$(1)'$ 円板の変位を$x$とおくと回転の変位は$r\theta$ゆえ，$x=r\theta$ ばねは$2x$伸びることになる $(a)$ \[ \begin{align} T&=\frac{1}{2}m\dot{x}^2+\frac{1}{2}J\dot{\theta}^2 \\ &=\frac{1}{2}m\dot{x}^2+\frac{1}{2}\left(\frac{mr^2}{2}\right)\left(\frac{\dot{x}}{r}\right)^2\\ &=\frac{1}{2}(m+\frac{m}{2})\dot{x}^2=\frac{1}{2}\left(\frac{3}{2}m\right)\dot{x}^2\\ \end{align}\] $(b)$ \[ U=\frac{1}{2}k(2x)^2=\frac{1}{2}(4k)x^2\\ \] $(c)$ \[ x=X\sin\omega_n t とおくと \dot{x}=X\omega_n\cos\omega_n t\\ x_{max}=X,\dot{x}_{max}=X\omega_nより\\ T_{max}=\frac{1}{2}(\frac{3}{2}m)(X\omega_n)^2 \] \[ U_{max}=\frac{1}{2}(4k)X^2\\ T_{max}=U_{max}より\\ \frac{1}{2}(\frac{3}{2}m)X^2{\omega_n}^2=\frac{1}{2}(4k)X^2\\ \omega_n=\sqrt{\frac{4k}{\frac{3}{2}m}}=\sqrt{\frac{8}{3}\cdot\frac{k}{m}}\\ \therefore \omega_n=\sqrt{\frac{8}{3}\cdot\frac{k}{m}}\\ \therefore f_n=\frac{\omega_n}{2\pi}=\frac{1}{2\pi}\cdot\sqrt{\frac{8}{3}\cdot\frac{k}{m}} \] $(2)'$ ばねの変位を$x$とおくと円板の変位は$\frac{x}{2}$となり，$\frac{x}{2}=r\theta$ $(a)$ \[ \begin{align} T&=\frac{1}{2}m(\frac{\dot{x}}{2})^2+\frac{1}{2}J\dot{\theta}^2\\ &=\frac{1}{2}m\left(\frac{\dot{x}}{2}\right)^2+\frac{1}{2}\left(\frac{mr^2}{2}\right)\left(\frac{\dot{x}}{2r}\right)^2\\ &=\frac{1}{2}\left(\frac{m}{4}+\frac{m}{8}\right)\dot{x}^2=\frac{1}{2}\left(\frac{3}{8}m\right)\dot{x}^2\\ \end{align}\] $(b)$ \[U=\frac{1}{2}kx^2\\ (1)'と同様に\\ T_{max}=\frac{1}{2}\left(\frac{3}{8}m\right)\left(X\omega_n\right)^2\\ U_{max}=\frac{1}{2}kX^2\\ \] \[ T_{max}=U_{max}より\\ \frac{1}{2}\left(\frac{3}{8}m\right)X^2{\omega_n}^2=\frac{1}{2}kX^2\\ \therefore \omega_n=\sqrt{\frac{k}{\frac{3}{8}m}}=\sqrt{\frac{8}{3}\cdot\frac{k}{m}}\\ \therefore f_n=\frac{\omega_n}{2\pi}=\frac{1}{2\pi}\cdot\sqrt{\frac{8}{3}\cdot\frac{k}{m}} \]