Problem Statement
This task investigates the dynamics of coupled mechanical oscillators, eigenvalue formulation, and normal modal energy transfer.
Two identical carts of mass $m$ oscillate horizontally on a frictionless track. Cart 1 has displacement $x_1(t)$ from equilibrium and is connected to a fixed wall by an elastic spring of stiffness $k$. Cart 2 has displacement $x_2(t)$ and is connected to an opposite fixed wall by another spring of stiffness $k$. The two carts are coupled together by a central spring of stiffness $k_c = 2k$.
**(a)** [4 marks] By applying Newton's second law $\sum F = m\ddot{x}$ to each cart, show that the system of coupled differential equations can be expressed in matrix form as:
$$\begin{pmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{pmatrix} = -\frac{k}{m} \begin{pmatrix} 3 & -2 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$
**(b)** [6 marks] A normal mode is defined by harmonic oscillations of the form $\mathbf{x}(t) = \mathbf{v} \cos(\omega t + \phi)$, where $\omega > 0$ is the angular frequency and $\mathbf{v} \neq \mathbf{0}$ is a constant mode shape vector.
Show that the angular frequencies must satisfy $\det\left( \mathbf{A} - \frac{m\omega^2}{k} \mathbf{I} \right) = 0$, where $\mathbf{A} = \begin{pmatrix} 3 & -2 \\ -2 & 3 \end{pmatrix}$, and hence find the two normal frequencies $\omega_1$ and $\omega_2$ (with $\omega_1 < \omega_2$) in terms of $\omega_0 = \sqrt{\frac{k}{m}}$.
**(c)** [5 marks] Determine an associated eigenvector $\mathbf{v}_1$ and $\mathbf{v}_2$ for each normal mode. Provide a clear physical interpretation of the motion of the carts in Mode 1 versus Mode 2.
**(d)** [6 marks] At $t = 0$, Cart 1 is displaced by a positive distance $d$ while Cart 2 is held at equilibrium ($x_1(0) = d, x_2(0) = 0$). Both carts are released from rest ($\dot{x}_1(0) = \dot{x}_2(0) = 0$).
Find the general solution for $x_1(t)$ and $x_2(t)$ for all $t \ge 0$.
**(e)** [6 marks] Using the sum-to-product trigonometric identities, show that:
$$x_1(t) = d \cos\left(\frac{\omega_2 - \omega_1}{2} t\right) \cos\left(\frac{\omega_2 + \omega_1}{2} t\right)$$
Hence find the beat period $T_{\text{beat}}$ and explain the mechanism by which energy is periodically exchanged between the two carts.
Verified Solution & Marking Scheme
Part (a): Equations of Motion [M1, A1, AG]
For Cart 1:
$$m\ddot{x}_1 = -k x_1 + k_c(x_2 - x_1) = -k x_1 + 2k(x_2 - x_1) = -3k x_1 + 2k x_2$$
For Cart 2:
$$m\ddot{x}_2 = -k x_2 - k_c(x_2 - x_1) = -k x_2 - 2k(x_2 - x_1) = 2k x_1 - 3k x_2$$
Dividing by $m$ and converting to matrix notation:
$$\begin{pmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{pmatrix} = -\frac{k}{m} \begin{pmatrix} 3 & -2 \\ -2 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$ [AG]
Part (b): Characteristic Equation & Normal Frequencies [M1, A1, A1]
Substituting $\mathbf{x}(t) = \mathbf{v} \cos(\omega t)$ gives $\ddot{\mathbf{x}}(t) = -\omega^2 \mathbf{v} \cos(\omega t)$.
$$-m\omega^2 \mathbf{v} = -k \mathbf{A}\mathbf{v} \implies \left(\mathbf{A} - \frac{m\omega^2}{k} \mathbf{I}\right)\mathbf{v} = \mathbf{0}$$
Let $\lambda = \frac{m\omega^2}{k}$. The characteristic equation is:
$$\det(\mathbf{A} - \lambda \mathbf{I}) = (3 - \lambda)^2 - (-2)^2 = \lambda^2 - 6\lambda + 9 - 4 = \lambda^2 - 6\lambda + 5 = 0$$
$$(\lambda - 1)(\lambda - 5) = 0 \implies \lambda_1 = 1, \quad \lambda_2 = 5$$
Since $\omega = \sqrt{\frac{k\lambda}{m}}$:
$$\omega_1 = \sqrt{\frac{k}{m}} = \omega_0, \quad \omega_2 = \sqrt{\frac{5k}{m}} = \sqrt{5}\omega_0$$
Part (c): Eigenvectors & Physical Mode Shapes [M1, A1, R1]
For $\lambda_1 = 1$: $(\mathbf{A} - \mathbf{I})\mathbf{v}_1 = \begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix} \begin{pmatrix} v_x \\ v_y \end{pmatrix} = \mathbf{0} \implies v_x = v_y$. Thus $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
*Physical Interpretation*: Carts oscillate in phase with identical amplitudes ($x_1 = x_2$). The central coupling spring is neither stretched nor compressed; oscillation occurs at the single spring natural frequency $\omega_0$.
For $\lambda_2 = 5$: $(\mathbf{A} - 5\mathbf{I})\mathbf{v}_2 = \begin{pmatrix} -2 & -2 \\ -2 & -2 \end{pmatrix} \begin{pmatrix} v_x \\ v_y \end{pmatrix} = \mathbf{0} \implies v_x = -v_y$. Thus $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
*Physical Interpretation*: Carts oscillate in anti-phase ($x_1 = -x_2$). The central spring is vigorously compressed and stretched, adding restorative force and elevating frequency to $\sqrt{5}\omega_0$.
Part (d): Initial Value Problem Solution [M1, M1, A1]
General solution is a linear superposition of normal modes:
$$\begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix} = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} \cos(\omega_1 t) + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} \cos(\omega_2 t)$$
Applying initial conditions at $t = 0$:
$$x_1(0) = c_1 + c_2 = d, \quad x_2(0) = c_1 - c_2 = 0 \implies c_1 = c_2 = \frac{d}{2}$$
Hence:
$$x_1(t) = \frac{d}{2} \left( \cos(\omega_1 t) + \cos(\omega_2 t) \right)$$
$$x_2(t) = \frac{d}{2} \left( \cos(\omega_1 t) - \cos(\omega_2 t) \right)$$
Part (e): Beat Phenomenon & Energy Transfer [M1, A1, R1]
Using $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$:
$$x_1(t) = d \cos\left(\frac{\omega_2 - \omega_1}{2} t\right) \cos\left(\frac{\omega_2 + \omega_1}{2} t\right)$$
Similarly for Cart 2, using $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$:
$$x_2(t) = d \sin\left(\frac{\omega_2 - \omega_1}{2} t\right) \sin\left(\frac{\omega_2 + \omega_1}{2} t\right)$$
The envelope modulating the rapid vibration is $\cos\left(\frac{\omega_2 - \omega_1}{2} t\right)$.
The amplitude of Cart 1 reaches zero when $\frac{\omega_2 - \omega_1}{2} t = \frac{\pi}{2}$, at which point Cart 2 has amplitude $d$ (full mechanical energy transferred to Cart 2).
The beat frequency is $\omega_{\text{beat}} = \omega_2 - \omega_1 = (\sqrt{5} - 1)\omega_0$, yielding beat period:
$$T_{\text{beat}} = \frac{2\pi}{\omega_2 - \omega_1} = \frac{2\pi}{(\sqrt{5} - 1)}\sqrt{\frac{m}{k}}$$