# 经典力学代写|CLASSICAL MECHANICS MATH228 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool CLASSICAL MECHANICS MATH228经典力学代写代考辅导服务！

## Instructions:

Classical mechanics is a branch of physics that deals with the study of the motion of objects under the influence of forces. It was first developed by Sir Isaac Newton in the 17th century and is based on three fundamental laws of motion:

1. The law of inertia: An object at rest will remain at rest and an object in motion will continue in a straight line at a constant speed unless acted upon by an external force.
2. The law of acceleration: The acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. This is expressed mathematically as F = ma, where F is the force applied to the object, m is its mass, and a is its acceleration.
3. The law of action-reaction: For every action, there is an equal and opposite reaction.

Classical mechanics also includes the study of energy and momentum, and how they are conserved in various physical processes. It has numerous applications in engineering, physics, and astronomy, and is the foundation for the development of modern physics.

A driven oscillator is described by
$$\ddot{x}+\omega_o^2 x=\frac{F}{m} \cos (\gamma t+\alpha) .$$
We found that the solution off resonance is
$$x(t)=B \cos \left(\omega_o t+\beta\right)+\frac{F / m}{\omega_o^2-\gamma^2} \cos (\gamma t+\alpha) .$$
which we can rearrange to
$$x(t)=C \cos \left(\omega_o t+\kappa\right)+\frac{F / m}{\omega_o^2-\gamma^2}\left(\cos (\gamma t+\alpha)-\cos \left(\omega_o t+\alpha\right)\right) .$$
with new constants $C$ and $\kappa$.
a) If the oscillator is driven close to the natural frequency $\omega_o$, we can write $\omega_o=\gamma+\epsilon$ with $\epsilon \ll \omega_o$. Keeping terms only linear in $\epsilon$ (i.e. set any $\epsilon$ with higher power to zero), show that we can write
$$x(t)=C \cos \left(\omega_o t+\kappa\right)+\frac{F / m}{2 \omega_o \epsilon}\left(\cos \left(\omega_o t+\alpha-\epsilon t\right)-\cos \left(\omega_o t+\alpha\right)\right)$$

a) Starting from the expression for $x(t)$ given in the problem, we substitute $\omega_o=\gamma+\epsilon$ and keep only terms linear in $\epsilon$. Using the identity $\cos(A+B)=\cos A\cos B-\sin A\sin B$, we have

\begin{align} x(t) &= C \cos(\gamma t + \epsilon t + \kappa) + \frac{F/m}{(\gamma+\epsilon)^2 – \gamma^2}\left[\cos(\gamma t + \alpha) – \cos(\gamma t + \epsilon t + \alpha)\right]\ &= C \cos(\gamma t + \kappa)\cos(\epsilon t) – C \sin(\gamma t + \kappa)\sin(\epsilon t)\ &\quad + \frac{F/m}{2\gamma\epsilon + \epsilon^2}\left[\cos(\gamma t + \alpha – \epsilon t) – \cos(\gamma t + \alpha)\right] + O(\epsilon^2)\ &= C \cos(\gamma t + \kappa)\cos(\epsilon t) + \frac{F/m}{2\gamma\epsilon}\sin(\gamma t + \kappa)\epsilon t\ &\quad + \frac{F/m}{2\gamma\epsilon + \epsilon^2}\left[\cos(\gamma t + \alpha)\cos(\epsilon t) + \sin(\gamma t + \alpha)\sin(\epsilon t) – \cos(\gamma t + \alpha)\right] + O(\epsilon^2) \end{align}

The term proportional to $\sin(\gamma t + \kappa)\epsilon t$ comes from expanding the sine term to first order in $\epsilon$ and keeping only the linear term. We can simplify the expression by using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ and discarding terms of order $\epsilon^2$ or higher. This gives

\begin{align} x(t) &= C \cos(\gamma t + \kappa)\cos(\epsilon t) + \frac{F/m}{2\gamma\epsilon}\sin(\gamma t + \kappa)\epsilon t\ &\quad + \frac{F/m}{2\gamma\epsilon}\sin(\alpha)\sin(\epsilon t) + O(\epsilon^2)\ &= C \cos(\omega_o t + \kappa) + \frac{F/m}{2\gamma\epsilon}\left[\cos(\omega_o t + \alpha – \epsilon t) – \cos(\omega_o t + \alpha)\right] + O(\epsilon^2) \end{align}

where we have used the definition of $\omega_o$ and the fact that $\sin(\gamma t + \kappa) = \sin(\omega_o t + \alpha)$ and $\cos(\gamma t + \kappa) = \cos(\omega_o t + \alpha)$.

b) Show that this evolves to the on resonance solution (LL 22.5) for $\epsilon \rightarrow 0$. Note: you may carry out the calculation using trigonometric identities or complex notation. Note: to compare with LL 22.5 , convert the above as follows: $$C \rightarrow a, \quad F \rightarrow f, \omega_0 \rightarrow \omega, \quad \kappa \rightarrow \alpha, \alpha \rightarrow \beta$$

To show that the off-resonance solution evolves to the on-resonance solution for $\epsilon \rightarrow 0$, we need to take the limit of the off-resonance solution as $\epsilon \rightarrow 0$ and show that it matches the on-resonance solution given by LL 22.5.

