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.

15.22
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The kinetic energy of the system is given by

T=12mν=14s˙ν2.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=12kν=14(sν+1sν)2,     s4+1=s1.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 sν=Rφνs_{ν} = R\varphi_{ν} , and take the angles φν\varphi_{ν} as generalized coordinates. Then the Lagrangian is

L=TV=12mR2ν=14φ˙ν212kR2ν=14(φν+1φν)2.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

ddtLφ˙ν=Lφν,\frac{d}{dt} \frac{∂L}{∂ \dot{\varphi}_{ν}} = \frac{∂L}{∂\varphi_{ν}},                               (15.47)

we find the equations of motion:

ddtLφ˙ν=mRφ¨ν\frac{d}{dt} \frac{∂L}{∂ \dot{\varphi}_{ν}} = mR\ddot{\varphi}_{ν}

 

=12kR2[2(φνφν+1)+2(φνφν1)]=−\frac{1}{2} kR^{2}[2(\varphi_{ν} − \varphi_{ν+1})+2(\varphi_{ν} − \varphi_{ν−1})]

 

=Lφν.= \frac{∂L}{∂\varphi_{ν}}.                                   (15.48)

For the case of four mass points, we then obtain

φ¨1=km(φ22φ1+φ4),\ddot{\varphi}_{1} = \frac{k}{m} (\varphi_{2} −2\varphi_{1} +\varphi_{4}),

 

φ¨2=km(φ32φ2+φ1),\ddot{\varphi}_{2} = \frac{k}{m} (\varphi_{3} −2\varphi_{2} +\varphi_{1}),

 

φ¨3=km(φ42φ3+φ2),\ddot{\varphi}_{3} = \frac{k}{m} (\varphi_{4} −2\varphi_{3} +\varphi_{2}),

 

φ¨4=km(φ12φ4+φ3),\ddot{\varphi}_{4} = \frac{k}{m} (\varphi_{1} −2\varphi_{4} +\varphi_{3}),

With the ansatz φν=Aνcosωt,φ¨ν=Aνω2cosωt\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:

(2kmω2km0kmkm2kmω2km00km2kmω2kmkm0km2kmω2)(A1A2A3A4)=0\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:

(2kmω2)2(4kmω2)(ω2)=0.\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

ω12=0,     ω22=4km,     ω32=ω42=2km.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) ω12=0:A1=A2=A3=A4ω^{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) ω22=4k/m:A1=A3=A2=A4ω^{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) ω32=ω42=2k/m:A1=A2=A3=A4  or     A1=A4=A2=A3ω^{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)).

 

15.23
15.24

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