(a) The appropriate generalized coordinates are the two angles ϑ 1 a n d ϑ 2 ϑ_{1} and ϑ_{2} ϑ 1 a n d ϑ 2
x 1 = l 1 c o s ϑ 1 , y 1 = l 1 s i n ϑ 1 , x_{1} = l_{1} cos ϑ_{1}, y_{1} = l_{1} sin ϑ_{1}, x 1 = l 1 c o s ϑ 1 , y 1 = l 1 s i n ϑ 1 ,
x 2 = l 1 c o s ϑ 1 + l 2 c o s ϑ 2 , y 2 = l 1 s i n ϑ 1 + l 2 s i n ϑ 2 . x_{2} = l_{1} cos ϑ_{1} + l_{2} cos ϑ_{2}, y_{2} = l_{1} sin ϑ_{1} +l_{2} sin ϑ_{2}. x 2 = l 1 c o s ϑ 1 + l 2 c o s ϑ 2 , y 2 = l 1 s i n ϑ 1 + l 2 s i n ϑ 2 .
(b) From (15.28), it follows by differentiation that
x ˙ 1 = − l 1 ϑ ˙ 1 s i n ϑ 1 , y ˙ 1 = l 1 ϑ ˙ 1 c o s ϑ 1 , \dot{x}_{1} =−l_{1} \dot{ϑ}_{1} sin ϑ_{1}, \dot{y}_{1} = l_{1} \dot{ϑ}_{1} cos ϑ_{1}, x ˙ 1 = − l 1 ϑ ˙ 1 s i n ϑ 1 , y ˙ 1 = l 1 ϑ ˙ 1 c o s ϑ 1 ,
x ˙ 2 = − l 1 ϑ ˙ 1 s i n ϑ 1 − l 2 ϑ ˙ 2 s i n ϑ 2 , y ˙ 2 = l 1 ϑ ˙ 1 c o s ϑ 1 + l 2 ϑ ˙ 2 c o s ϑ 2 . \dot{x}_{2} =−l_{1}\dot{ϑ}_{1} sin ϑ_{1} −l_{2} \dot{ϑ}_{2} sin ϑ_{2}, \dot{y}_{2} = l_{1} \dot{ϑ}_{1} cos ϑ_{1} +l_{2} \dot{ϑ}_{2} cos ϑ_{2}. x ˙ 2 = − l 1 ϑ ˙ 1 s i n ϑ 1 − l 2 ϑ ˙ 2 s i n ϑ 2 , y ˙ 2 = l 1 ϑ ˙ 1 c o s ϑ 1 + l 2 ϑ ˙ 2 c o s ϑ 2 . The kinetic energy of the system is
T = 1 2 m 1 ( x ˙ 1 2 + y ˙ 1 2 ) + 1 2 m 2 ( x ˙ 2 2 + y ˙ 2 2 ) T = \frac{1}{2} m_{1}(\dot{x}^{2}_{1} + \dot{y}^{2}_{1} )+ \frac{1}{2} m_{2}(\dot{x}^{2}_{2} + \dot{y}^{2}_{2} ) T = 2 1 m 1 ( x ˙ 1 2 + y ˙ 1 2 ) + 2 1 m 2 ( x ˙ 2 2 + y ˙ 2 2 )
= 1 2 m 1 l 1 2 ϑ ˙ 1 2 + 1 2 m 2 ( l 1 2 ϑ ˙ 1 2 + l 2 2 ϑ ˙ 2 2 + 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 c o s ( ϑ 1 − ϑ 2 ) ) . = \frac{1}{2} m_{1}l^{2}_{1} \dot{ϑ}^{2}_{1} + \frac{1}{2} m_{2}\left(l^{2}_{1} \dot{ϑ}^{2}_{1} +l^{2}_{2} \dot{ϑ}^{2}_{2}+2l_{1}l_{2} \dot{ϑ}_{1} \dot{ϑ}_{2} cos(ϑ_{1} −ϑ_{2})\right). = 2 1 m 1 l 1 2 ϑ ˙ 1 2 + 2 1 m 2 ( l 1 2 ϑ ˙ 1 2 + l 2 2 ϑ ˙ 2 2 + 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 c o s ( ϑ 1 − ϑ 2 ) ) . (Addition theorem!)
