(i) Derive the equations for the uniaxial motion of the spring-mass system shown in Fig. 11.1. The supporting surface is smooth and all springs are linearly elastic and unstretched initially. (ii) Determine the motion of the system for the special symmetric case when m1 = m2 = m and k1 = k2 = k.

Question

(i) Derive the equations for the uniaxial motion of the spring-mass system shown in Fig. 11.1. The supporting surface is smooth and all springs are linearly elastic and unstretched initially. (ii) Determine the motion of the system for the special symmetric case when [latex]m_1=m_2=m ext { and } k_1=k_2=k[/latex].

Accepted Answer

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

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

Notice that this has the form (11.24) in which [latex]\left[M_{j k} ight]=\operatorname{diag}\left[m_1, m_2 ight][/latex] 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

[latex]T=\frac{1}{2} M_{j k}\left(q_r ight) \dot{q}_j \dot{q}_k[/latex] (11.24)

[latex]V=\frac{1}{2} k_1 x_1^2 + \frac{1}{2} k_2\left(x_2 - x_1 ight)^2 + \frac{1}{2} k_1 x_2^2[/latex] (11.40b)

Therefore, the Lagrangian (11.34) is given by

[latex]L\left(\dot{q}_r, q_r, t ight) \equiv T\left(\dot{q}_r, q_r, t ight) - V\left(q_r ight)[/latex] (11.34)

[latex]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 ight) - \frac{1}{2} k_2\left(x_2 - x_1 ight)^2,[/latex] (11.40c)

and hence

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}_1} ight)=m_1 \ddot{x}_1, \quad \frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}_2} ight)=m_2 \ddot{x}_2[/latex] (11.40d)

[latex]\frac{\partial L}{\partial x_1}=-k_1 x_1 + k_2\left(x_2 - x_1 ight), \quad \frac{\partial L}{\partial x_2}=-k_1 x_2 - k_2\left(x_2 - x_1 ight) .[/latex] (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:

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_k} ight)-\frac{\partial L}{\partial q_k}=0[/latex] (11.35)

[latex]m_1 \ddot{x}_1 + \left(k_1 + k_2 ight) x_1 - k_2 x_2=0, \quad m_2 \ddot{x}_2 + \left(k_1 + k_2 ight) x_2 - k_2 x_1=0.[/latex] (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 [latex]m_1=m_2=m[/latex] and [latex]k_1=k_2=k[/latex]. Then with [latex]p^2=k / m[/latex], the equations in (11.40f) simplify to

[latex]\ddot{x}_1 + p^2\left(2 x_1 - x_2 ight)=0, \quad \ddot{x}_2 + p^2\left(2 x_2 - x_1 ight)=0[/latex]. (11.40g)

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

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

in which [latex]C_1[/latex] and [latex]C_2[/latex] 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 [latex]C_1 / C_2[/latex]:

[latex]\begin{aligned}\left(2 p^2 - \alpha^2 ight) C_1 - p^2 C_2 &=0, \-p^2 C_1 + \left(2 p^2 - \alpha^2 ight) C_2 &=0\end{aligned},[/latex] (11.40i)

For nontrivial amplitudes [latex]C_k[/latex], we must have

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

This leads to the quadratic equation [latex]\left(2 p^2 - \alpha^2 ight)^2 - p^4=0[/latex] with the following two solutions for the circular frequency in (11.40h):

[latex]\alpha_1=p, \quad \alpha_2=\sqrt{3} p[/latex]. (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 [latex]C_{jk}[/latex] the different amplitudes associated with each generalized coordinate [latex]x_j[/latex], frequency [latex]\alpha_k[/latex] and phase angle [latex]\phi_k[/latex]. In particular, [latex]x_1^T=C_{11} \sin \left(\alpha_1 t + \phi_1 ight)[/latex] and [latex]x_1^T=C_{12} \sin \left(\alpha_2 t + \phi_2 ight)[/latex], and hence the general solution for [latex]x_1[/latex] is provided by their sum. Similarly for [latex]x_2[/latex]. Hence, with the aid of (11.40h) and (11.40k), the general solution of the linear system (11.40g) is given by

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

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

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

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

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

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

[latex]x_1=x_2=C_{11} \sin \left(p t + \phi_1 ight)[/latex]. (11.40o)

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

[latex]-x_1=x_2=C_{22} \sin \left(\sqrt{3} p t + \phi_2 ight)[/latex]. (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 , [latex]\sqrt{3} p[/latex]. In the case (11.40o), the masses move in the same direction with the same circular frequency p , initial phase [latex]\phi_1[/latex] and amplitude [latex]C_{11}[/latex]. In the case (11.40p), the masses move symmetrically, in opposite directions with the same circular frequency [latex]\sqrt{3} p[/latex], initial phase [latex]\phi_2,[/latex] and amplitude [latex]C_{22}[/latex].

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

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

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

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

The two natural frequencies of the system are the characteristic frequencies given by (11.40k). By introduction of new coordinates [latex]\xi_k[/latex], 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 [latex]\xi_k[/latex] which may be defined in terms of the physical coordinates [latex]x_k[/latex] by

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

has only one frequency . These are the natural modes of vibration of the system. The coordinates [latex]\xi_k[/latex] 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 [latex]x_1=\xi_1 + \xi_2 ext { and }x_2=\xi_1 - \xi_2[/latex]. These relations uncouple the original equations of motion. Let the reader use these results to show that (11.40g) may be written as

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

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

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

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