Question 15.15: Four mass points of mass m move on a circle of radius R. Eac...
Four mass points of mass m move on a circle of radius R. Each mass point is coupled to its two neighboring points by a spring with spring constant k (Fig. 15.22). Find the Lagrangian of the system, and derive the equations of motion of the system. Calculate the eigenfrequencies of the system, and discuss the related eigenvibrations.
The kinetic energy of the system is given by
T = \frac{1}{2} m \sum\limits_{ν=1}^{4}{\dot{s}^{2}_{ν}} . (15.44)
For small displacements from the equilibrium position, the potential reads
V = \frac{1}{2} k \sum\limits_{ν=1}^{4}{(s_{ν+1} −s_{ν} )^{2}}, \ \ \ \ \ s_{4+1} = s_{1}. (15.45)
We set s_{ν} = R\varphi_{ν} , and take the angles \varphi_{ν} as generalized coordinates. Then the Lagrangian is
L = T −V = \frac{1}{2} mR^{2} \sum\limits_{ν=1}^{4}{\dot{\varphi}^{2}_{ν}}− \frac{1}{2} kR^{2} \sum\limits_{ν=1}^{4} {(\varphi_{ν+1} −\varphi_{ν} )^{2}}. (15.46)
From the Lagrange equations
\frac{d}{dt} \frac{∂L}{∂ \dot{\varphi}_{ν}} = \frac{∂L}{∂\varphi_{ν}}, (15.47)
we find the equations of motion:
\frac{d}{dt} \frac{∂L}{∂ \dot{\varphi}_{ν}} = mR\ddot{\varphi}_{ν}=−\frac{1}{2} kR^{2}[2(\varphi_{ν} − \varphi_{ν+1})+2(\varphi_{ν} − \varphi_{ν−1})]
= \frac{∂L}{∂\varphi_{ν}}. (15.48)
For the case of four mass points, we then obtain
\ddot{\varphi}_{1} = \frac{k}{m} (\varphi_{2} −2\varphi_{1} +\varphi_{4}),\ddot{\varphi}_{2} = \frac{k}{m} (\varphi_{3} −2\varphi_{2} +\varphi_{1}),
\ddot{\varphi}_{3} = \frac{k}{m} (\varphi_{4} −2\varphi_{3} +\varphi_{2}),
\ddot{\varphi}_{4} = \frac{k}{m} (\varphi_{1} −2\varphi_{4} +\varphi_{3}),
With the ansatz \varphi_{ν} = A_{ν} cos ωt, \ddot{\varphi}_{ν} =−A_{ν}ω^{2} cos ωt, we are led to the following linear system of equations:
\begin{pmatrix} 2\frac{k}{m}− ω^{2} & -\frac{k}{m} & 0 & -\frac{k}{m} \\ -\frac{k}{m} & 2\frac{k}{m}− ω^{2} & -\frac{k}{m} &0 \\ 0 &-\frac{k}{m} &2\frac{k}{m}− ω^{2} & -\frac{k}{m} \\ -\frac{k}{m}& 0 & -\frac{k}{m} & 2\frac{k}{m}− ω^{2}\end{pmatrix} \begin{pmatrix} A_{1}\\ A_{2} \\ A_{3} \\ A_{4} \end{pmatrix}=0 (15.50)
For the nontrivial solutions, the determinant of the coefficient matrix must vanish. This condition leads to the determining equation for the eigenfrequencies:
\left(2 \frac{k}{m} − ω^{2}\right)^{2} \left(4 \frac{k}{m} −ω^{2}\right) (−ω^{2}) = 0. (15.51)
The frequencies are
ω^{2}_{1} = 0, \ \ \ \ \ ω^{2}_{2}= 4 \frac{k}{m}, \ \ \ \ \ ω^{2}_{3}= ω^{2}_{4}= 2 \frac{k}{m}. (15.52)
To calculate the related eigenvibrations, we insert these frequencies into the system of equations (15.50).
(1) ω^{2}_{1}= 0: A_{1} = A_{2} = A_{3} = A_{4}. The system does not vibrate but performs a uniform rotation (Fig. 15.23(a)).
(2) ω^{2}_{2} = 4k/m: A_{1} = A_{3} =−A_{2} =−A_{4}. Two neighboring mass points perform an out-of-phase vibration (Fig. 15.23(b)).
(c) ω^{2}_{3}= ω^{2}_{4}= 2k/m: A_{1} = A_{2} = −A_{3} = −A_{4} or A_{1} = A_{4} = −A_{2} = −A_{3}. Two neighboring mass points vibrate in phase (Fig. 15.24(a,b)).
