Advanced Classical Mechanics/Linear Motion

We will examine the dynamics of a particle that is free to travel along only one dimension. It could be restricted to travel along a line or in a circle for example. The results for one-dimensional motion of a single particle provide an foundation to understand motion in more than one dimension and the dynamics of many-particle systems.

General Results

Before looking at some specific examples let's look at a rather general case. A particle moves along one dimension subject to a force that only depends on position along that dimension. We have

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

Let's define the kinetic energy of the particle to be

$T=\frac{1}{2}m{\dot{x}}^2.$

The rate of change of the kinetic energy is

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

and let's calculate the total change in the kinetic energy from one time to another

$T_2-T_1=\underset{t_1}{\overset{t_2}{\int }}\dot{T}dt=\underset{t_1}{\overset{t_2}{\int }}\frac{dx}{dt}F(x)dt=\underset{x_1}{\overset{x_2}{\int }}F(x)dx=V_1-V_2$

where we have defined the potential energy

$V(x)=-\underset{x_0}{\overset{x}{\int }}F(x)dx.$

If we rearrange things we get

$T_2+V_2=T_1+V_1$

so $E=T+V$ is a constant over the motion. This is the conservation of energy. We can invert the integral to calculate the force in terms of the potential energy

$F(x)=-\frac{dV}{dx}$

or we can calculate the velocity of the particle as a function of position using the fact that energy is conserved $E=m{\dot{x}}^2/2+V(x)$ so

$\frac{1}{2}m{\dot{x}}^2=E-V(x).$

In particular the motion is restricted to the region where $V(x)\le E$.

The Pendulum

Let's consider a small ball of mass $m$ suspended from a pivot by a light rod of length $l$. The ball is restricted to move in a vertical circle. It is useful to write the position of the ball in terms of the angle $\theta$ that the rod makes with the vertical direction.

The position of the bob is

$x=l\sin \theta ,y=-l\cos \theta$

using Cartesian coordinates. The force is $F_y=-mg$. Let's construct the potential energy

$V=-\underset{0}{\overset{y}{\int }}{F}_{y}d{y}^{\prime }=\underset{0}{\overset{y}{\int }}mgd{y}^{\prime }=mgy=-mgl\cos \theta$

and the kinetic energy

$T=\frac{1}{2}m{v}^2=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)=\frac{1}{2}m(l^2{\cos }^2\theta {\dot{\theta }}^2+l^2{\sin }^2\theta {\dot{\theta }}^2)=\frac{1}{2}m{l}^2{\dot{\theta }}^2.$

The Period

A plot of the ratio between the actual period of a pendulum and the approximate value obtained for small angles, as a function of the amplitude.

Let's say the bob is released at an angle ${\theta }_0$, how long does in take to reach the bottom, $\theta =0$. This is one quarter of the period of the pendulum. The total energy of the bob is

$E=T+V=-mgl\cos {\theta }_0=\frac{1}{2}m{l}^2{\dot{\theta }}^2-mgl\cos \theta .$

We can solve this equation for $\dot{\theta }$ to get

$\dot{\theta }=\frac{d\theta }{dt}={\left[\frac{2g}{l}(\cos \theta -\cos {\theta }_0)\right]}^{1/2}$

We can solve this differential equation by the separation of variables

$-\sqrt{\frac{l}{2g}}\underset{{\theta }_0}{\overset{0}{\int }}\frac{1}{\sqrt{\cos \theta -\cos {\theta }_0}}d\theta =\underset{0}{\overset{P/4}{\int }}dt=\frac{P}{4}$

where $P$ is the period of the pendulum. The integral is not elementary but we can do it numerically or using special functions. However, if $\theta$ is small we can use $\cos \theta \approx 1-{\theta }^2/2$ to get another integral

$P\approx 4\sqrt{\frac{l}{g}}\underset{0}{\overset{{\theta }_0}{\int }}\frac{1}{\sqrt{{\theta }_0^2-{\theta }^2}}d\theta =4\sqrt{\frac{l}{g}}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-u^2}}du$

    $\approx {-4\sqrt{\frac{l}{g}}\arccos \theta |}_0^1=2\pi \sqrt{\frac{l}{g}}$.

We can also calculate the period as a function of amplitude numerically or using special functions

Bounded and Unbounded Motion

The motion of the pendulum has two different characters depending on the initial energy of the bob. In the previous example we considered a situation where the bob starts without any motion, so the initial energy is purely potential. On the other hand, the energy of the bob could be more arbitrary. Let's write

$E=\frac{1}{2}m{\dot{\theta }}^2-mgl\cos \theta .$

Remember that the total energy of the particle is defined up to an additive constant. The cosine ranges from -1 to 1, so if

$E<mgl$

the motion of the bob is periodic because at some value of $\theta$ the velocity of the bob vanishes and then changes sign. Specifically,

${\theta }_{max}=\arccos \left(\frac{E}{mgl}\right).$

On the other hand if $E>mgl$ the equation has no real solution. In this case the velocity of the bob doesn't change sign and the bob goes round and round, slowing as it goes over the top.

