Question 11.6: (i) Derive the equations for the uniaxial motion of the spri...

Solution of (i). The holonomic constraints for the uniaxial motion are  evident. Let each mass  m_1  and  m_2  be displaced uniaxially an amount  x_1  and  x_2,  respectively. The system has two degrees of freedom with independent generalized coordinates  \left(q_1, q_2\right)=\left(x_1, x_2\right).  The total kinetic energy for the system of particles is

T=\frac{1}{2} m_1 \dot{x}_1^2  +  \frac{1}{2} m_2 \dot{x}_2^2 .                     (11.40a)

Notice that this has the form (11.24) in which  \left[M_{j k}\right]=\operatorname{diag}\left[m_1, m_2\right]  is a diagonal matrix . The spring forces are conservative, and all other forces are workless. In consequence, the system is conservative with total potential energy

T=\frac{1}{2} M_{j k}\left(q_r\right) \dot{q}_j \dot{q}_k                         (11.24)

V=\frac{1}{2} k_1 x_1^2  +  \frac{1}{2} k_2\left(x_2  –  x_1\right)^2  +  \frac{1}{2} k_1 x_2^2                  (11.40b)

Therefore, the Lagrangian (11.34) is given by

L\left(\dot{q}_r, q_r, t\right) \equiv T\left(\dot{q}_r, q_r, t\right)  –  V\left(q_r\right)                      (11.34)

L=\frac{1}{2} m_1 \dot{x}_1^2  +  \frac{1}{2} m_2 \dot{x}_2^2  –  \frac{1}{2} k_1\left(x_1^2  +  x_2^2\right)  –  \frac{1}{2} k_2\left(x_2  –  x_1\right)^2,                    (11.40c)

and hence

\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}_1}\right)=m_1 \ddot{x}_1, \quad \frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}_2}\right)=m_2 \ddot{x}_2                      (11.40d)

\frac{\partial L}{\partial x_1}=-k_1 x_1  +  k_2\left(x_2  –  x_1\right), \quad \frac{\partial L}{\partial x_2}=-k_1 x_2  –  k_2\left(x_2  –  x_1\right) .                                 (11.40e)

Use of (11.40d) and (11.40e) in (11.35) for k=1,2 delivers two coupled, ordinary, linear differential equations of motion for the conservative system:

\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k}\right)-\frac{\partial L}{\partial q_k}=0                             (11.35)

m_1 \ddot{x}_1  +  \left(k_1  +  k_2\right) x_1  –  k_2 x_2=0, \quad m_2 \ddot{x}_2  +  \left(k_1  +  k_2\right) x_2  –  k_2 x_1=0.                    (11.40f)

Solution of (ii). The general solution of this coupled system of  equations may be readily obtained; however, to simplify the analysis and preserve the essential methodology, the motion will be determined for the special symmetrical case when  m_1=m_2=m  and  k_1=k_2=k.  Then with  p^2=k / m,  the equations in (11.40f) simplify to

\ddot{x}_1  +  p^2\left(2 x_1  –  x_2\right)=0, \quad \ddot{x}_2  +  p^2\left(2 x_2  –  x_1\right)=0.                   (11.40g)

The typical procedure for solving this class of problems is to  assume a trial solution for  x_1  and  x_2  having the same circular  frequency  \alpha  and initial phase  \phi. We thus consider a trial solution of the form

x_1^T=C_1 \sin (\alpha t  +  \phi), \quad x_2^T=C_2 \sin (\alpha t  +  \phi).                      (11.40h)

in which  C_1  and  C_2  are constant amplitudes. Substitution of (11.40h) into (11.40g) yields a system of two homogeneous algebraic equations that determine ex and the amplitude ratio  C_1 / C_2:

\begin{aligned}\left(2 p^2  –  \alpha^2\right) C_1  –  p^2 C_2 &=0, \\-p^2 C_1  +  \left(2 p^2  –  \alpha^2\right) C_2 &=0\end{aligned},                             (11.40i)

For nontrivial amplitudes  C_k,  we must have

\operatorname{det}\left[\begin{array}{cc}2 p^2  –  \alpha^2 & -p^2 \\-p^2 & 2 p^2  –  \alpha^2\end{array}\right]=0.                            (11.40j)

This leads to the quadratic equation   \left(2 p^2  –  \alpha^2\right)^2  –  p^4=0  with the following two solutions for the circular frequency in (11.40h):

\alpha_1=p, \quad \alpha_2=\sqrt{3} p.                    (11.40k)

Equation (11.40j) is called the characteristic equation for the system, and  its positive roots (11.40k) are called characteristic frequencies. The latter also are known as eigenfrequencies, normal mode, or natural frequencies.

