \(\mathbf{A}\), we find: \begin{align*} && \det \begin{pmatrix} -2-\lambda & 5 \\ -a & 2 -\lambda \end{pmatrix} &= (\lambda^2-4)+5a\\ &&&= \lambda^2 +5a-4 \end{align*} Therefore if \(a = 1\), \(\lambda = \pm i\). In which case we should expect the complementary solutions to be of the form \(\mathbf{x} = \begin{pmatrix} A \sin t + B \cos t \\ C \sin t + D \cos t \end{pmatrix}\). The first equation tells us that \((A-5D+B)\cos t + (-B+5C)\sin t=0\) so the complementary solution is:\(\mathbf{x} = \begin{pmatrix} 5(D-C) \sin t + 5C \cos t \\ C \sin t + D \cos t \end{pmatrix}\). Looking for a particular integral, we should expect to try something like \(\mathbf{x} = \begin{pmatrix} Et\cos t+Ft\sin t\\ Gt\cos t+Ht \sin t\end{pmatrix}\) and we find