Posted on Categories:Kinesiology, 物理代写, 运动学

# 物理代写|声学代写Acoustics代考|ME474 Equilibrium, Stability, and Hooke’s Law

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 物理代写|声学代写Acoustics代考|Equilibrium, Stability, and Hooke’s Law

Pause for a moment as you are reading and look around. Close your eyes, count to ten, and then open your eyes. Probably not much has changed. This is because most of the material we observe visually is in a state of stable equilibrium. ${ }^4$ For an object to be in equilibrium, the vector sum of all the forces acting on that body (i.e., the net force) must be zero, and the first derivative of the object’s potential energy with respect to its position must be zero. For that equilibrium state to be stable, the second derivative of the object’s potential energy must be positive.

Figure $1.2$ illustrates three possible equilibrium conditions based on the rate of change of a body’s potential energy with position. For this illustration, let us assume that the solid curve represents the height, $z$, above some reference height, $z_0$, in a uniform gravitational field. The (gravitational) potential energy, $P E(z)$, of each of the three balls shown in Fig. $1.2$ is therefore proportional to their height, $z$, where the mass of each ball is $m$ and $g$ is the acceleration due to gravity that is assumed to be independent of $z: P E(z)=m g\left(z-z_o\right)$.

The three balls in Fig. $1.2$ each respond to a small displacement from their equilibrium positions differently. We can think of the solid curve as representing a flat surface on the left, two peaks at the center and the right, and a valley between two peaks. All three balls are in a state of mechanical equilibrium because the vector sum of the force of gravity (down) and of the surface (up) is zero. In all three cases, the first derivative of the potential energy vanishes at all three locations: $d(P E) / d x=0$. The ball on the left is in a state of neutral equilibrium because it can be moved to the left or to the right by a small distance and it will still be at equilibrium, even at its displaced position. The curve at that location is flat and horizontal.

## 物理代写|声学代写Acoustics代考|Potentials and Forces

The relationship between forces and potential energy can be extended beyond our simple stability example. In general, the net (vector) force, $\vec{F}{\text {net, }}$, is the negative of the gradient of the (scalar) potential energy: $\vec{\nabla}(P E)=-\vec{F}{\text {net. }}{ }^6$ This is consistent with the definitions of work and energy.
$$W_{1,2} \equiv \int_1^2 \vec{F} \cdot d \vec{x}=(P E)1-(P E)_2=-\Delta(P E)$$ The right-hand side assumes that $\vec{F}$ is a conservative force ${ }^7$ and the work, $W{1,2}$, done in moving an object from position 1 to position 2, over some distance along the direction of the force (indicated by the “dot product” under the integral), leads to a change in potential energy, $-\Delta(P E)$. Again, application of the Fundamental Theorem of Calculus leads to the desired relationship between the gradients of the potential energy and the net force: $\vec{\nabla}(P E)=-\vec{F}_{\text {net }}$.

If we limit ourselves to the current example of a ball in the valley, we can expand the potential energy about the stable equilibrium position that is identified as $x_o$.

$$P E\left(x_o+d x\right)=P E\left(x_o\right)+\left.\frac{d^2(P E)}{d x^2}\right|{x_o} \frac{(d x)^2}{2}+\left.\frac{d^3(P E)}{d x^3}\right|{x_o} \frac{(d x)^3}{6}+\ldots$$
Note that the first derivative of $P E$ is missing from Eq. (1.23) because it is zero if $x_o$ is the equilibrium position. ${ }^8$ The term proportional to $(d x)^3$ corresponds to the contribution to the difference between the actual curve and the dashed parabola in Fig. 1.2. If the deviation between the two curves is symmetric, then the leading correction term would be proportional to $(d x)^4$. If the deviation is not symmetric, the leading correction term will be proportional to $(d x)^3$.

In our one-dimensional example, the gradient of the potential energy will simply be the derivative of Eq. (1.23) with respect to $x$.
$$F_{n e t}(d x)=-\frac{d(P E)}{d x}=-\left.\frac{d^2(P E)}{d x^2}\right|{x_o}(d x)-\left.\frac{d^3(P E)}{d x^3}\right|{x_o} \frac{(d x)^2}{2}-\cdots$$
For sufficiently small displacements from equilibrium, the parabolic approximation to the potential energy curve (i.e., the dashed parabola in Fig. 1.2) provides an adequate representation, and therefore the series of Eq. (1.24) that describes the force can be truncated after the first term.
$$F_{n e t}(d x) \cong-\left.\frac{d^2(P E)}{d x^2}\right|_{x_o}(d x) \equiv-\mathrm{K}(d x)$$

## 物理代写|声学代写声学代考|电位和力

$$W_{1,2} \equiv \int_1^2 \vec{F} \cdot d \vec{x}=(P E)1-(P E)2=-\Delta(P E)$$右边假设$\vec{F}$是一个保守力${ }^7$，将一个物体从位置1移动到位置2所做的功$W{1,2}$，沿力的方向移动了一段距离(用积分下面的“点积”表示)，导致势能$-\Delta(P E)$的变化。同样，应用微积分基本定理得到了势能梯度和合力之间的期望关系:$\vec{\nabla}(P E)=-\vec{F}{\text {net }}$ .

$$P E\left(x_o+d x\right)=P E\left(x_o\right)+\left.\frac{d^2(P E)}{d x^2}\right|{x_o} \frac{(d x)^2}{2}+\left.\frac{d^3(P E)}{d x^3}\right|{x_o} \frac{(d x)^3}{6}+\ldots$$注意。的一阶导数 $P E$ 在式(1.23)中缺失，因为如果 $x_o$ 是平衡位置。 ${ }^8$ 这一项与 $(d x)^3$ 对应于图1.2中实际曲线与虚线抛物线之差的贡献。如果两条曲线之间的偏差是对称的，那么前导修正项将成比例 $(d x)^4$。如果偏差不对称，前导修正项将成正比 $(d x)^3$.

$$F_{n e t}(d x)=-\frac{d(P E)}{d x}=-\left.\frac{d^2(P E)}{d x^2}\right|{x_o}(d x)-\left.\frac{d^3(P E)}{d x^3}\right|{x_o} \frac{(d x)^2}{2}-\cdots$$

$$F_{n e t}(d x) \cong-\left.\frac{d^2(P E)}{d x^2}\right|_{x_o}(d x) \equiv-\mathrm{K}(d x)$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。