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

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## 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:

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