物理代写|理论力学作业代写Theoretical Mechanics代考|PHYSICS2532

物理代写|理论力学作业代写Theoretical Mechanics代考|Damped Linear Oscillator

Because of the unavoidable friction every oscillating process is exponentially damped unless an additional external force acts. We will now include the latter in our considerations. The equation of motion (2.169) is then to be replaced by
$$\ddot{x}+2 \beta \dot{x}+\omega_0^2 x=\frac{1}{m} F(t)$$
We choose the same denotations as in the last section and restrict ourselves to the important special case of a periodic force:
$$F(t)=f \cos \bar{\omega} t$$
One can realize the periodic force by, for instance, a wheel spinning with constant angular velocity and being connected via a drive rod to the oscillating body (Fig. 2.31).

Here again we have an exact non-mechanical realization (Fig. 2.32) by the electrical oscillator circuit if one applies to it a periodic alternating voltage $U_0 \sin \bar{\omega} t$ :
$$L \ddot{I}+R \dot{I}+\frac{1}{C} I=U_0 \bar{\omega} \cos \bar{\omega} t .$$

物理代写|理论力学作业代写Theoretical Mechanics代考|Arbitrary One-Dimensional Space-Dependent Force

In such a case a general procedure for solving the equation of motion
$$m \ddot{x}=F(x)$$
can be developed that ultimately reduces the problem to so-called ‘quadratures’, i.e. to the explicit evaluation of well-defined integrals. This method leads at first to purely mathematically defined auxiliary quantities (e.g. constants of integration), which, however, later will acquire fundamental physical meanings, such as energy, potential, work, power, ….
We multiply (2.199) with $\dot{x}$ :
$$m \ddot{x} \dot{x}=F(x) \dot{x}$$
This can then obviously also be written in the following form:
$$\frac{d}{d t}\left(\frac{m}{2} x^2\right)=-\frac{d}{d t} V(x)$$
if one understands by $V(x)$ the following indefinite integral:
$$V(x)=-\int^x F\left(x^{\prime}\right) d x^{\prime}$$
$V(x)$ is in a certain sense the antiderivative of the force $F(x)$ being therefore a known quantity except for an additive constant. The minus sign is simply a convention without any deeper physical meaning.

By the integration process the Eq. (2.200) provides a new constant which we want to denote by $E$ :
$$\frac{m}{2} \dot{x}^2=E-V(x)$$

物理代写|理论力学作业代写Theoretical Mechanics代考|Damped Linear Oscillator

$$\ddot{x}+2 \beta \dot{x}+\omega_0^2 x=\frac{1}{m} F(t)$$

$$F(t)=f \cos \bar{\omega} t$$

$$L \ddot{I}+R \dot{I}+\frac{1}{C} I=U_0 \bar{\omega} \cos \bar{\omega} t .$$

物理代写|理论力学作业代写Theoretical Mechanics代考|Arbitrary One-Dimensional Space-Dependent Force

$$m \ddot{x}=F(x)$$

$$m \ddot{x} \dot{x}=F(x) \dot{x}$$

$$\frac{d}{d t}\left(\frac{m}{2} x^2\right)=-\frac{d}{d t} V(x)$$

$$V(x)=-\int^x F\left(x^{\prime}\right) d x^{\prime}$$
$V(x)$ 在某种意义上是力的反导数 $F(x)$ 因此是一个已知量，除了一个附加営数。减号只是一种约定，没有 任何更深层次的物理意义。

$$\frac{m}{2} \dot{x}^2=E-V(x)$$

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