Let's break down the problem step by step: ### **Problem Restatement** Three beads with masses \(m\), \(m\), and \(\mu\) are free to slide on a frictionless ring of radius \(a\). They are connected by springs of stiffness \(k\) and equilibrium length \(l = \sqrt{3}a\). Find the normal frequencies and normal modes of vibration. --- ## **1. Geometry and Coordinates** Since the beads are constrained to a ring, each bead’s position can be described by an angle \(\theta_i\) (for \(i = 1,2,3\)), measured from some reference direction. Let the masses be at angular positions: - \(\theta_1\) (mass \(m\)) - \(\theta_2\) (mass \(m\)) - \(\theta_3\) (mass \(\mu\)) The equilibrium positions are at \(120^\circ\) from each other because the springs are at their equilibrium length \(l = \sqrt{3}a\) (which is the chord length subtending \(120^\circ\) at the center of the ring). --- ## **2. Small Oscillations: Set Up the Equations** Let the small angular displacements from equilibrium be: - \(\theta_1 = \theta_{1,0} + \eta_1\) - \(\theta_2 = \theta_{2,0} + \eta_2\) - \(\theta_3 = \theta_{3,0} + \eta_3\) where \(\eta_i\) are small. --- ### **Kinetic Energy** \[ T = \frac{1}{2} m a^2 (\dot{\eta}_1^2 + \dot{\eta}_2^2) + \frac{1}{2} \mu a^2 \dot{\eta}_3^2 \] --- ### **Potential Energy** For small displacements, the extension of each spring connecting beads \(i\) and \(j\) is: \[ \Delta l_{ij} \approx a |\eta_i - \eta_j| \] Thus, the potential energy due to the springs is: \[ V = \frac{1}{2} k a^2 \left[ (\eta_1 - \eta_2)^2 + (\eta_2 - \eta_3)^2 + (\eta_3 - \eta_1)^2 \right] \] --- ## **3. Equations of Motion** Write the Lagrangian \( L = T - V \): \[ L = \frac{1}{2} m a^2 (\dot{\eta}_1^2 + \dot{\eta}_2^2) + \frac{1}{2} \mu a^2 \dot{\eta}_3^2 - \frac{1}{2} k a^2 \left[ (\eta_1 - \eta_2)^2 + (\eta_2 - \eta_3)^2 + (\eta_3 - \eta_1)^2 \right] \] Calculate the equations of motion for each \(\eta_i\): #### For \(\eta_1\): \[ m a^2 \ddot{\eta}_1 = -k a^2 [2\eta_1 - \eta_2 - \eta_3] \] \[ m \ddot{\eta}_1 = -k [2\eta_1 - \eta_2 - \eta_3] \] #### For \(\eta_2\): \[ m \ddot{\eta}_2 = -k [2\eta_2 - \eta_3 - \eta_1] \] #### For \(\eta_3\): \[ \mu \ddot{\eta}_3 = -k [2\eta_3 - \eta_1 - \eta_2] \] --- ## **4. Matrix Form** Express in vector form: \[ \begin{pmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & \mu \end{pmatrix} \begin{pmatrix} \ddot{\eta}_1 \\ \ddot{\eta}_2 \\ \ddot{\eta}_3 \end{pmatrix} = -k \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} \eta_1 \\ \eta_2 \\ \eta_3 \end{pmatrix} \] --- ## **5. Normal Modes and Frequencies** Assume solutions of the form \(\eta_i = A_i e^{i\omega t}\): \[ \text{(Mass matrix)} \cdot (-\omega^2 \mathbf{A}) = -k \cdot (\text{Spring matrix}) \cdot \mathbf{A} \] \[ \left[ k (\text{Spring matrix}) - \omega^2 (\text{Mass matrix}) \right] \mathbf{A} = 0 \] Nontrivial solutions exist when: \[ \det \left| k \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix} - \omega^2 \begin{pmatrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & \mu \end{pmatrix} \right| = 0 \] --- ## **6. Eigenvalue Problem** Let \(\omega^2 = \lambda\), divide by \(k\): \[ \begin{pmatrix} 2-\frac{\lambda}{k}m & -1 & -1 \\ -1 & 2-\frac{\lambda}{k}m & -1 \\ -1 & -1 & 2-\frac{\lambda}{k}\mu \end{pmatrix} \] Set determinant to zero: --- ### **Special Case:** \(\mu = m\) (All masses equal) The eigenvalues are well known for a symmetric 3-mass ring: - Zero mode (all move together): \(\omega_0 = 0\) - Doubly-degenerate mode: \(\omega^2 = 3k/m\) ### **General Case:** \(\mu \neq m\) Let’s find the eigenvalues for general \(\mu\). Let - \(x = \omega^2 / k\) The characteristic equation is: \[ \left| \begin{matrix} 2 - m x & -1 & -1 \\ -1 & 2 - m x & -1 \\ -1 & -1 & 2 - \mu x \end{matrix} \right| = 0 \] Expand the determinant: \[ (2-mx)\begin{vmatrix} 2-mx & -1\\ -1 & 2-\mu x \end{vmatrix} - [-1]\begin{vmatrix} -1 & -1\\ -1 & 2-\mu x \end{vmatrix} + [-1]\begin{vmatrix} -1 & 2-mx\\ -1 & -1 \end{vmatrix} \] Calculate each minor: First minor: \[ \begin{vmatrix} 2-mx & -1\\ -1 & 2-\mu x \end{vmatrix} = (2-mx)(2-\mu x) - (-1)(-1) = (2-mx)(2-\mu x) - 1 \] Second minor: \[ \begin{vmatrix} -1 & -1\\ -1 & 2-\mu x \end{vmatrix} = (-1)(2-\mu x) - (-1)(-1) = -2 + \mu x - 1 = \mu x - 3 \] Third minor: \[ \begin{vmatrix} -1 & 2-mx\\ -1 & -1 \end{vmatrix} = (-1)(-1) - (-1)(2-mx) = 1 + 2 - mx = 3 - mx \] So the full determinant: \[ (2-mx)\left[(2-mx)(2-\mu x) - 1\right] + (\mu x - 3) + (3 - mx) = 0 \] Let’s expand further: \[ (2-mx)[(2-mx)(2-\mu x) - 1] + (\mu x - 3) + (3 - mx) = 0 \] \[ (2-mx)[(2-mx)(2-\mu x) - 1] + (\mu x - mx) \] \[ (2-mx)[(2-mx)(2-\mu x) - 1] + x(\mu - m) \] Let’s expand \((2-mx)(2-\mu x)\): \[ (2-mx)(2-\mu x) = 4 - 2\mu x - 2mx + m\mu x^2 \] Now, plug in: \[ (2-mx)[(4 - 2\mu x - 2mx + m\mu x^2) - 1] + x(\mu - m) = 0 \] \[ (2-mx)[3 - 2\mu x - 2mx + m\mu x^2] + x(\mu - m) = 0 \] Let’s expand: \[ (2-mx)(3 - 2\mu x - 2mx + m\mu x^2) + x(\mu - m) = 0 \] Expand the first term: \[ = (2)(3 - 2\mu x - 2mx + m\mu x^2) - mx(3 - 2\mu x - 2mx + m\mu x^2) \] \[ = [6 - 4\mu x - 4mx + 2m\mu x^2] - [3mx - 2m\mu x^2 - 2m^2 x^2 + m^2\mu x^3] \] So sum: \[ 6 - 4\mu x - 4mx + 2m\mu x^2 - 3mx + 2m\mu x^2 + 2m^2 x^2 - m^2\mu x^3 + x(\mu - m) = 0 \] Group terms: - Constant: \(6\) - \(x\) terms: \(-4\mu x - 4mx - 3mx + x(\mu - m)\) - \(x^2\) terms: \(2m\mu x^2 + 2m\mu x^2 + 2m^2 x^2\) - \(x^3\) term: \(-m^2\mu x^3\) So: - \(x\) terms: \(-4\mu x - 7mx + \mu x - mx = (-4\mu x + \mu x) + (-7mx - mx) = (-3\mu x) + (-8mx)\) - \(x^2\) terms: \(2m\mu x^2 + 2m\mu x^2 + 2m^2 x^2 = 4m\mu x^2 + 2m^2 x^2\) Therefore: \[ 6 - 3\mu x - 8mx + 4m\mu x^2 + 2m^2 x^2 - m^2\mu x^3 = 0 \] Or, rearranged: \[ - m^2\mu x^3 + (4m\mu + 2m^2)x^2 + (-3\mu - 8m)x + 6 = 0 \] This cubic equation gives the three normal mode frequencies, as \(\omega^2 = xk\). --- ## **7. Physical Interpretation** - **Zero mode:** All beads move together around the ring (rotation); this is the translation symmetry of the system (\(\omega_0 = 0\)). - **Other two modes:** Relative oscillations of the beads (their patterns depend on the mass values, but for \(\mu = m\), they correspond to the two degenerate modes of the symmetric triangle). --- ## **Summary** ### **Normal frequencies:** Find the roots \(x_i\) of: \[ - m^2\mu x^3 + (4m\mu + 2m^2)x^2 + (-3\mu - 8m)x + 6 = 0 \] Normal frequencies: \(\omega_i^2 = x_i k\), \(i = 1,2,3\). ### **Normal modes:** - Zero mode: all beads move together. - Other modes: beads move in relative opposition, their nature depending on the values of \(m\) and \(\mu\). --- If you want explicit forms for the normal modes for a particular value of \(\mu\), let me know!