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# 数学代写|应用数学代考APPLIED MATHEMATICS代写|EPMATH110 Case study

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## 数学代写|应用数学代考APPLIED MATHEMATICS代写|Case study

Setup. To illustrate the preceding results on dimensional methods, and the process of modelling a simple mechanical system, we study the motion of a pendulum released from rest. Figure $1.1$ illustrates the system, which consists of a string of length $\ell$, with one end attached to a fixed support point, and the other end attached to a ball of mass $m$. We assume the string is always in tension and hence straight, and we let $\theta$ denote the angle between the string and a vertical line through the support point, and arbitrarily take the positive direction to be counter-clockwise. We assume that gravitational acceleration $g$ is directed in the downward, vertical direction. When the ball is raised and released from the rest conditions $\theta=\theta_0$ and $\frac{d \theta}{d t}=0$ at time $t=0$, the ball will swing back-and-forth in a periodic motion. We seek to understand various aspects of this motion; for example, how the period depends on the parameters $m, g, \ell$, and $\theta_0$.
Outline of model. We assume that the motion occurs in a plane and introduce an origin and $x, y$ coordinates as shown. The standard unit vectors in the positive $x$ and $y$ directions are denoted by $\vec{i}$ and $\vec{j}$, and the position vector for the ball is denoted by $\vec{r}$. It will be convenient to introduce unit vectors $\vec{e}r$ and $\vec{e}\theta$ that are parallel and perpendicular to $\vec{r}$. For any angle $\theta$, the components of these vectors are $\vec{r}=\ell \sin \theta \vec{i}+$ $\ell \cos \theta \vec{j}, \vec{e}r=\sin \theta \vec{i}+\cos \theta \vec{j}$, and $\vec{e}\theta=\cos \theta \vec{i}-\sin \theta \vec{j}$. By differentiating the position with respect to time, we obtain the velocity and acceleration vectors
$$\begin{gathered} \frac{d \vec{r}}{d t}=\ell \cos \theta \frac{d \theta}{d t} \vec{i}-\ell \sin \theta \frac{d \theta}{d t} \vec{j} \ \frac{d^2 \vec{r}}{d t^2}=\left[\ell \cos \theta \frac{d^2 \theta}{d t^2}-\ell \sin \theta\left(\frac{d \theta}{d t}\right)^2\right] \vec{i}-\left[\ell \sin \theta \frac{d^2 \theta}{d t^2}+\ell \cos \theta\left(\frac{d \theta}{d t}\right)^2\right] \vec{j} \end{gathered}$$

## 数学代写|应用数学代考APPLIED MATHEMATICS代写|Reduced equation for period

Reduced equation for period. The quantities $P, g, \ell, \theta_0$ have dimensions $[P]=$ $T,[g]=L / T^2,[\ell]=L$ and $\left[\theta_0\right]=1$. A dimensional basis is ${T, L}$, and the dimensional exponent matrix in this basis is
$$A=\left(\Delta_P, \Delta_g, \Delta_{\ell}, \Delta_{\theta_0}\right)=\left(\begin{array}{rrrr} 1 & -2 & 0 & 0 \ 0 & 1 & 1 & 0 \end{array}\right) .$$
An arbitrary power product has the form $\pi=P^{b_1} g^{b_2} \ell^{b_3} \theta_0^{b_4}$. The equation $A v=0$, where $v=\left(b_1, \ldots, b_4\right)$, has two free variables, and the general solution is
$$b_1=-2 b_3, \quad b_2=-b_3, \quad b_3 \text { and } b_4 \text { free. }$$
Since there are two free variables, there are two independent solutions. For the first solution, we choose $b_3=-1 / 2$ and $b_4=0$, which gives $\pi_1=P \sqrt{g / \ell}$. For the second solution, we choose $b_3=0$ and $b_4=1$, which gives $\pi_2=\theta_0$. This is a full set of independent dimensionless power products, and is normalized with respect to $P$.
By the $\pi$-theorem, the period equation $P=F\left(g, \ell, \theta_0\right)$ must be equivalent to
$$\pi_1=\phi\left(\pi_2\right) \quad \text { or } \quad P=\sqrt{\frac{\ell}{g}} \phi\left(\theta_0\right),$$
for some function $\phi$. Thus the relation between the quantities $P, g, \ell, \theta_0$ is not characterized by an unknown function of three quantities $F\left(g, \ell, \theta_0\right)$, but is instead characterized by an unknown function of one quantity $\phi\left(\theta_0\right)$. Equivalently, the dependence of $F\left(g, \ell, \theta_0\right)$ on the quantities $g$ and $\ell$ is completely dictated by dimensional considerations.

# 应用数学代考

## 数学代写|应用数学代考APPLIED MATHEMATICS代写|Case study

$$\frac{d \vec{r}}{d t}=\ell \cos \theta \frac{d \theta}{d t} \vec{i}-\ell \sin \theta \frac{d \theta}{d t} \vec{j} \frac{d^2 \vec{r}}{d t^2}=\left[\ell \cos \theta \frac{d^2 \theta}{d t^2}-\ell \sin \theta\left(\frac{d \theta}{d t}\right)^2\right] \vec{i}-\left[\ell \sin \theta \frac{d^2 \theta}{d t^2}+\ell \cos \theta\left(\frac{d \theta}{d t}\right)^2\right] \vec{j}$$

## 数学代写|应用数学代考APPLIED MATHEMATICS代写|Reduced equation for period

$$A=\left(\Delta_P, \Delta_g, \Delta_{\ell}, \Delta_{\theta_0}\right)=\left(\begin{array}{llllllll} 1 & -2 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right) .$$

$b_1=-2 b_3, \quad b_2=-b_3, \quad b_3$ and $b_4$ free.

$$\pi_1=\phi\left(\pi_2\right) \quad \text { or } \quad P=\sqrt{\frac{\ell}{g}} \phi\left(\theta_0\right),$$

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## 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。