Motion near Equilibrium

The motion of a particle can only be in equilibrium if the total forces on it vanish. If the force only depends on the position of the particle (as it does for a conservative force), then the slope of potential energy must vanish at the equilibrium point because $F(x)=-V^{\prime }(x).$

For simplicity (and without loss of generality) we can take the point of equilibrium to be $x=0$ and take the potential energy to be zero at this point. We can expand the potential energy as a Taylor series about any point. About the point of equilibrium we have

$V(x)=V(0)+x{V}^{\prime }(0)+\frac{1}{2}{x}^2{V}^{″}(0)+\cdots .$

For the equilibrium point the first two terms vanish so we have

$V(x)\approx \frac{1}{2}k{x}^2\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}k=V^{″}(0).$

The motion of a particle near almost any point of equilibrium is described by this potential energy, so it appears often in all branches of physics and it is useful to analyze in some detail. If the constant $k$ is positive, the potential energy curve is concave up. The motion of a particle of any energy is periodic between the points where $E=V(x)$,

$x=\pm \sqrt{\frac{2E}{k}}.$

On the other hand, if $k$ is negative, the particle approaches the origin and slows down as it gets closer. If $E<0$, the particle turns around at

$x=\pm \sqrt{-\frac{2E}{k}}.$

If the energy of the particle is positive it surmounts the barrier and continues travelling on the other side.

The force corresponding to the potential energy function is $F=-kx$, yielding an equation of motion

$m\ddot{x}+kx=0.$

This is a linear differential equation for $x(t)$ with constant coefficients. Such equations can generally be solved by linear combinations of exponential functions. Let's try the substitution

$x(t)=A{e}^{at}+B{e}^{bt}$

to get

$mA{a}^2{e}^{at}+kA{e}^{at}+mB{b}^2{e}^{bt}+kB{e}^{bt}=0$

which we can solve for any values of $A$ and $B$ as long as

$a=\sqrt{-\frac{k}{m}}$ and $b=-\sqrt{-\frac{k}{m}}.$

If $k$ is negative, it is handy to use the related hyperbolic functions to look at the motion

$x(t)=C\cosh pt+D\sinh pt$

where $p=\sqrt{-k/m}$. From the discussion earlier if the energy of the particle is less than zero, then the motion is restricted to one side of the barrier and the variable $x$ always has the same sign. Looking at the solution, this means that $|C|>|D|.$ On the other hand, if the energy is positive we have $|D|>|C|$.

If $k$ is positive, the constants $a$ and $b$ are imaginary. In this case we can use

$x(t)=C\cos \omega t+D\sin \omega t$

where $\omega =\sqrt{k/m}$.

Alternative Solution

We can also solve for the motion using the same techniques as we used for the pendulum. For conservative linear motion this is always possible, but for motion in several dimensions, one can use this technique only if there are additional conserved quantities. Using the conservation of energy has the advantage that it doesn't appear that we are guessing for the solution. The energy of the particle is conserved

$E=T+V=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}k{x}^2.$

We can solve this equation for the velocity of the particle as a function of position

$\dot{x}=\frac{dx}{dt}=\pm \sqrt{\frac{2E}{m}-\frac{k}{m}{x}^2}.$

We can solve this differential equation by the separation of variables

$\underset{t_0}{\overset{t_1}{\int }}dt=\pm \underset{x_0}{\overset{x_1}{\int }}\frac{1}{\sqrt{\frac{2E}{m}-\frac{k}{m}{x}^2}}dx=\pm \sqrt{\frac{m}{2E}}\underset{x_0}{\overset{x_1}{\int }}\frac{1}{\sqrt{1-\frac{k}{2E}{x}^2}}dx$

We have four possibilities depending on the signs of $E$ and $k$. If $E$ and $k$ are both positive the integral is

$t_1-t_0=\pm \sqrt{\frac{m}{k}}[\arcsin \left(\sqrt{\frac{k}{2E}}{x}_1\right)-\arcsin \left(\sqrt{\frac{k}{2E}}{x}_0\right)]$

and if we invert this formula we get

$x_1=\sqrt{\frac{2E}{k}}\sin [\sqrt{\frac{k}{m}}(t_1-t_0)+\arcsin \left(\sqrt{\frac{k}{2E}}{x}_0\right)]$

where we have dropped the $\pm$ sign for clarity.