To get the potential energy, we adopt a plane as a reference height, at the distance l 1 + l 2 l_{1} +l_{2} l 1 + l 2
V = m 1 g [ l 1 + l 2 − l 1 c o s ϑ 1 ] + m 2 g [ l 1 + l 2 − ( l 1 c o s ϑ 1 + l 2 c o s ϑ 2 ) ] V = m_{1}g[l_{1} +l_{2} − l_{1} cos ϑ_{1}] +m_{2}g \left[l_{1} +l_{2} − (l_{1} cos ϑ_{1} +l_{2} cos ϑ_{2})\right] V = m 1 g [ l 1 + l 2 − l 1 c o s ϑ 1 ] + m 2 g [ l 1 + l 2 − ( l 1 c o s ϑ 1 + l 2 c o s ϑ 2 ) ]
The Lagrangian then becomes
L = T − V
= 1 2 m 1 l 1 2 ϑ ˙ 1 2 + 1 2 m 2 [ l 1 2 ϑ ˙ 1 2 + l 2 2 ϑ ˙ 2 2 + 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 c o s ( ϑ 1 − ϑ 2 ) ] = \frac{1}{2} m_{1}l^{2}_{1} \dot{ϑ}^{2}_{1} + \frac{1}{2} m_{2}\left[l^{2}_{1} \dot{ϑ}^{2}_{1} +l^{2}_{2} \dot{ϑ}^{2}_{2}+2l_{1}l_{2} \dot{ϑ}_{1} \dot{ϑ}_{2} cos(ϑ_{1} −ϑ_{2})\right] = 2 1 m 1 l 1 2 ϑ ˙ 1 2 + 2 1 m 2 [ l 1 2 ϑ ˙ 1 2 + l 2 2 ϑ ˙ 2 2 + 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 c o s ( ϑ 1 − ϑ 2 ) ]
− m 1 g [ l 1 + l 2 − l 1 c o s ϑ 1 ] + m 2 g [ l 1 + l 2 − ( l 1 c o s ϑ 1 + l 2 c o s ϑ 2 ) ] -m_{1}g[l_{1} +l_{2} − l_{1} cos ϑ_{1}] +m_{2}g \left[l_{1} +l_{2} − (l_{1} cos ϑ_{1} +l_{2} cos ϑ_{2})\right] − m 1 g [ l 1 + l 2 − l 1 c o s ϑ 1 ] + m 2 g [ l 1 + l 2 − ( l 1 c o s ϑ 1 + l 2 c o s ϑ 2 ) ]
(c) The Lagrange equations with ϑ 1 a n d ϑ 2 ϑ_{1} and ϑ_{2} ϑ 1 a n d ϑ 2
d d t ( ∂ L ∂ ϑ ˙ 1 ) − ∂ L ∂ ϑ 1 = 0 , d d t ( ∂ L ∂ ϑ ˙ 2 ) − ∂ L ∂ ϑ 2 = 0. \frac{d}{dt} \left(\frac{∂L}{∂ \dot{ϑ}_{1}}\right) − \frac{∂L}{∂ϑ_{1}}= 0, \frac{d}{dt} \left(\frac{∂L}{∂ \dot{ϑ}_{2}}\right) − \frac{∂L}{∂ϑ_{2}}= 0. d t d ( ∂ ϑ ˙ 1 ∂ L ) − ∂ ϑ 1 ∂ L = 0 , d t d ( ∂ ϑ ˙ 2 ∂ L ) − ∂ ϑ 2 ∂ L = 0 . One has
∂ L ∂ ϑ 1 = − m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 1 g l 1 s i n ϑ 1 − m 2 g l 1 s i n ϑ 1 , \frac{∂L}{∂ϑ_{1}} =−m_{2}l_{1}l_{2} \dot{ϑ}_{1} \dot{ϑ}_{2} sin(ϑ_{1} −ϑ_{2})−m_{1}gl_{1} sin ϑ_{1} −m_{2}gl_{1} sin ϑ_{1}, ∂ ϑ 1 ∂ L = − m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 1 g l 1 s i n ϑ 1 − m 2 g l 1 s i n ϑ 1 ,
