ランダウ・リフシッツ 力学 §24 問題 3

ランダウ・リフシッツの力学(増補第3版) §24 問題2のメモです。

今回も長いです。こんなに長くかかるとは思いませんでした(涙目)。

ランダウ先生には怒られそうですが、今回もMaximaを使って検算しましたよ。Straightforward と言われればそれまでだけど、式変形はしんどい。

方程式

式(24.1)から

\begin{align} m_A x_1 + m_B x_2 + m_C x_3 &= 0 \\ m_A y_1 + m_B y_2 + m_C y_3 &= 0 . \end{align} 式(24.2)から(\(x_{01} = l_1, x_{02} = 0, x_{03}= -l_2\)) \begin{align} m_A (x_{01} y_1 – y_{01}x_1) + m_B(x_{02}y_2 – y_{02} x_2) + m_C(x_{03} y_3 – y_{03}x_3) & = 0 \\ m_A l_1 y_1 + m_C (-l_2) y_3 & = 0 \\ \end{align}

これから

\[ y_3 = \cfrac{m_A l_1}{m_C l_2} y_1 \] \begin{align} m_A y_1 + m_B y_2 + m_C \cfrac{m_A l_1}{m_C l_2} y_1 &= 0 \\ m_A \left( 1 + \cfrac{l_1}{ l_2} \right) y_1 + m_B y_2 &= 0 \\ y_2 & = -\cfrac{m_A}{m_B} \left( 1 + \cfrac{l_1}{ l_2} \right) y_1 \\ & = – 2\cfrac{m_A}{m_B} \cfrac{l}{ l_2} y_1 \\ \end{align}

屈曲のポテンシャルエネルギーは(紛らわしいので\(\delta l = \Delta l\)と書きます)

\[ \cfrac{k_1}{2} (\Delta l_1)^2 + \cfrac{k’_1}{2} (\Delta l_2)^2 + \cfrac{k_2}{2} \delta^2 \]

横振動

縦振動の運動エネルギー\(T\)は

\begin{align} T = \cfrac{m_A}{2} \dot{y}_1^2 + \cfrac{m_B}{2} \dot{y}_2^2 + \cfrac{m_C}{2} \dot{y}_3^2. \end{align}

ラグランジアンは

\begin{align} L = \cfrac{m_A}{2} \dot{y}_1^2 + \cfrac{m_B}{2} \dot{y}_2^2 + \cfrac{m_C}{2} \dot{y}_3^2 – \cfrac{k_2}{2} l^2 \delta^2 \end{align} \begin{align} \delta = \cfrac{y_1 – y_2}{l_1} + \cfrac{y_3 – y_2}{l_2} \end{align} \begin{align} y_1 – y_2 & = y_1 \left[1 + \cfrac{m_A}{m_B} \left( 1 + \cfrac{l_1}{ l_2} \right)\right] = y_1 \left(1+ 2\cfrac{m_A}{m_B} \cfrac{l}{ l_2} \right) \end{align} \begin{align} y_3 – y_2 & = \cfrac{m_A l_1}{m_C l_2} y_1 + 2\cfrac{m_A}{m_B} \cfrac{l}{ l_2} y_1 = y_1 \cfrac{m_A}{l_2} \left(\cfrac{l_1}{m_C} + 2 \cfrac{l}{m_B} \right) \end{align} \begin{align} \cfrac{\delta}{y_1} = & \cfrac{1}{l_1} \cfrac{m_B l_2 + 2 m_A l}{m_B l_2} + \cfrac{m_A}{l_2^2} \left[\cfrac{l_1}{m_C} + \cfrac{l}{m_B}\right] \\ = & \cfrac{1}{l_1l_2^2 m_B m_C} \left( m_B m_C l_2^2 + 2m_A m_C l l_2 + m_A m_B l_1^2 + 2 m_A m_C l l_1 \right) \\ = & \cfrac{1}{l_1l_2^2 m_B m_C} \left( m_A m_B l_1^2 + m_B m_C l_2^2 + 4 m_A m_C l^2 \right) \end{align}

