$(a)$ \[\begin{align} \ k_e&=k_1+\frac{1}{\frac{1}{k_2}+\frac{1}{k_3}}\\ &=k_1+\frac{1}{\frac{k_3+k_2}{k_2k_3}} \end{align}\] \[\therefore k_e=k_1+\frac{k_2k_3}{k_2+k_3} \] $(b)$ \[\begin{align} k_e&=\frac{1}{\frac{1}{k_1}+\frac{1}{k_2+k_3}}\\&=\frac{1}{\frac{k_2+k_3+k_1}{k_1(k_2+k_3)}} \end{align}\] \[ \therefore k_e=\frac{k_1(k_2+k_3)}{k_1+k_2+k_3} \] $(c)$ \[ k_e=k_1+k_2 \] \[ \therefore k_e=k_1+k_2 \]

$(1)'$ 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，$\frac{x}{2}=r\theta$ $(a)$ \[ \begin{align} T&=\frac{1}{2}m(\frac{\dot{x}}{2})^2+\frac{1}{2}J\dot{\theta}^2\\ &=\frac{1}{2}m\left(\frac{\dot{x}}{2}\right)^2+\frac{1}{2}\left(\frac{mr^2}{2}\right)\left(\frac{\dot{x}}{2r}\right)^2\\ &=\frac{1}{2}\left(\frac{m}{4}+\frac{m}{8}\right)\dot{x}^2=\frac{1}{2}\left(\frac{3}{8}m\right)\dot{x}^2\\ \end{align}\] $(b)$ \[U=\frac{1}{2}kx^2\\ (1)'と同様に\\ T_{max}=\frac{1}{2}\left(\frac{3}{8}m\right)\left(X\omega_n\right)^2\\ U_{max}=\frac{1}{2}kX^2\\ \] \[ T_{max}=U_{max}より\\ \frac{1}{2}\left(\frac{3}{8}m\right)X^2{\omega_n}^2=\frac{1}{2}kX^2\\ \therefore \omega_n=\sqrt{\frac{k}{\frac{3}{8}m}}=\sqrt{\frac{8}{3}\cdot\frac{k}{m}}\\ \therefore f_n=\frac{\omega_n}{2\pi}=\frac{1}{2\pi}\cdot\sqrt{\frac{8}{3}\cdot\frac{k}{m}} \]

$(a)$ \[ J=m\cdot\frac{(D/2)^2}{2}=200\,\rm{kg}\times\frac{(1\,\rm{m}/2)^2}{2}\\ =25.0\,\kgsqm\hspace{20mm}\\ \underline{\therefore J=25.0\ \kgsqm} \] $(b)$ \[ \begin{align} k_{te}&=\frac{1}{\frac{1}{k_{t1}}+\frac{1}{k_{t2}}}=\frac{k_{t1}\cdot k_{t2}}{k_{t1}+k_{t2}}\\ k_{t1}&=\frac{GI_{p}}{L_1}=\frac{G\pi d_1^4} {32L_1}=\frac{81\times10^9\,\rm{Pa}\times\pi\times(0.02\,\rm{m})^4}{32\times0.1\,\rm{m}}\\ &=12.72\times10^3\,\rm{N・m/rad}\\ k_{t2}&=\frac{GI_{p}}{L_2}=\frac{G\pi d_2^4}{32L_2} =\frac{81\times10^9\,\rm{Pa}\times\pi\times(0.03\,\rm{m})^4}{32\times0.2\,\rm{m}}\\ &=32.20\times10^3\,\rm{N・m/rad}\\ k_{te}&=\frac{12.72\times10^3\times32.20\times10^3}{12.72\times10^3+32.20\times10^3}=9.118\times10^3\,\rm{N・m/rad}\\ \therefore k_{te} &=9.12\,\rm{kN・m/rad} \end{align} \] $(c)$ \[J\ddot\theta=(-k_{te}\cdot\theta)より\\ \therefore \hspace{10mm}\underline{J\ddot\theta+k_{te}\theta=0}\] $(d)$ \[\omega_n=\sqrt{\frac{k_{te}}{J}}=\sqrt \frac{9.118\times10^3\,\rm{N・m/rad}}{25.0\,\rm{kg・m}^2}=19.09\,\rm{rad/s}\hspace{10mm}\\ \underline{\therefore \omega _n=19.1\,\rm{rad/s}}\] \[f_n=\frac{\omega_n}{2\pi}=\frac{19.09\,\rm{rad/s}}{2\pi}=3.038\,\rm{Hz}\hspace{10mm}\\ \underline{\therefore f_n=3.04\,\rm{Hz}} \]