∂ L ∂ ϑ ˙ 1 = m 1 l 1 2 ϑ ˙ 1 + m 2 l 1 2 ϑ ˙ 1 + m 2 l 1 l 2 ϑ ˙ 2 c o s ( ϑ 1 − ϑ 2 ) , \frac{∂L}{∂ \dot{ϑ}_{1}} = m_{1}l^{2}_{1} \dot{ϑ}_{1}+m_{2}l^{2}_{1} \dot{ϑ}_{1} +m_{2}l_{1}l_{2} \dot{ϑ}_{2} cos(ϑ_{1} −ϑ_{2}), ∂ ϑ ˙ 1 ∂ L = m 1 l 1 2 ϑ ˙ 1 + m 2 l 1 2 ϑ ˙ 1 + m 2 l 1 l 2 ϑ ˙ 2 c o s ( ϑ 1 − ϑ 2 ) ,
∂ L ∂ ϑ 2 = m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 2 g l 2 s i n ϑ 2 , \frac{∂L}{∂ϑ_{2}}= m_{2}l_{1}l_{2} \dot{ϑ}_{1} \dot{ϑ}_{2} sin(ϑ_{1} −ϑ_{2})−m_{2}gl_{2} sin ϑ_{2}, ∂ ϑ 2 ∂ L = m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 2 g l 2 s i n ϑ 2 ,
∂ L ∂ ϑ ˙ 2 = m 2 l 2 2 ϑ ˙ 2 + m 2 l 1 l 2 ϑ ˙ 1 c o s ( ϑ 1 − ϑ 2 ) . \frac{∂L}{∂ \dot{ϑ}_{2}}= m_{2}l^{2}_{2} \dot{ϑ}_{2}+m_{2}l_{1}l_{2} \dot{ϑ}_{1} cos(ϑ_{1} −ϑ_{2}). ∂ ϑ ˙ 2 ∂ L = m 2 l 2 2 ϑ ˙ 2 + m 2 l 1 l 2 ϑ ˙ 1 c o s ( ϑ 1 − ϑ 2 ) . Thus, the Lagrange equations read
m 1 l 1 2 ϑ ¨ 1 + m 2 l 1 2 ϑ ¨ 1 + m 2 l 1 l 2 ϑ ¨ 2 c o s ( ϑ 1 − ϑ 2 ) − m 2 l 1 l 2 ϑ ˙ 2 ( ϑ ˙ 1 − ϑ ˙ 2 ) s i n ( ϑ 1 − ϑ 2 ) m_{1}l^{2}_{1} \ddot{ϑ}_{1} +m_{2}l^{2}_{1} \ddot{ϑ}_{1} + m_{2}l_{1}l_{2} \ddot{ϑ}_{2} cos(ϑ_{1} − ϑ_{2})− m_{2}l_{1}l_{2} \dot{ϑ}_{2}(\dot{ϑ}_{1} − \dot{ϑ}_{2}) sin(ϑ_{1} −ϑ_{2}) m 1 l 1 2 ϑ ¨ 1 + m 2 l 1 2 ϑ ¨ 1 + m 2 l 1 l 2 ϑ ¨ 2 c o s ( ϑ 1 − ϑ 2 ) − m 2 l 1 l 2 ϑ ˙ 2 ( ϑ ˙ 1 − ϑ ˙ 2 ) s i n ( ϑ 1 − ϑ 2 )
= − m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 1 g l 1 s i n ϑ 1 − m 2 g l 1 s i n ϑ 1 =−m_{2}l_{1}l_{2} \dot{ϑ}_{1} \dot{ϑ}_{2} sin(ϑ_{1} −ϑ_{2})−m_{1}gl_{1} sin ϑ_{1} −m_{2}gl_{1} sin ϑ_{1} = − m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 1 g l 1 s i n ϑ 1 − m 2 g l 1 s i n ϑ 1 and
m 2 l 2 2 ϑ ¨ 2 + m 2 l 1 l 2 ϑ ¨ 1 c o s ( ϑ 1 − ϑ 2 ) − m 2 l 1 l 2 ϑ ˙ 1 ( ϑ ˙ 1 − ϑ ˙ 2 ) s i n ( ϑ 1 − ϑ 2 ) m_{2}l^{2}_{2} \ddot{ϑ}_{2} +m_{2}l_{1}l_{2} \ddot{ϑ}_{1} cos(ϑ_{1}− ϑ_{2})− m_{2}l_{1}l_{2} \dot{ϑ}_{1}(\dot{ϑ}_{1} − \dot{ϑ}_{2}) sin(ϑ_{1} −ϑ_{2}) m 2 l 2 2 ϑ ¨ 2 + m 2 l 1 l 2 ϑ ¨ 1 c o s ( ϑ 1 − ϑ 2 ) − m 2 l 1 l 2 ϑ ˙ 1 ( ϑ ˙ 1 − ϑ ˙ 2 ) s i n ( ϑ 1 − ϑ 2 )