これから

\begin{align} y_1 = & \delta\,\,\,\, \cfrac{l_1\, l_2^2\, m_B m_C}{m_A m_B l_1^2 + m_B m_C l_2^2 + 4 m_A m_C l^2} \\ y_2 = & \delta\,\,\,\, (-2) \cfrac{l_1\, l_2\, l\, m_A m_C} {m_A m_B l_1^2 + m_B m_C l_2^2 + 4 m_A m_C l^2}\\ y_3 = & \delta\,\,\,\, \cfrac{l_1^2\, l_2\, m_A m_B }{m_A m_B l_1^2 + m_B m_C l_2^2 + 4 m_A m_C l^2} \\ \end{align}

この表式を使ってラグランジアンを\(\delta\)で表現します:

\begin{align} L = & \cfrac{l_1^2 l_2^2 m_A m_B m_C}{2(m_A m_B l_1^2 + m_B m_C l_2^2 + 4 m_A m_C l^2)^2} (l_2^2\, m_B m_C + 4 l^2 m_A m_C + l_1^2 m_A m_B) \dot{\delta}^2 – \cfrac{k_2}{2} l^2 \delta^2 \\ = & \cfrac{l_1^2 l_2^2 m_A m_B m_C}{2(m_A m_B l_1^2 + m_B m_C l_2^2 + 4 m_A m_C l^2)} \,\,\, \dot{\delta}^2 – \cfrac{k_2}{2} l^2 \delta^2 \\ \end{align}

これから振動数\(\omega\)は

\begin{align} \omega^2 = & k_2 l^2 \,\, \cfrac{(m_A m_B l_1^2 + m_B m_C l_2^2 + 4 m_A m_C l^2)} {l_1^2 l_2^2 m_A m_B m_C} \\ = & \cfrac{k_2 l^2}{l_1^2 l_2^2} \,\, \left( \cfrac{ l_1^2}{m_C} + \cfrac{ l_2^2}{m_A} + 4 \cfrac{ l^2}{m_B} \right). \end{align}

縦振動

本文ではさらりと述べていますが、導出はとても長いです。(実は一部maximaを使いながら解きました。)

\begin{align} x_1 = & \cfrac{1}{2}\left(Q_a + Q_s \right) \\ x_3 = & \cfrac{1}{2}\left(Q_a – Q_s \right) \\ x_2 = & -\cfrac{1}{m_B}\left(m_A x_1 + m_C x_3 \right) \\ = & -\cfrac{1}{2m_B}\left(m_A \left(Q_a + Q_s \right) + m_C \left(Q_a – Q_s \right) \right) \\ = & -\cfrac{1}{2m_B}\left((m_A + m_C) Q_a + (m_A – m_C)Q_s \right) \\ \end{align}

運動エネルギーを計算します:

