A system shown in Fig. 8.8a consists of two blocks of mass m1 and m2 connected by a spring S2 of stiffness k2, while m1 is fastened to a rigid wall by a spring S1 of stiffness k1, The system is displaced arbitrarily along its axis from its natural state and released to perform oscillations on a

Question

A system shown in Fig. 8.8a consists of two blocks of mass [latex]m_1[/latex] and [latex]m_2[/latex] connected by a spring [latex]S_2[/latex] of stiffness [latex]k_2[/latex], while m1 is fastened to a rigid wall by a spring [latex]S_1[/latex] of stiffness [latex]k_1[/latex], The system is displaced arbitrarily along its axis from its natural state and released to perform oscillations on a smooth horizontal surface. (i) Find the total energy of the system. (ii) Derive the differential equations of motion for the system. (iii) With reference to the free body diagram of each particle, derive each of the horizontal forces and their totals from their potential functions.

Accepted Answer

Solution of (i). We model the physical system in Fig. 8.8a as a system of two particles (center of mass objects) with mass [latex]m_1 ext{ and }m_2[/latex]. respectively. Their separate free body diagrams are shown in Fig. 8.8b. To find the total energy of the system [latex]\beta=\left\{m_1, m_2 ight\}[/latex], we first note that the weights [latex]W_1[/latex] and [latex]W_2[/latex], and the normal, smooth surface reaction forces [latex]N_1[/latex] and [latex]N_2[/latex] are external forces that do no work in the motion , and the elastic spring forces are conservative. As usual , the infinitesimal internal, mutual gravitational force between the particles is ignored. The system [latex]\beta=\left\{m_1, m_2 ight\}[/latex], therefore, is conservative.

Let [latex]x_1[/latex] and [latex]x_2[/latex] denote the respective displacements of [latex]m_1[/latex] and [latex]m_2[/latex] from the natural state of the springs [latex]S_1[/latex] and [latex]S_2[/latex] in Fig . 8.8a. There are no constraints relating these variables, so the system has two degrees of freedom. The spring [latex]S_1[/latex] exerts an external force on [latex]m_1[/latex] with external potential energy [latex]\phi_1[/latex]; but no relevant external forces act on [latex]m_2, \mathrm{so} \phi_2=0[/latex]. The total external potential energy (8.72) is thus given by

[latex]\Phi(\beta) \equiv \sum_{k=1}^n \phi_k\left(\mathbf{x}_k ight)[/latex] (8.72)

[latex]\Phi=\phi_1=\frac{1}{2} k_1 x_1^2 .[/latex] (8.95a)

The mutual internal force on [latex]m_1[/latex] and [latex]m_2[/latex] is due to the spring [latex]S_2[/latex]. These forces, shown in Fig. 8.8b, are equal but oppositely directed, so the total internal force is zero; but the total internal potential energy is not. The internal potential energy arising from the elastic force exerted on [latex]m_1[/latex] by [latex]m_2[/latex] is [latex]\beta_{12}=\frac{1}{2} k_2\left(x_2 - x_1 ight)^2[/latex], and the internal potential energy arising from the equal but oppositely directed elastic force exerted on [latex]m_2[/latex] by [latex]m_1[/latex] is [latex]\beta_{21}=\frac{1}{2} k_2\left(x_2 - x_1 ight)^2[/latex]. Clearly, the mutual internal potential energy [latex]\beta_{12}=\beta_{21}[/latex] is a symmetric function of the change in distance, hence , also the current distance between the particles, as indicated in (8.75) and (8.76). The total internal potential energy (8.79) is thus determined by [latex]B=\frac{1}{2}\left(\beta_{12} + \beta_{21} ight)[/latex], that is,

[latex]\beta_{j k}=\psi\left(\left|\mathbf{r}_{j k} ight| ight)[/latex] (8.75)

[latex]\beta_{j k}=\beta_{k j}[/latex] (8.76)

[latex]B=\beta_{12}=\frac{1}{2} k_2\left(x_2 - x_1 ight)^2[/latex]. (8.95b)

Consequently, the total internal potential energy of the system in Fig. 8.8a is simply the elastic potential energy due to the spring [latex]S_2[/latex]. Therefore, with (8.95a) and (8.95b), the total potential energy (8.84) of the system is

[latex]V(\beta) \equiv \Phi(\beta) + B(\beta)[/latex] (8.84)

[latex]V(\beta)=\Phi + B=\frac{1}{2} k_1 x_1^2 + \frac{1}{2} k_2\left(x_2 - x_1 ight)^2[/latex] (8.95c)

It is evident that the total potential energy function [latex]V(\beta)[/latex] may be written down immediately by inspection of the system in Fig. 8.8a. In fact, this is the usual procedure to follow.

The total kinetic energy (8.50) of the system in Fig. 8.8a is

[latex]K(\beta, t) \equiv \sum_{k=1}^n K_k(t)=\sum_{k=1}^n \frac{1}{2} m_k \mathbf{v}_k \cdot \mathbf{v}_k,[/latex] (8.50)

[latex]K(\beta, t)=\frac{1}{2} m_1 \dot{x}_1^2 + \frac{1}{2} m_2 \dot{x}_2^2 .[/latex] (8.95d)

The total energy of the conservative system [latex]\beta=\left\{m_1, m_2 ight\}[/latex] now follows from (8.86):

[latex]K + V=E ext {, a constant. }[/latex] (8.86)

[latex]K + V=\frac{1}{2} m_1 \dot{x}_1^2 + \frac{1}{2} m_2 \dot{x}_2^2 + \frac{1}{2} k_1 x_1^2 + \frac{1}{2} k_2\left(x_2 - x_1 ight)^2=E, ext { a constant. }[/latex] (8.95e)

