$(a)$
\[
\begin{equation}
\left \{
\begin{array}{l}
m_1\ddot{x}_1=-kx_1-k(x_1-x_2) \\
m_2\ddot{x}_2=-k(x_2-x_1)-kx_2
\end{array}
\right.
\end{equation}
\]
\[
\begin{equation}
\left \{
\begin{array}{l}
m_1\ddot{x}_1+2kx_1-kx_2=0 \\
m_2\ddot{x}_2-kx_1+2kx_2=0
\end{array}
\right.
\end{equation}
\]
\[
\begin{equation}
\left.
\begin{array}{l}
x_1=X_1\cos\omega t \\
x_2=X_2\cos\omega t
\end{array}
\right\}と仮定\\
\end{equation}
\]
\[
\ddot{x}_1=-X_1\omega^2\cos\omega t\\
\ddot{x}_1=-X_2\omega^2\cos\omega t
\]
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}
\[\omega_1=\sqrt{\frac{3-\sqrt{3}}{2}\cdot\frac{50\times10^3\,\rm{N/m}}{1\,\rm{kg}}}=178.0\,\rm{rad/s}\]
\[\omega_2=\sqrt{\frac{3+\sqrt{3}}{2}\cdot\frac{50\times10^3\,\rm{N/m}}{1\,\rm{kg}}}=343.9\,\rm{rad/s}\]
\[\underline{\therefore\omega_1=178\,\rm{rad/s},\omega_2=344\,\rm{rad/s}}\]
\[f_1=\frac{\omega_1}{2\pi}=\frac{178.0\,\rm{rad/s}}{2\pi}=28.32\,\rm{Hz}\]
\[f_2=\frac{\omega_2}{2\pi}=\frac{343.9\,\rm{rad/s}}{2\pi}=54.73\,\rm{Hz}\]
\[\underline{\therefore f_1=28.3\,\rm{Hz},f_2=54.7\,\rm{Hz}}\]