\begin{align} T = & \cfrac{m_A}{2} \dot{x}_1^2 + \cfrac{m_B}{2} \dot{x}_2^2 + \cfrac{m_C}{2} \dot{x}_3^2 \\ = & \cfrac{m_A}{2} \cfrac{1}{2^2} (\dot{Q}_a + \dot{Q}_s)^2 + \cfrac{m_B}{2} \cfrac{1}{(2m_B)^2} ((m_A + m_C) \dot{Q}_a + (m_A – m_C) \dot{Q}_s )^2 + \cfrac{m_C}{2} \cfrac{1}{2^2} (\dot{Q}_a – \dot{Q}_s)^2 \\ = & \cfrac{m_A m_B}{8m_B} (\dot{Q}_a^2 + \dot{Q}_s^2 + 2 \dot{Q}_a \dot{Q}_s) \\ + & \cfrac{1}{8m_B} ((m_A + m_C)^2 \dot{Q}_a^2 + (m_A – m_C)^2 \dot{Q}_s^2 + 2 (m_A + m_C) (m_A – m_C) \dot{Q}_a \dot{Q}_s ) \\ + & \cfrac{m_B m_C}{8 m_B} (\dot{Q}_a^2 + \dot{Q}_s^2 – 2 \dot{Q}_a \dot{Q}_s ) \\ = & \cfrac{1}{8m_B} ( m_A m_B + (m_A + m_C)^2 + m_B m_C) \dot{Q}_a^2 + \cfrac{1}{8m_B} (m_A m_B + (m_A – m_C)^2 + m_B m_C) \dot{Q}_s^2 \\ + & \cfrac{2}{8m_B} (m_A m_B + (m_A + m_C) (m_A – m_C) – m_B m_C ) \dot{Q}_a \dot{Q}_s \\ = & \cfrac{1}{8m_B} (m_A + m_C) ( m_A + m_C + m_B ) \dot{Q}_a^2 + \cfrac{1}{8m_B} ((m_A + m_C) m_B + (m_A – m_C)^2 ) \dot{Q}_s^2 \\ + & \cfrac{2}{8m_B} ((m_A -m_C)m_B + (m_A + m_C) (m_A – m_C) ) \dot{Q}_a \dot{Q}_s \\ = & \cfrac{1}{8m_B} (m_A + m_C) (m_A + m_B + m_C) \dot{Q}_a^2 + \cfrac{1}{8m_B} ((m_A + m_C) m_B + (m_A – m_C)^2 ) \dot{Q}_s^2 \\ + & \cfrac{2}{8m_B} (m_A -m_C) (m_B + m_A + m_C ) \dot{Q}_a \dot{Q}_s \\ = & \cfrac{\mu}{8m_B} (m_A + m_C) \dot{Q}_a^2 + \cfrac{1}{8m_B} ((m_A + m_C) m_B + (m_A – m_C)^2 ) \dot{Q}_s^2 + \cfrac{\mu}{4m_B} (m_A -m_C) \dot{Q}_a \dot{Q}_s \\ \end{align} \begin{align} \cfrac{d}{dt}\cfrac{\partial L}{\partial \dot{Q}_a} = & \cfrac{d}{dt}\cfrac{\partial (T-U)}{\partial \dot{Q}_a} \\ = & \cfrac{\mu}{4m_B} (m_A + m_C) \ddot{Q}_a + \cfrac{\mu}{4m_B} (m_A -m_C) \ddot{Q}_s \end{align} \begin{align} \cfrac{d}{dt}\cfrac{\partial L}{\partial \dot{Q}_s} = & \cfrac{d}{dt}\cfrac{\partial (T-U)}{\partial \dot{Q}_s} \\ = & \cfrac{1}{4m_B} ((m_A + m_C) m_B + (m_A – m_C)^2 ) \ddot{Q}_s + \cfrac{\mu}{4m_B} (m_A -m_C) \ddot{Q}_a \\ \end{align} \begin{align} x_1 – x_2 = & \cfrac{1}{2}\left(Q_a + Q_s \right) + \cfrac{1}{2m_B}\left((m_A + m_C) Q_a + (m_A – m_C)Q_s \right) \\ = & \cfrac{1}{2m_B}\left( (m_A + m_B + m_C) Q_a + (m_A + m_B – m_C)Q_s \right) \\ = & \cfrac{\mu }{2m_B} Q_a + \cfrac{1 }{2m_B} (m_A + m_B – m_C)Q_s \\ x_3 – x_2 = & \cfrac{1}{2}\left(Q_a – Q_s \right) + \cfrac{1}{2m_B}\left((m_A + m_C) Q_a + (m_A – m_C)Q_s \right) \\ = & \cfrac{1}{2m_B} (m_A + m_B + m_C) Q_a + \cfrac{1}{2m_B} (m_A -m_B – m_C)Q_s \\ = & \cfrac{\mu}{2m_B} Q_a + \cfrac{1}{2m_B} (m_A -m_B – m_C)Q_s \\ \end{align}