= m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 2 g l 2 s i n ϑ 2 =m_{2}l_{1}l_{2} \dot{ϑ}_{1} \dot{ϑ}_{2} sin(ϑ_{1} −ϑ_{2})−m_{2}gl_{2} sin ϑ_{2} = m 2 l 1 l 2 ϑ ˙ 1 ϑ ˙ 2 s i n ( ϑ 1 − ϑ 2 ) − m 2 g l 2 s i n ϑ 2 or
( m 1 + m 2 ) l 1 2 ϑ ¨ 1 + m 2 l 1 l 2 ϑ ¨ 2 c o s ( ϑ 1 − ϑ 2 ) + m 2 l 1 l 2 ϑ ˙ 2 2 s i n ( ϑ 1 − ϑ 2 ) (m_{1} + m_{2})l^{2}_{1} \ddot{ϑ}_{1} +m_{2}l_{1}l_{2}\ddot{ϑ}_{2} cos(ϑ_{1} − ϑ_{2})+ m_{2}l_{1}l_{2} \dot{ϑ}^{2}_{2} sin(ϑ_{1} − ϑ_{2}) ( m 1 + m 2 ) l 1 2 ϑ ¨ 1 + m 2 l 1 l 2 ϑ ¨ 2 c o s ( ϑ 1 − ϑ 2 ) + m 2 l 1 l 2 ϑ ˙ 2 2 s i n ( ϑ 1 − ϑ 2 )
= − ( m 1 + m 2 ) g l 1 s i n ϑ 1 =−(m_{1} +m_{2})gl_{1} sin ϑ_{1} = − ( m 1 + m 2 ) g l 1 s i n ϑ 1
and
m 2 l 2 2 ϑ ¨ 2 + m 2 l 1 l 2 ϑ ¨ 1 c o s ( ϑ 1 − ϑ 2 ) − m 2 l 1 l 2 ϑ ˙ 1 2 s i n ( ϑ 1 − ϑ 2 ) m_{2}l^{2}_{2} \ddot{ϑ}_{2} +m_{2}l_{1}l_{2} \ddot{ϑ}_{1} cos(ϑ_{1}− ϑ_{2})− m_{2}l_{1}l_{2} \dot{ϑ}^{2}_{1} sin(ϑ_{1} −ϑ_{2}) m 2 l 2 2 ϑ ¨ 2 + m 2 l 1 l 2 ϑ ¨ 1 c o s ( ϑ 1 − ϑ 2 ) − m 2 l 1 l 2 ϑ ˙ 1 2 s i n ( ϑ 1 − ϑ 2 )
= − m 2 g l 2 s i n ϑ 2 =−m_{2}gl_{2} sin ϑ_{2} = − m 2 g l 2 s i n ϑ 2 These are the desired equations of motion.
(d) For the case
m 1 = m 2 = m a n d l 1 = l 2 = l , m_{1} = m_{2} = m and l_{1} = l_{2} = l, m 1 = m 2 = m a n d l 1 = l 2 = l , (15.30) reduce to
2 l ϑ ¨ 1 + l ϑ ¨ 2 c o s ( ϑ 1 − ϑ 2 ) + l ϑ ˙ 2 2 s i n ( ϑ 1 − ϑ 2 ) = − 2 g s i n ϑ 1 , 2l\ddot{ϑ}_{1} + l\ddot{ϑ}_{2} cos(ϑ_{1} −ϑ_{2})+l\dot{ϑ}^{2}_{2} sin(ϑ_{1} −ϑ_{2})=−2g sin ϑ_{1}, 2 l ϑ ¨ 1 + l ϑ ¨ 2 c o s ( ϑ 1 − ϑ 2 ) + l ϑ ˙ 2 2 s i n ( ϑ 1 − ϑ 2 ) = − 2 g s i n ϑ 1 ,
l ϑ ¨ 1 c o s ( ϑ 1 − ϑ 2 ) + l ϑ ¨ 2 − l ϑ ˙ 1 2 s i n ( ϑ 1 − ϑ 2 ) = − g s i n ϑ 2 . l\ddot{ϑ}_{1} cos(ϑ_{1} −ϑ_{2})+l\ddot{ϑ}_{2} −l\dot{ϑ}^{2}_{1} sin(ϑ_{1} −ϑ_{2})=−g sin ϑ_{2}. l ϑ ¨ 1 c o s ( ϑ 1 − ϑ 2 ) + l ϑ ¨ 2 − l ϑ ˙ 1 2 s i n ( ϑ 1 − ϑ 2 ) = − g s i n ϑ 2 .