If $E$ is negative and $k$ is positive, there is no solution. If $k$ is negative, the solutions are unbounded. For positive $E$ we have

$x_1=\sqrt{-\frac{2E}{k}}\sinh [\sqrt{-\frac{k}{m}}(t_1-t_0)+\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\left(\sqrt{-\frac{k}{2E}}{x}_0\right)]$

and for negative $E$ we have

$x_1=\sqrt{\frac{2E}{k}}\cosh [\sqrt{-\frac{k}{m}}(t_1-t_0)+\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{h}\left(\sqrt{\frac{k}{2E}}{x}_0\right)].$

Complex Representation

We found earlier that the general solution for small oscillations was

$x=A{e}^{pt}+B{e}^{-pt}.$

When $k$ is positive, the constant $p$ is purely imaginary. It is customary to define $p^2=-{\omega }^2=-k/m$, yielding the periodic solution

$x=A{e}^{i\omega t}+B{e}^{-i\omega t}.$

This solution is of course a complex number. However, both the real and imaginary parts of this solution are also solutions of the differential equation, or one could let $A=\overline{B}$ yielding the real solution,

$x=(c-id)(\cos \omega t+i\sin \omega t)+(c+id)(\cos \omega t-i\sin \omega t)=2c\cos \omega t+2d\sin \omega t$

that we obtained earlier. In general we considering periodic phenomena, we will use solutions of the form $x(t)=A{e}^{i\omega t}$ with the proviso that since the physical quantity $x$ that we will take the real part of the solution.

Damped Linear Oscillator

We found that the motion of a particle near a position of stable equilibrium under a conservative force approximates that of a simple harmonic oscillator. There may be some energy loss as well, so there may be a force that depends on the velocity. If we take $x$ and $\dot{x}$ as small quantities, a good approximation to the total force is

$F=-kx-\lambda \dot{x}$

yielding an equation of motion

$m\ddot{x}+\lambda \dot{x}+kx=0.$

Using the earlier results we will try a solution of the form $x=A{e}^{pt}$ which yields an equation

$A{e}^{pt}(m{p}^2+\lambda p+k)=0.$

The roots of this equation are

$p=-\frac{\lambda }{2m}\pm \sqrt{{\left(\frac{\lambda }{2m}\right)}^2-\frac{k}{m}}.$

There are three regimes of the damping.

Large Damping

If

${\left(\frac{\lambda }{2m}\right)}^2>\frac{k}{m},$

then both roots for $p$ are real and negative. Specifically we have

${\gamma }_{\pm }=\frac{\lambda }{2m}\pm \sqrt{{\left(\frac{\lambda }{2m}\right)}^2-\frac{k}{m}}.$

and the solution

$x=A{e}^{-{\gamma }_{+}t}+B{e}^{-{\gamma }_{-}t}.$

At late times, the second term dominates and the damping time is approximately $1/{\gamma }_{-}$.

Small Damping

If

${\left(\frac{\lambda }{2m}\right)}^2<\frac{k}{m},$

then both roots for $p$ are complex and the general solution is

$x=\mathrm{R}\mathrm{e}\left(A{e}^{i\omega t-\gamma t}\right)=a{e}^{-\gamma t}\cos (\omega t-\theta )$

where $\gamma =\lambda /(2m)$,

$\omega =\sqrt{{\omega }_0^2-{\gamma }^2}$

and ${\omega }_0^2=k/m$.

We can define the quality factor or 'Q' of the oscillator to be

$Q=\frac{m{\omega }_0}{\lambda }=\frac{{\omega }_0}{2\lambda }.$

Critical Damping

In the limiting case we have ${\omega }_0=\lambda$ and the two roots for $p$ coincide. The resulting solution $x=a{e}^{-\gamma t}$ cannot be general because a second-order differential equation but have two constants of integration. Here we have

$x=(a+bt)e^{-\gamma t}$

Driven Oscillators

We can imagine some force driving the oscillator. Perhaps someone is pushing on a child on a swing. We get the following equation of motion,

$m\ddot{x}+\lambda \dot{x}+kx=F(t).$

This is an inhomogeneous equation of motion. The general solution to such an equation is the solution to the homogeneous equation (which has two constants of integration) plus a particular solution to the equation. To find the particular let's consider a periodic driving force

$F(t)=F_1{e}^{i{\omega }_{1}t}.$

Since the equation of motion is linear, the particular solution for the sum of several driving forces is the sum of the solutions for each driving force. Let's look for a solution that varies at the same frequency as the driving force

$x(t)=A_1{e}^{i{\omega }_{1}t}.$

Substituting this into the equation of motion yields an equation for the amplitude of the oscillation

$A_1(-m{\omega }_1^2+i\lambda {\omega }_1+k)=F_1$

and solving for the amplitude yields

$A_1=\frac{F_1}{m}\frac{1}{{\omega }_0^2-{\omega }_1^2+2i\gamma {\omega }_1}.$

The amplitude is complex and we can divide it into a magnitude and a phase $a{e}^{-i{\theta }_1}$. Using the fact that $1/A_1=1/a{e}^{i{\theta }_1}$ we have

$a=\frac{1}{\sqrt{{({\omega }_0^2-{\omega }_1^2)}^2+4{\gamma }^2{\omega }_1^2}}\mathrm{a}\mathrm{n}\mathrm{d}\tan {\theta }_1=\frac{2\gamma {\omega }_1}{{\omega }_0^2-{\omega }_1^2}.$

This yields the real general solution to our equation of the form

$x=a_1\cos ({\omega }_{1}t-{\theta }_1)+a{e}^{-\gamma t}\cos (\omega t-\theta ).$

The second term that depends on the initial conditions decreases exponentially. After a long time this transcient can be neglected and the oscillations are solely governed by the driving force. Template:Subpage navbar