ポテンシャル・エネルギーを計算します:

\begin{align} U = & \cfrac{k_1}{2} (x_1 – x_2)^2 + \cfrac{k_1′}{2}(x_3 – x_2)^2 \\ = & \cfrac{k_1}{2}\left[\cfrac{\mu }{2m_B} Q_a + \cfrac{1 }{2m_B} (m_A + m_B – m_C)Q_s \right]^2 \\ + & \cfrac{k’_1}{2}\left[\cfrac{\mu}{2m_B} Q_a + \cfrac{1}{2m_B} (m_A -m_B – m_C)Q_s \right]^2 \\ = & \cfrac{k_1}{8m_B^2}\left[\mu Q_a + (m_A + m_B – m_C)Q_s \right]^2 \\ + & \cfrac{k’_1}{8m_B^2}\left[\mu Q_a + (m_A -m_B – m_C)Q_s \right]^2. \end{align}

座標に関する微分を計算します:

\begin{align} \cfrac{\partial L}{\partial Q_a} = & \cfrac{\partial (T-U)}{\partial Q_a} \\ = & – \cfrac{k_1}{4m_B^2} \mu \left[\mu Q_a + (m_A + m_B – m_C)Q_s \right] – \cfrac{k’_1}{4m_B^2} \mu \left[\mu Q_a + (m_A -m_B – m_C)Q_s \right] \\ = & – \cfrac{\mu^2}{4m_B^2}(k_1 + k’_1) Q_a – \cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C) k’_1\right] Q_s \end{align} \begin{align} \cfrac{\partial L}{\partial Q_s} = & \cfrac{\partial (T-U)}{\partial Q_s} \\ = & -\cfrac{k_1}{4m_B^2} (m_A + m_B – m_C) \left[\mu Q_a + (m_A + m_B – m_C)Q_s \right] – \cfrac{k’_1}{4m_B^2} (m_A -m_B – m_C) \left[\mu Q_a + (m_A -m_B – m_C)Q_s \right] \\ = & – \cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C)k’_1 \right] Q_a – \cfrac{1}{4m_B^2} \left[\ (m_A + m_B – m_C)^2k_1 + (m_A -m_B – m_C)^2k’_1 \right] Q_s \end{align}

運動方程式

\begin{align} Q_a = & A_a e^{i\omega t} \\ Q_s = & A_s e^{i\omega t} \\ \end{align}

と置いて運動方程式を立てます。

\[ \cfrac{d}{dt}\cfrac{\partial L}{\partial \dot{Q}_a} – \cfrac{\partial L}{\partial Q_a} = 0 \]

については

\begin{align} & -\omega^2 \cfrac{\mu}{4m_B} (m_A + m_C) A_a e^{i\omega t} -\omega^2 \cfrac{\mu}{4m_B} (m_A – m_C) A_s e^{i\omega t} \\ & + \cfrac{\mu^2}{4m_B^2}(k_1 + k’_1) A_a e^{i\omega t} + \cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C) k’_1\right] A_s e^{i\omega t} \\ & = 0. \end{align} \[ \cfrac{d}{dt}\cfrac{\partial L}{\partial \dot{Q}_s} – \cfrac{\partial L}{\partial Q_s} = 0 \]

については

\begin{align} & -\omega^2 \cfrac{1}{4m_B} ((m_A + m_C) m_B + (m_A – m_C)^2 ) A_s e^{i\omega t} – \omega^2 \cfrac{\mu}{4m_B} (m_A -m_C) A_a e^{i\omega t} \\ & + \cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C)k’_1 \right] A_a e^{i\omega t} \\ & + \cfrac{1}{4m_B^2} \left[\ (m_A + m_B – m_C)^2k_1 + (m_A -m_B – m_C)^2k’_1 \right] A_s e^{i\omega t} \\ & = 0. \end{align}