The on-resonance solution given by LL 22.5 is:

$x(t)=a \cos (\omega t+\alpha)+\frac{f}{2 m \omega} t \sin (\omega t+\alpha)$

where $\omega = \omega_0$ and $\alpha = \beta$.

Substituting the constants given in the note, we have:

$x(t)=a \cos \left(\omega_0 t+\beta\right)+\frac{f}{2 m \omega_0} t \sin \left(\omega_0 t+\beta\right)$

Now, let’s take the limit of the off-resonance solution as $\epsilon \rightarrow 0$: \begin{align*} x(t) &= C \cos(\omega_0 t + \kappa) + \frac{F/m}{\omega_0^2 – \gamma^2}(\cos(\gamma t + \alpha) – \cos(\omega_0 t + \alpha)) \ &= C \cos(\omega_0 t + \kappa) + \frac{F/m}{\omega_0^2 – \gamma^2}\cos(\gamma t + \alpha) – \frac{F/m}{\omega_0^2 – \gamma^2}\cos(\omega_0 t + \alpha) \end{align*}

We can use $a, b, c$ and $d$ from the previous problem to make the matrix $M$ such that $\vec{x}(t+\Delta t)=M \vec{x}(t)$. Find the eigenvalues of $M$. Take $\Delta t=4 \pi / \omega_o$ and find the eigenvectors.

Recall that the system of differential equations is given by:

\begin{aligned} \dot{x}_1 & =x_2 \ \dot{x}_2 & =-\frac{k}{m} x_1-\frac{c}{m} x_2 \ \dot{x}_3 & =x_4 \ \dot{x}_4 & =-\frac{k}{m} x_3-\frac{c}{m} x_4+\frac{F_0}{m} \cos \left(\omega_d t\right)\end{aligned}

We can rewrite this system of differential equations in matrix form as:

$\frac{d}{d t}\left(\begin{array}{l}x_1 \ x_2 \ x_3 \ x_4\end{array}\right)=\left(\begin{array}{cccc}0 & 1 & 0 & 0 \ -\frac{k}{m} & -\frac{c}{m} & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & -\frac{k}{m} & -\frac{c}{m}\end{array}\right)\left(\begin{array}{l}x_1 \ x_2 \ x_3 \ x_4\end{array}\right)+\left(\begin{array}{c}0 \ 0 \ 0 \ \frac{F_0}{m} \cos \left(\omega_d t\right)\end{array}\right)$

Let’s define the matrix $M$ as:

$M=\left(\begin{array}{cccc}0 & 1 & 0 & 0 \ -\frac{k}{m} & -\frac{c}{m} & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & -\frac{k}{m} & -\frac{c}{m}\end{array}\right)$

# 经典力学 Classical Mechanics|PHYS7711Boston College Assignment

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Assignment-daixieTM为您提供波士顿学院Boston College PHYS7711 Classical Mechanics 经典力学代写代考辅导服务！

## Instructions:

Classical mechanics is a branch of physics that deals with the motion of objects under the influence of forces. It provides a framework for describing and predicting the behavior of physical systems ranging from the motion of planets and satellites to the behavior of atoms and molecules.

Classical mechanics is based on three fundamental laws of motion, known as Newton’s laws. The first law states that an object at rest will remain at rest, and an object in motion will continue to move in a straight line with constant velocity, unless acted upon by an external force. The second law relates the force acting on an object to its acceleration: F = ma, where F is the force, m is the mass of the object, and a is its acceleration. The third law states that for every action, there is an equal and opposite reaction.

In addition to these laws, classical mechanics also includes the concept of energy, both kinetic and potential. Kinetic energy is the energy possessed by an object due to its motion, while potential energy is the energy associated with the position or configuration of an object.

Classical mechanics is essential for understanding many areas of science and technology, including astronomy, engineering, and mechanics. It forms the foundation for many other branches of physics, such as thermodynamics, electromagnetism, and quantum mechanics.

Consider the motion of an object close to the surface of the Earth moving under the influence of Earth’s gravity.
a) The gravitational force law
$$\vec{F}=-\frac{G M m}{r^2} \hat{r}$$
reduces to
$$\vec{F}=-g m \hat{z}$$
at the surface with $z$ pointing away from the center of the Earth (i.e., “up”). Express $g$ in terms of $G$, $M_{\oplus}$ and $R_{\oplus}$. Compute $g$ in SI units.