$(a)$ \[ \nu^2=\frac{|a_1|}{|a_2|}\times\frac{|a_2|}{|a_3|}=e^{\varepsilon T_d}\] 両辺の対数をとると \[\ln\nu^2=\ln\frac{|a_1|}{|a_3|}=\varepsilon T_d\] \[2\ln\nu=2\delta=\ln\frac{|a_1|}{|a_3|}= \varepsilon T_d\] \[\delta=\frac{1}{2}\ln\frac{|a_1|}{|a_3|}=\frac{1}{2}\ln\frac{1}{0.0226}=1.894\\ \underline{\therefore \delta =1.89}\] $(b)$ \[\delta=\frac{\varepsilon T_d}{2}=\frac{\omega_n\cdot\zeta\cdot\frac{2\pi}{\omega_d}}{2}=\frac{\omega_n\cdot \zeta\cdot\pi}{\omega_n\sqrt{1-\zeta^2}}\] \begin{align}\delta^2(1-\zeta^2)&=(\pi\zeta)^2\\\delta^2-\delta^2\zeta^2&=(\pi\delta)^2\\\delta^2&=(\pi\zeta)^2+\delta^2\zeta^2\\&=(\pi^2+\delta^2)\zeta^2\end{align} \[ \begin{align} \zeta&=\sqrt{\frac{\delta^2}{\pi^2+\delta^2}}\\&=\sqrt{\frac{1.894^2}{\pi^2+1.894^2}}\\&=0.5163\\ \therefore\zeta&=\underline{0.516} \end{align} \] $(c)$ \begin{align}\omega_d&=\omega_n\sqrt{1-\zeta^2}\\&=\sqrt{\frac{k}{m}}\cdot\sqrt{1-\zeta^2}\\&=\sqrt{\frac{1200\,\rm{N/m}}{5\,\rm{kg}}}\cdot\sqrt{1-0.5163^2}\\&=13.26\,\rm{rad/s}\end{align}\[\underline{\therefore\omega_d=13.3\,\rm{rad/s}}\] \begin{align}f_d&=\frac{w_d}{2\pi}\\ &=\frac{13.26\,\rm{rad/s}}{2\pi}\\ &=2.110\,\rm{Hz}\end{align} \[\underline{\therefore f_d=2.11\,\rm{Hz}}\] $(d)$ \begin{align}c&=c_{cr}\cdot\zeta\\&=2\sqrt{mk}\cdot \zeta\\ &=2\sqrt{5\,\rm{kg}\times1200\,\rm{N/m}}\times0.5163\\ &=79.98\,\rm{N/(m/s)}\end{align} \[\underline{\therefore c=80.0\,\rm{N/(m/s)}}\]

\[伝達率：T_R=\frac{|F_T|}{F}=\frac{A\sqrt{k^2+(c\omega)^2}}{m_u\cdot e\cdot\omega^2}\] \[\underline{\therefore 遠心力：F=m_u\cdot e\cdot\omega^2}\] \[ここで，\sqrt{k^2+(c\omega)^2}=k\sqrt{1+(\frac{c\omega}{k})^2}=k\sqrt{1+(\frac{c}{m}\cdot\frac{\omega}{k/m})^2}\\ =k\sqrt{1+(2\varepsilon\cdot\frac{\omega}{\omega_n^2})^2}=k\sqrt{1+(\frac{2\varepsilon}{\omega_n}\cdot\frac{\omega}{\omega_n})^2}=k\sqrt{1+(2\zeta Z)^2}\] \begin{align}ゆえに，T_R&=\frac{A\cdot k\sqrt{1+(2\zeta Z^2)}}{m_u\cdot e\cdot\omega^2}\\ &=\frac{(\frac{m_ue}{m})Z^2}{\sqrt{(1-Z^2)^2+(2\zeta Z)^2}}\cdot\frac{k\sqrt{1+(2\zeta Z)^2}}{m_u\cdot e\cdot \omega^2}\\ &=\frac{Z^2}{m}\cdot\frac{k}{\omega^2}\cdot\frac{\sqrt{1+(2\zeta Z^2)}}{\sqrt{(1-Z^2)^2+(2\zeta Z)^2}}\\ &=\frac{Z^2}{m}\cdot\frac{k}{\omega^2}\cdot\frac{\sqrt{1+(2\zeta Z)^2}}{\sqrt{(1-Z^2)^2}+(2\zeta Z)^2}\end{align} \[ここで，\frac{Z^2}{m}\cdot\frac{k}{ \omega^2 }=Z^2\cdot\frac{k}{m}\cdot\frac{1}{\omega^2}=Z^2\cdot\frac{\omega_n^2}{\omega^2}=\frac{\omega^2}{\omega_n^2}\cdot\frac{\omega_n^2}{\omega^2}=1\] \[\underline{\therefore T_R=\frac{\sqrt{1+(2\zeta Z)^2}}{\sqrt{(1-Z^2)^2+(2\zeta Z)^2}}}\]