運動方程式は

\begin{align} & \left[-\omega^2 \cfrac{\mu}{4m_B} (m_A + m_C) + \cfrac{\mu^2}{4m_B^2}(k_1 + k’_1) \right] A_a \\ & + \left[ -\omega^2 \cfrac{\mu}{4m_B} (m_A – m_C)+\cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C) k’_1\right] \right] A_s \\ & = 0 \end{align} \begin{align} & \left[- \omega^2 \cfrac{\mu}{4m_B} (m_A -m_C) + \cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C)k’_1 \right] \right] A_a \\ & + \left[ -\omega^2 \cfrac{1}{4m_B} ((m_A + m_C) m_B + (m_A – m_C)^2 ) + \cfrac{1}{4m_B^2} \left[\ (m_A + m_B – m_C)^2k_1 + (m_A -m_B – m_C)^2k’_1 \right] \right] A_s \\ & = 0 \end{align} これだと見にくいので \begin{align} a_{11} A_a + a_{12} A_s & = 0 \\ a_{21} A_a + a_{22} A_s & = 0 \end{align} と置いて特有方程式は \[ a_{11} a_{22} – a_{12} a_{21} = 0 \] です。

それぞれ係数は

\begin{align} a_{11} & = -\omega^2 \cfrac{\mu}{4m_B} (m_A + m_C) + \cfrac{\mu^2}{4m_B^2}(k_1 + k’_1) \\ a_{12} & = -\omega^2 \cfrac{\mu}{4m_B} (m_A – m_C)+\cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C) k’_1\right] \\ a_{21} & = – \omega^2 \cfrac{\mu}{4m_B} (m_A -m_C) + \cfrac{\mu}{4m_B^2} \left[ (m_A + m_B – m_C)k_1 + (m_A -m_B – m_C)k’_1 \right] \\ a_{22} & = -\omega^2 \cfrac{1}{4m_B} ((m_A + m_C) m_B + (m_A – m_C)^2 ) + \cfrac{1}{4m_B^2} \left[\ (m_A + m_B – m_C)^2k_1 + (m_A -m_B – m_C)^2k’_1 \right] \end{align} です。

\(\omega\)についての方程式

\(\omega\)の係数を求めるうえで、特有方程式を

\[ C_1 \omega^4 + C_2 \omega + C_3 = 0 \]

の形に書きます。

\(\omega^4\)の係数

\(\omega^4\)の係数\(C_1\)は \begin{align} C_1 = & \cfrac{1}{(4m_B)^2}\mu (m_A+m_C) ((m_A+m_C) m_B + (m_A-m_C)^2) – \mu^2 (m_A-m_C)^2 \\ =& \cfrac{\mu}{(4m_B)^2} \left[(m_A+m_C) ((m_A+m_C) m_B + (m_A-m_C)^2) – (m_A + m_B + m_C) (m_A-m_C)^2\right] \\ =& \cfrac{\mu}{(4m_B)^2} \left[ ((m_A+m_C)^2 m_B + (m_A^2 – m_C^2)(m_A-m_C)) – (m_A + m_B + m_C) (m_A^2 -2m_A m_C + m_C^2)\right]\\ =& \cfrac{\mu}{(4m_B)^2} (m_A^2m_B + 2 m_Am_Bm_C + m_Bm_C^2 +m_A^3 + m_C^3 -m_Am_C^2 – m_A^2m_C \\ & – (m_A^3 -2m_A^2 m_C + m_A m_C^2 + m_A^2m_B -2m_A m_B m_C + m_B m_C^2 + m_A^2m_C -2m_A m_C^2 + m_C^3 ))\\ =& \cfrac{1}{(4m_B)^2} \,\, 4 \mu m_Am_Bm_C \\ =& \cfrac{1}{16 m_B^2} \,\, 4 \mu m_Am_Bm_C . \end{align}