(e) If moreover the oscillations are small, then s i n ϑ = ϑ , c o s ϑ = 1 sin ϑ = ϑ, cos ϑ = 1 s i n ϑ = ϑ , c o s ϑ = 1 ϑ ˙ 2 \dot{ϑ}_{2} ϑ ˙ 2
2 l ϑ ¨ 1 + l ϑ ¨ 2 = − 2 g ϑ 1 , l ϑ ¨ 1 + l ϑ ¨ 2 = − g ϑ 2 . 2l\ddot{ϑ}_{1} + l\ddot{ϑ}_{2} =−2gϑ_{1}, l\ddot{ϑ}_{1} +l\ddot{ϑ}_{2} =−gϑ_{2}. 2 l ϑ ¨ 1 + l ϑ ¨ 2 = − 2 g ϑ 1 , l ϑ ¨ 1 + l ϑ ¨ 2 = − g ϑ 2 .
(f) With the ansatz
ϑ 1 = A 1 e i ω t , ϑ 2 = A 2 e i ω t ϑ_{1} = A_{1} e^{iωt}, ϑ_{2} = A_{2}e^{iωt} ϑ 1 = A 1 e i ω t , ϑ 2 = A 2 e i ω t
we then obtain
2 ( g − l ω 2 ) A 1 − l ω 2 A 2 = 0 , − l ω 2 A 1 + ( g − l ω 2 ) A 2 = 0. 2(g −lω^{2})A_{1} −lω^{2}A_{2} = 0, −lω^{2}A_{1} +(g −lω^{2})A_{2} = 0. 2 ( g − l ω 2 ) A 1 − l ω 2 A 2 = 0 , − l ω 2 A 1 + ( g − l ω 2 ) A 2 = 0 .
To ensure that A 1 a n d A 2 A_{1} and A_{2} A 1 a n d A 2
∣ 2 ( g − l ω 2 ) − l ω 2 − l ω 2 g − l ω 2 ∣ = 0 \begin{vmatrix} 2(g − lω^{2}) & −lω^{2} \\ −lω^{2} & g − lω^{2} \end{vmatrix}=0 ∣ ∣ ∣ ∣ ∣ 2 ( g − l ω 2 ) − l ω 2 − l ω 2 g − l ω 2 ∣ ∣ ∣ ∣ ∣ = 0 and therefore,
l 2 ω 4 − 4 l g ω 2 + 2 g 2 = 0 l^{2}ω^{4} −4lgω^{2} + 2g^{2} = 0 l 2 ω 4 − 4 l g ω 2 + 2 g 2 = 0 with the solutions
ω 2 = 4 l g ± 16 l 2 g 2 − 8 l 2 g 2 2 l 2 = ( 2 ± 2 ) g l ; ω^{2} = \frac{4lg ± \sqrt{16l^{2}g^{2} −8l^{2}g^{2}}}{2l^{2}} = (2 ±\sqrt{2})\frac{g}{l} ; ω 2 = 2 l 2 4 l g ± 1 6 l 2 g 2 − 8 l 2 g 2 = ( 2 ± 2 ) l g ; i.e.,
ω 1 2 = ( 2 + 2 ) g l , ω 2 2 = ( 2 − 2 ) g l . ω^{2}_{1} = (2 +\sqrt{2})\frac{g}{l}, ω^{2}_{2} = (2−\sqrt{2})\frac{g}{l} . ω 1 2 = ( 2 + 2 ) l g , ω 2 2 = ( 2 − 2 ) l g .
By inserting (15.34) into (15.33), we obtain
ω 1 2 : A 2 = − 2 A 1 ω^{2}_{1} : A_{2} =−\sqrt{2} A_{1} ω 1 2 : A 2 = − 2 A 1
ω 2 2 : A 2 = 2 A 1 ω^{2}_{2} : A_{2} =\sqrt{2} A_{1} ω 2 2 : A 2 = 2 A 1