$(1)$ ばねの静的たわみは \[ \delta＝l_0-l_x ＝ 0.40\,\rm{m}－0.15\,\rm{m} ＝ 0.25\,\rm{m} \] ばねこわさの定義(単位長さ伸縮させるために必要な力)より \[ k ＝ \frac{w}{\delta}＝ \frac{2000\,\rm{N}}{0.25\,\rm{m}} ＝ 8000\,\rm{N/m}＝ 8\,\rm{kN/m} \] $(2)$ 質量は \[ m ＝\frac{ 2000\,\rm{N}}{9.81\,\rm{m/s^2}} ＝ \frac{2000\,\rm{kg\cdot m/s^2}} {9.81\,\rm{m/s^2}} ＝ 203.8\,\rm{kg} \] 物体が$y$軸方向に直線運動しており，時刻$t$で加速度$\ddot{x}$を得るためには，質量$m$に動的力が作用している． そのとき変位$x$ならば，ばねの復元力は$－kx$ゆえ，運動方程式は， \[ m\ddot{x} ＝－kx\\ m\ddot{x} ＋kx ＝0\\ \ddot{x} ＋\left(\frac{k}{m}\right)x ＝０\\ \ddot{x}＋\omega_n^2 x ＝０ \] 固有角振動数は \[\begin{align} \omega_n＝\sqrt{\frac{k}{m}} &=\sqrt{\frac{8000\,\rm{N/m}}{203.8\,\rm{kg}}}\\ &＝\sqrt{\frac{8000 \,\rm{\left(kg\cdot m/s^2 \right)/m}}{203.8\,\rm{kg}}} \\ &=\sqrt{39.25\,\rm{1/s^2}} \\ &= 6.265\,\rm{rad/s}\\ &＝ 6.27\,\rm{rad/s} \end{align}\] 固有振動数$f_n$は \[ f_n ＝\frac{\omega_n}{2\pi} ＝ \frac{6.265\,\rm{rad/s}}{2\pi\,\rm{rad}} ＝0.9971\,\rm{Hz}＝ 0.997 \,\rm{Hz} \] $(2)$別解 \[ (1)より \ \delta＝\frac{w}{k}=\frac{mg}{k} \] \[\begin{align} \omega_n &=\sqrt{\frac{k}{m}}\\ &=\sqrt{\frac{g}{\delta}}\\ &=\sqrt{\frac{9.81\,\rm{m/s^2}}{0.25\,\rm{m}}} \\ &=\sqrt{39.24\,\rm{m/s^2}} \\ &＝6.264\,\rm{rad/s} \end{align}\] 静的知識を混同して勘違いする例が多いので，この考え方は使わないこと．

$(1)$ \[\begin{align} k_1=\frac{T}{\theta} &=\frac{5000\,\rm{N\cdot m}}{\left\{3\,\rm{deg}\cdot \frac{\pi\,\rm{rad}}{180\,\rm{deg}} \right\}}\\ &=\frac{5000\,\rm{N\cdot m}}{0.05235\,\rm{rad}}\\ &=95490\,\rm{N\cdot m/rad}\\ &＝ 95.5\,\rm{kN\cdot m/rad} \end{align}\] $(2)$ \[\begin{align} \omega_n＝\sqrt{\frac{k_t}{J}} &=\sqrt{\frac{95490\,\rm{N\cdot m/rad}}{2\,\rm{kg\cdot m^2}}}\\ &=\sqrt{\frac{95490\,\rm{kg\cdot m\cdot m/s^2\cdot rad}}{2\,\rm{kgm^2}}}\\ &=218.5 \,\rm{rad/s}\\ &＝ 219 \,\rm{rad/s}\\\end{align}\] \[ f_n ＝\frac{\omega_n}{2\pi} ＝\frac{218.5 \,\rm{rad/s}}{2\pi\,\rm{rad}}＝ 34.77\,\rm{Hz} ＝ 34.8\,\rm{Hz} \]

$(a)$ \[\omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{200\times10^3\,\rm{N/m}}{5\,\rm{kg}}}=200.0\,\rm{rad/s}\] \[\underline{\therefore W_n=200\,\rm{rad/s}}\] \[f_n=\frac{\omega n}{2\pi}=\frac{200.0\,\rm{rad/s}}{2x}=31.83\,\rm{Hz}\] \[\underline{\therefore f_n=31.8\,\rm{Hz}}\] $(b)$ \[Z=\frac{f}{f_n}=\frac{20\,\rm{Hz}}{31.83\,\rm{Hz}}=0.6283<1\] \[\zeta=0よりA=\frac{\delta st}{|1-Z^2|}\] \begin{align}Z<1よりA&=\frac{\delta st}{1-Z^2}=\frac{F/k}{1-Z^2}\\ &=\frac{10\,\rm{N}/200\times10^3\,\rm{N/m}}{1-0.6283^2}\\ &=0.08261\times10^{-3}\,\rm{m}\end{align} \[\underline{\therefore A=0.0826\,\rm{mm}}\] $(c)$ \[\begin{align}A&=\frac{10\,\rm{N}/200\times10^3\,\rm{N/m}}{\sqrt{(1ー0.6283^2)^2+(2\times0.1\times0.6283)^2}}\\ &=0.08088\times10^{-3}\,\rm{m}\\ \end{align} \] \[\underline{\therefore A=0.0809\,\rm{mm}}\] $(d)$ \begin{align}C&=\zeta・2\sqrt{mr}\\ &=0.1\times2\times\sqrt{5\,\rm{kg}\times200\times10^3\,\rm{N/m}}=200.0\,\rm{N/(m/s)}\\ \end{align} \[\underline{\therefore C=200\,\rm{N/(m/s)}}\]