\(\omega^2\)の係数

\(\omega^2\)の係数\(C_2\)は %-(ma+mb+mc)*(ma+mc)*((ma+mb-mc)^2*k1+(ma-mb-mc)^2*k2) -(ma+mb+mc)^2*(k1+k2)*((ma+mc)*mb+(ma-mc)^2) + 2*(ma+mb+mc)^2*(ma-mc)*((ma+mb-mc)*k1+(ma-mb-mc)*k2) ; \begin{align} 4^2 m_B^3 C_2 = & – \mu (m_A+m_C)((m_A+m_B-m_C)^2 k_1 + (m_A-m_B-m_C)^2 k’_1) – \mu^2 (k_1+k’_1) ((m_A+m_C)m_B+(m_A-m_C)^2) \\ +& 2 \mu^2 (m_A-m_C) ((m_A+m_B-m_C)k_1 + (m_A-m_B-m_C) k’_1) \\ = & \mu \left[2\mu (m_A-m_C) (m_A+m_B-m_C) – (m_A + m_C)(m_A+m_B-m_C)^2 – \mu ((m_A+m_C)m_B+(m_A-m_C)^2)\right] k_1 \\ + & \mu \left[ 2\mu (m_A – m_C) (m_A – m_B – m_C) – (m_A + m_C) (m_A-m_B-m_C)^2 – \mu (m_A+m_C)m_B+(m_A-m_C)^2 \right] k’_1 \\s. \end{align}

\(\mu\)を除いた\(k_1\)の係数は

\begin{align} &2\mu (m_A-m_C) (m_A+m_B-m_C) – (m_A + m_C)(m_A+m_B-m_C)^2 – \mu ((m_A+m_C)m_B+(m_A-m_C)^2) \\ = & 2 m_C^3-2 m_A m_C^2-2 m_B^2 m_C-4 m_A m_B m_C-2 m_A^2 m_C+2 m_A m_B^2+4 m_A^2 m_B+2 m_A^3 \\ & -m_C^3+2 m_B m_C^2+m_A m_C^2-m_B^2 m_C+m_A^2 m_C-m_A m_B^2-2 m_A^2 m_B-m_A^3 \\ & -m_C^3-2 m_B m_C^2+m_A m_C^2-m_B^2 m_C+m_A^2 m_C-m_A m_B^2-2 m_A^2 m_B-m_A^3 \\ = & – 4 m_B^2 m_C – 4 m_A m_B m_C \\ = & -4 m_B(m_A m_C + m_B m_C) \end{align}

同様に\(\mu\)を除いた\(k’_1\)の係数は

\begin{align} & 2\mu (m_A – m_C) (m_A – m_B – m_C) – (m_A + m_C) (m_A-m_B-m_C)^2 – \mu (m_A+m_C)m_B+(m_A-m_C)^2 \\ = & 2 m_C^3 + 4 m_B m_C^2 – 2 m_A m_C^2 + 2 m_B^2 m_C – 4 m_A m_B m_C – 2 m_A^2 m_C – 2 m_A m_B^2 + 2 m_A^3 \\ & – m_C^3 – 2 m_B m_C^2 + m_A m_C^2 – m_B^2 m_C + m_A^2 m_C – m_A m_B^2 + 2 m_A^2 m_B – m_A^3 \\ & – m_C^3 – 2 m_B m_C^2 + m_A m_C^2 – m_B^2 m_C + m_A^2 m_C – m_A m_B^2 – 2 m_A^2 m_B – m_A^3 \\ = & – 4 m_A m_B m_C – 4 m_A m_B^2 \\ = & – 4 m_B (m_A m_C + m_A m_B) \\ \end{align} \begin{align} C_2 = & \cfrac{-4 m_B \mu}{16m_B^3} (k_1 (m_B m_C + m_A m_C) + k’_1 (m_A m_C + m_A m_B)) \\ = & \cfrac{-4 \mu}{16m_B^2} (k_1 (m_B m_C + m_A m_C) + k’_1 (m_A m_C + m_A m_B)) \end{align}

