$(a)$ \[ \begin{equation} \left \{ \begin{array}{l} m_1\ddot{x}_1=-kx_1-k(x_1-x_2) \\ m_2\ddot{x}_2=-k(x_2-x_1)-kx_2 \end{array} \right. \end{equation} \] \[ \begin{equation} \left \{ \begin{array}{l} m_1\ddot{x}_1+2kx_1-kx_2=0 \\ m_2\ddot{x}_2-kx_1+2kx_2=0 \end{array} \right. \end{equation} \] \[ \begin{equation} \left. \begin{array}{l} x_1=X_1\cos\omega t \\ x_2=X_2\cos\omega t \end{array} \right\}と仮定\\ \end{equation} \] \[ \ddot{x}_1=-X_1\omega^2\cos\omega t\\ \ddot{x}_1=-X_2\omega^2\cos\omega t \] 運動方程式に代入 \begin{equation} \left \{ \begin{array}{l} -m_1X_1\omega^2+2kX_1-kX_2=0 \\ -m_2X_2\omega^2-kX_1+2kX_2=0 \end{array} \right. \end{equation} \begin{equation} \left \{ \begin{array}{l} (-m_1\omega^2+2k)X_1-kX_2=0\\ -kX_1+(-m_2\omega^2+2k)X_2=0 \end{array} \right. \end{equation} \begin{equation} \left | \begin{array}{cc} -m_1\omega^2+2k & -k \\ -k & -m_2\omega^2+2k \end{array} \right|\left | \begin{array}{l} X_1\\ X_2 \end{array} \right|=0 \end{equation} $(b)$ \begin{equation} \left | \begin{array}{cc} -m_1\omega^2+2k & -k \\ -k & -m_2\omega^2+2k \end{array} \right|=0 \end{equation} $(c)$ \[(-m_1\omega^2+2k)(-m_2\omega^2+2k)-k^2=0\] \[m_1m_2\omega^4-2m_1k\omega^2-2m_2k\omega^2+4k^2-k^2=0\] \[m_1m_2\omega^4-2(m_1+m_2)kw^2+3k^2=0\] ここで，$m_1=2m_2$を代入 \[2m_2^2\omega^4-2\cdot3m_2k\omega^2+3k^2=0\] \[2m_2^2\omega^4-6m_2k\omega^2+3k^2=0\] 解の公式より \begin{align} \omega^2&=\frac{6m_2k\pm\sqrt{36m_2^2k^2-4\times2m_2^2\times 3k^2}}{2\times2m_2^2}\\ &=\frac{6m_2k\pm\sqrt{36m_2^2k^2-24m_2^2k^2}}{4m_2^2}\\ &=\frac{6m_2k\pm\sqrt{12}m_2k}{4m_2^2}\\ &=\frac{3\pm\sqrt{3}}{2}\cdot\frac{k}{m_2}\end{align} \[\omega_1=\sqrt{\frac{3-\sqrt{3}}{2}\cdot\frac{50\times10^3\,\rm{N/m}}{1\,\rm{kg}}}=178.0\,\rm{rad/s}\] \[\omega_2=\sqrt{\frac{3+\sqrt{3}}{2}\cdot\frac{50\times10^3\,\rm{N/m}}{1\,\rm{kg}}}=343.9\,\rm{rad/s}\] \[\underline{\therefore\omega_1=178\,\rm{rad/s}，\omega_2=344\,\rm{rad/s}}\] \[f_1=\frac{\omega_1}{2\pi}=\frac{178.0\,\rm{rad/s}}{2\pi}=28.32\,\rm{Hz}\] \[f_2=\frac{\omega_2}{2\pi}=\frac{343.9\,\rm{rad/s}}{2\pi}=54.73\,\rm{Hz}\] \[\underline{\therefore f_1=28.3\,\rm{Hz}，f_2=54.7\,\rm{Hz}}\]