For each of these frequencies, the system (11.40i) determines a corresponding amplitude ratio and separate trial solutions of the type (11.40h). It is useful, therefore, to denote by  C_{jk}  the different amplitudes associated with each generalized coordinate  x_j,  frequency   \alpha_k  and phase angle \phi_k.  In  particular,  x_1^T=C_{11} \sin \left(\alpha_1 t  +  \phi_1\right)  and  x_1^T=C_{12} \sin \left(\alpha_2 t  + \phi_2\right),  and hence the general solution for  x_1  is provided by their sum. Similarly for  x_2.  Hence, with the aid of (11.40h) and (11.40k), the general solution of the linear system (11.40g) is given by

\begin{aligned}&x_1=C_{11} \sin \left(p t  +  \phi_1\right)  +  C_{12} \sin \left(\sqrt{3} p t  +  \phi_2\right), \\&x_2=C_{21} \sin \left(p t  +  \phi_1\right)  +  C_{22} \sin \left(\sqrt{3} p t  +  \phi_2\right) .\end{aligned}                     (11.40l)

When  \alpha_k  is used in (11.40i), the former amplitudes  C_1,  C_2  are replaced  by the respective amplitudes   C_{1 k}, C_{2 k}  Successive use of (11.40k) in (11.40i) yields the following amplitude ratios corresponding to  \alpha_1  and  \alpha_2:

\frac{C_{11}}{C_{21}}=1, \quad \frac{C_{22}}{C_{12}}=-1 .                  (11.40m)

In consequence, the general solution (11.40l) may be written as

\begin{aligned}&x_1=C_{11} \sin \left(p t+\phi_1\right)-C_{22} \sin \left(\sqrt{3} p t  +  \phi_2\right), \\&x_2=C_{11} \sin \left(p t  +  \phi_1\right)  +  C_{22} \sin \left(\sqrt{3} p t  +  \phi_2\right).\end{aligned}                   (11.40n)

The four constants  C_{11}, C_{22}, \phi_1,  and  \phi_2  are determined by assigned  initial data. Suppose we specify the initial data so that  C_{22}=0,  then the solution (11.40n) has the form

x_1=x_2=C_{11} \sin \left(p t  +  \phi_1\right).                            (11.40o)

On the other hand, if we specify initial data so that  C_{11}=0,  we obtain

-x_1=x_2=C_{22} \sin \left(\sqrt{3} p t  +  \phi_2\right).                     (11.40p)

Each of these motions is described by a single amplitude, frequency, and  phase. In general, a motion of a multidegree of freedom vibrating system that can be described by a single frequency is called a mode. The solutions (11.40o) and (11.40p) correspond to modes having the distinct natural frequencies p and ,  \sqrt{3} p.  In the case (11.40o), the masses move in the same direction with the same circular frequency p , initial phase  \phi_1  and amplitude  C_{11}.  In the case (11.40p), the masses move symmetrically, in opposite directions with the same circular frequency  \sqrt{3} p,  initial phase  \phi_2,  and amplitude  C_{22}.

Indeed, when the masses are equally displaced in the same direction an  amount B and released from rest, the initial data  x_1(0)=x_2(0)=B,  \dot{x}_1(0)=\dot{x}_2(0)=0  applied to the system (11.40n)  provides four algebraic equations that are satisfied with   C_{22}=0, \phi_1=\pi / 2,  and   B=C_{11}.  Thus, (11.40o) has the explicit form

x_1=x_2=C_{11} \cos p t .                          (11.40q)

Similarly, when initially the masses are equally but oppositely displaced an  amount D and released from rest so that   -x_1(0)=x_2(0)=D  and   \dot{x}_1(0)=\dot{x}_2(0)=0,  we find   C_{11}=0, \phi_2=\pi / 2,  and  D=C_{22}.  So, the explicit solution is

-x_1=x_2=C_{22} \cos \sqrt{3} p t.                    (11.40r)

The two natural frequencies of the system are the characteristic frequencies given by (11.40k). By introduction of new coordinates   \xi_k,  the most general motion of the system (11.40n)for arbitrary initial data may be viewed as the superposition of normal mode motions corresponding to these natural frequencies. Indeed, we see from (11.40n)that each of the coordinates  \xi_k  which may be defined in terms of the physical coordinates   x_k  by

\begin{aligned}&\xi_1 \equiv \frac{1}{2}\left(x_1  +  x_2\right)=C_{11} \sin \left(p t  +  \phi_1\right), \\&\xi_2 \equiv \frac{1}{2}\left(x_1  –  x_2\right)=-C_{22} \sin \left(\sqrt{3} p t  +  \phi_2\right)\end{aligned},                     (11.40s)

has only one frequency . These are the natural modes of vibration of the system. The coordinates  \xi_k  are called normal coordinates and their corresponding modes given on the right-hand side of equations (11.40s) are known as the normal modes of vibration. It is now evident that the general motion (11.40n) is a superposition of normal mode motions described by   x_1=\xi_1  +  \xi_2 \text { and }x_2=\xi_1  –  \xi_2.  These relations uncouple the original equations of motion. Let the reader use these results to show that (11.40g) may be written as

\ddot{\xi}_1  +  p^2 \xi_1=0, \quad \ddot{\xi}_2  +  3 p^2 \xi_2=0.                      (11.40t)

These are the normal equations of motion for which the normal mode  frequencies, now evident, are given in (11.40k).