定数項

定数項\(C_3\)は

%(k1+k2)*((ma+mb-mc)^2*k1+(ma-mb-mc)^2*k2) – ((ma+mb-mc)*k1+(ma-mb-mc)*k2)*((ma+mb-mc)*k1+(ma-mb-mc)*k2); \begin{align} C_3 = & \cfrac{\mu^2}{(4m_B^2)^2} (k_1+k’_1)((m_A+m_B-m_C)^2k_1+(m_A-m_B-m_C)^2k’_1) – ((m_A+m_B-m_C)k_1+(m_A-m_B-m_C)k’_1)^2\\ = & \cfrac{\mu^2}{16m_B^4}(((m_A+m_B-m_C)^2 k_1^2 + (m_A-m_B-m_C)^2 {k’_1}^2 + ((m_A+m_B-m_C)^2 + (m_A-m_B-m_C)^2 ) k_1 k’_1 ) \\ – & ((m_A+m_B-m_C)^2k_1^2 + 2(m_A + m_B – m_C)(m_A – m_B – m_C) k_1 k’_1 + (m_A – m_B – m_C)^2 {k’_1}^2)^2 )\\ = & \cfrac{\mu^2}{16m_B^4} 2(m_A^2 + m_B^2 + m_C^2 -2m_Am_C) k_1 k’_1 – 2((m_A – m_C)^2 – m_B^2) k_1 k’_1 \\ = & \cfrac{\mu^2}{16m_B^4} 2(m_A^2 + m_B^2 + m_C^2 -2m_Am_C) k_1 k’_1 – 2(m_A^2 – 2m_A m_C + m_C^2 – m_B^2) k_1 k’_1 \\ = & \cfrac{\mu^2}{16m_B^4} 4 m_B^2 k_1 k’_1 \\ = & \cfrac{1}{16m_B^2} \,\,\,\, 4 \mu^2 k_1 k’_1 . \\ \end{align}

最終的な\(\omega\)の方程式

最終的に\(\omega\)の方程式は

\begin{align} \cfrac{1}{16 m_B^2} \,\, 4 \mu m_Am_Bm_C \omega^4- \cfrac{4 \mu}{16m_B^2} (k_1 (m_B m_C + m_A m_C) + k’_1 (m_A m_C + m_A m_B))\omega^2 + \cfrac{4 \mu^2 k_1 k’_1}{16 m_B^2} & = 0 \\ m_Am_Bm_C \omega^4- (k_1 (m_B m_C + m_A m_C) + k’_1 (m_A m_C + m_A m_B))\omega^2 + \mu k_1 k’_1 & = 0 \\ \omega^4- \left[ k_1 \left( \cfrac{1}{m_A} + \cfrac{1}{m_B} \right) + k’_1 \left( \cfrac{1}{m_B} + \cfrac{1}{m_C} \right) \right] \omega^2 + \cfrac{\mu k_1 k’_1}{m_A m_B m_C} & = 0 \end{align}

§24 問題1の結果との比較

この方程式で\(m_C = m_A\), \(k’_1 = k_1 = k\)と置くと

\begin{align} \omega^4- 2\left[ k \left( \cfrac{1}{m_A} + \cfrac{1}{m_B} \right) \right] \omega^2 + \cfrac{\mu k^2 }{m_A^2 m_B } & = 0 \\ \omega^4- \left[ k \left( \cfrac{2m_A +2 m_B}{m_A m_B} \right) \right] \omega^2 + \cfrac{\mu k }{m_A m_B } \cfrac{k}{m_A}& = 0 \\ \omega^4- \left[ k \left( \cfrac{2m_A + m_B}{m_A m_B} + \cfrac{m_B}{m_A m_B} \right) \right] \omega^2 + \cfrac{\mu k }{m_A m_B } \cfrac{k}{m_A}& = 0 \\ \omega^4- \left[ k \left( \cfrac{\mu}{m_A m_B} + \cfrac{1}{m_A} \right) \right] \omega^2 + \cfrac{\mu k }{m_A m_B } \cfrac{k}{m_A}& = 0 \\ \left(\omega^2 – \cfrac{k\mu}{m_A m_B} \right) \left(\omega^2 – \cfrac{k}{m_A} \right) & = 0. \end{align} これから\(\omega\)が \begin{align} \omega_a & = \sqrt{\cfrac{k\mu}{m_A m_B} } \\ \omega_{s1} & = \sqrt{\cfrac{k}{m_A}} \end{align} が得られます。これは§24 問題1の結果と一致します。

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