At the surface of the Earth, the distance $r$ from the center of the Earth to the object is equal to the radius of the Earth, $R_{\oplus}$. Thus, we have

$\vec{F}=-\frac{G M_{\oplus} m}{R_{\oplus}^2} \hat{r} \approx-\frac{G M_{\oplus} m}{R_{\oplus}^2} \hat{z}$

where we have approximated $\hat{r}$ by $\hat{z}$ because the surface of the Earth is approximately a plane perpendicular to the $z$-axis.

The magnitude of the force is given by

$F=|\vec{F}|=\frac{G M_{\oplus} m}{R_{\oplus}^2}$

By Newton’s second law, the force is related to the object’s mass $m$ and acceleration $a$ via

$\vec{F}=m \vec{a}$

Since the object is moving only vertically, we have $\vec{a}=a\hat{z}$ and $|\vec{a}|=a$. Therefore,

F=ma=mg

where $g=|\vec{a}|$ is the acceleration due to gravity. Combining these equations, we obtain

$m g=\frac{G M_{\oplus} m}{R_{\oplus}^2}$

which can be solved for $g$ to give

$g=\frac{G M_{\oplus}}{R_{\oplus}^2}$

Substituting the values for $G$, $M_{\oplus}$, and $R_{\oplus}$ in SI units, we have

$g=\frac{\left(6.674 \times 10^{-11} \mathrm{~m}^3 \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\right) \times\left(5.9722 \times 10^{24} \mathrm{~kg}\right)}{\left(6.371 \times 10^6 \mathrm{~m}\right)^2} \approx 9.81 \mathrm{~m} / \mathrm{s}^2$

Therefore, the acceleration due to gravity at the surface of the Earth is approximately $9.81 ,\mathrm{m/s^2}$.

The range, $R$, of a projectile launched at an angle $\theta$ above the horizon with initial velocity $v_0$ is given by:

$$R = \frac{v_0^2\sin(2\theta)}{g},$$

where $g$ is the acceleration due to gravity.

On the Moon, the acceleration due to gravity is $g_{\text{Moon}} = 1.62 \text{ m/s}^2$. Therefore, the range of a cannonball fired at velocity $v_0$ and angle $\theta$ above the horizon on the Moon is:

$$R = \frac{v_0^2\sin(2\theta)}{g_{\text{Moon}}}.$$

Note that since there is no air on the Moon, we do not need to worry about air resistance affecting the motion of the cannonball.

Hooke’s law for a spring of constant $k$ is $F=-k x$. A mass $m$ is pushed from position $x_o$ with velocity $v_o$ at $t=0$. Find the subsequent motion, $x(t)$.

The equation of motion for a mass on a spring is given by Newton’s second law: $F = ma$, where $F$ is the force exerted by the spring, $m$ is the mass of the object attached to the spring, and $a$ is its acceleration. In this case, the force exerted by the spring is given by Hooke’s law, $F = -kx$, where $x$ is the displacement of the mass from its equilibrium position.

Assuming that the mass is only moving along the $x$-axis, we can write the equation of motion as:

ma=−kx

Since $a = \frac{d^2x}{dt^2}$, we can rewrite this as:

$m \frac{d^2 x}{d t^2}=-k x$

This is a second-order ordinary differential equation, which we can solve using the characteristic equation method. We assume that the solution takes the form:

$x(t)=A \cos (\omega t)+B \sin (\omega t)$

where $A$ and $B$ are constants that depend on the initial conditions, and $\omega$ is the angular frequency of the oscillation. Substituting this into the differential equation, we get:

$-\omega^2 m(A \cos (\omega t)+B \sin (\omega t))=-k(A \cos (\omega t)+B \sin (\omega t))$

Dividing both sides by $A\cos(\omega t) + B\sin(\omega t)$, we get:

$-\omega^2 m=-k \Rightarrow \omega=\sqrt{\frac{k}{m}}$

So the general solution is:

$x(t)=A \cos \left(\sqrt{\frac{k}{m}} t\right)+B \sin \left(\sqrt{\frac{k}{m}} t\right)$

To find the constants $A$ and $B$, we use the initial conditions. At $t=0$, the mass is at position $x_0$ with velocity $v_0$. So we have:

$x(0)=x_0 \Rightarrow A=x_0$

and

$v(0)=\left.\frac{d x}{d t}\right|_{t=0}=v_0=-\sqrt{\frac{k}{m}} x_0+B \sqrt{\frac{k}{m}}$

Solving for $B$, we get:

$B=\frac{v_0+\sqrt{\frac{k}{m}} x_0}{\sqrt{\frac{k}{m}}}$

Therefore, the solution for $x(t)$ is:

$x(t)=x_0 \cos \left(\sqrt{\frac{k}{m}} t\right)+\frac{v_0}{\sqrt{\frac{k}{m}}} \sin \left(\sqrt{\frac{k}{m}} t\right)$

This is the position of the mass as a function of time, given its initial position and velocity.