Solution of (ii). Application of the law of motion to each particle shown separately in the free body diagrams of Fig. 8.8b yields the following equations of motion for the system :

[latex]m_1 \ddot{x}_1=-k_1 x_1 + k_2\left(x_2 - x_1 ight), \quad m_2 \ddot{x}_2=-k_2\left(x_2 - x_1 ight)[/latex] (8.95f)

Notice that the sum of these equations is [latex]m_1 \ddot{x}_1 + m_2 \ddot{x}_2=-k_1 x_1[/latex], or [latex]m(\beta) \ddot{x}^*= F(\beta),[/latex] which is the equation of motion (8.5) of the system with total external force [latex]F(\beta)=-k_1 x_1[/latex]. This system equation is not useful. The motions [latex]x_1[/latex] and [latex]x_2[/latex] of the particles are coupled, but independent-the motion of one particle influences but does not determine the motion of the other. As a consequence, these equations cannot be separately integrated. The solution of coupled equations of this kind is considered in Chapter 11.

Solution of (iii). We next determine the horizontal forces and their totals from their potential functions. The total external force acting on each particle is determined by use of (8.95a) in (8.70):

[latex]\mathbf{f}_k=-\boldsymbol{ abla}_k \phi_k\left(\mathbf{x}_k ight), \quad k=1,2, \ldots, n[/latex] (8.70)

[latex]\mathbf{f}_1=- abla_1 \phi_1=-\frac{\partial \phi_1}{\partial x_1} \mathbf{i}=-k_1 x_1 \mathbf{i}, \quad \mathbf{f}_2=- abla_2 \phi_2=-\frac{\partial \phi_2}{\partial x_2} \mathbf{i}=\mathbf{0}[/latex] (8.95g)

Hence, the total external force on the system, by (8.4), is [latex]\mathbf{F}(\beta, t)=\mathbf{f}_1 + \mathbf{f}_2=-k_1 x_1 \mathbf{i}[/latex] , noted earlier. All other forces that act must be the equal, oppositely directed external and internal contributions. Specifically, the equal and oppositely directed external forces are [latex]N_1 = -W_1, N_2 = -W_2[/latex]; these have no influence on the motion. The total internal force acting on each particle is obtained by use of (8.95b) in (8.80):

[latex]\mathbf{b}_j=\sum_{\substack{k=1 \ k eq j}}^n \mathbf{b}_{j k}=-\sum_{\substack{k=1 \ k eq j}}^n abla_j \beta_{j k}\left(\left|\mathbf{r}_{j k} ight| ight)[/latex] (8.80)

[latex]\begin{aligned}&\mathbf{b}_1=- abla_1 \beta_{12}=-\frac{\partial \beta_{12}}{\partial x_1} \mathbf{i}=k_2\left(x_2 - x_1 ight) \mathbf{i}, \&\mathbf{b}_2=- abla_2 \beta_{21}=-\frac{\partial \beta_{21}}{\partial x_2} \mathbf{i}=-k_2\left(x_2 - x_1 ight) \mathbf{i} .\end{aligned}[/latex] (8.95h)

Of course, the total internal force acting on the system is [latex]b_1 + b_2 = 0[/latex]. From (8.95g) and (8.95h), the total force [latex]F_k = f_k + b_k[/latex] acting on each particle separately is

[latex]\mathbf{F}_1=\left[-k_1 x_1 + k_2\left(x_2 - x_1 ight) ight] \mathbf{i}, \quad \mathbf{F}_2=-k_2\left(x_2 - x_1 ight) \mathbf{i}[/latex] (8.95i)

These are the totals of the forces on the right-hand side of the equations in (8.95f) and shown in the free body diagrams of Fig. 8.8b.

Finally, we apply equations (8.87), (8.89) , and (8.90). Using (8.95c) in the first relation of (8.87), we have

[latex]\mathbf{F}_k=- abla_k V\left(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_n ight)=- abla_k \Phi - abla_k B[/latex] (8.87)

[latex] abla_k B=\sum_{\substack{j=1 \ k eq j}}^n abla_k \beta_{k j}=-\mathbf{b}_k[/latex] (8.89)

[latex]\mathbf{F}(\beta, t)=\sum_{k=1}^n \mathbf{F}_k=-\sum_{k=1}^n abla_k V(\beta)=-\sum_{k=1}^n abla_k \Phi=\sum_{k=1}^n \mathbf{f}_k[/latex] (8.90)

[latex]\mathbf{F}_1=-\boldsymbol{ abla}_1 V=\left[-k_1 x_1 + k_2\left(x_2 - x_1 ight) ight] \mathbf{i}, \quad \mathbf{F}_2=-\boldsymbol{ abla}_2 V=-k_2\left(x_2 - x_1 ight) \mathbf{i}[/latex], (8.95j)

in agreement with (8.95i). It is seen immediately that use of these relations in
(8.91) returns the separate equations on motion (8.95f) . With (8.95b) in (8.89), we find

[latex]m_k \ddot{\mathbf{x}}_k=- abla_k V(\beta), \quad k=1,2, \ldots, n .[/latex] (8.91)

[latex]\mathbf{b}_1=- abla_1 B=k_2\left(x_2 - x_1 ight) \mathbf{i}, \quad \mathbf{b}_2=- abla_2 B=-k_2\left(x_2 - x_1 ight) \mathbf{i}[/latex] (8.95k)

in accord with (8.95h). And, finally, substitution of (8.95j) into (8.90) yields

[latex]\mathbf{F}(\beta, t)=- abla_1 V - abla_2 V=-k_1 x_1 \mathbf{i}[/latex] (8.95l)

the total external force on the system .

Question 8.7: A system shown in Fig. 8.8a consists of two blocks of mass m...

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