Hooke's Law

Review the chain rule, or how to take the derivative of composite functions.

Attach a spring obeying Hooke's law to one end of a little 1-d rolling cart and the force on the cart is conservative, given by $$F_x= - k(x-c)=-\frac{dU(x)}{dx},$$ where $c$ is the equilibrium position.

The potential energy for this mass-on-a-spring is: $$U(x) = \frac{1}{2}k(x-c)^2+C$$

Other potentials may not be purely quadratic, but all continuous 'potential energy valleys' are quadratic at small enough distances from the bottom of the valley as we shall now show.

You can see this by doing a Taylor expansion around the potential energy minimum at 'c' in the diagram of some potential energy $U$ at right. $$U(r) = U(c) + U'(c) (r-c) + \frac{1}{2}U''(c) (r-c)^2+\frac{1}{6}U'''(c)(r-c)^3+...$$

The slope of the potential energy is zero at the equilibrium position $c$: $U'(c)=(dU)/(dr)=\left.0\right|^{r=c}$ at the bottom of the valley.

Close to $r=c$, it's a pretty good approximation to the potential to say that, $$U(r)\approx U(c) +\frac 12 U''(c)\,(r-c)^2.$$

Taking the derivative of this approximate potential wrt to $r$, the restoring force is $$-\frac{\del U}{\del r}=-\left[U''(c)\right](r-c)$$ which looks just like Hooke's law for springs with a restoring force $-k(r-c)$ with $k=\frac{\del^2}{\del r^2} \left.U(r)\right|^{r=c}$, for deviations from equilibrium.

Let's define a new coordinate $\epsilon \equiv r-c$, which is the deviation away from the equilibrium position at $r=c$. Taking the second derivative of both sides of this definition, you can perhaps readily see that $\ddot \epsilon = \frac{\del^2}{\del t^2}(x-c)=\ddot x$ since the second derivative of a constant vanishes.

If the equation of motion looks like $m\ddot{x}=-\frac{\del U}{\del r}$, then we could find the equilibrium position $c$, and then write the equation of motion for small deviations from the equilibrium in terms of the relative coordinate as $$m\ddot\epsilon\approx-k\epsilon$$ and we know that the frequency of oscillations for this differential equation is $$\omega=\sqrt{\frac km}.$$

Example 4.7: Cube on cylinder (or sphere)

For a cube any flat-bottomed object with a center of mass which is $b$ units above the bottom balancing on a cylinder (or sphere) the potential energy is: $$U(\theta) = mg[(r+b)\cos \theta + r \theta \sin\theta)$$

Take the derivative wrt $\theta$, set it equal to 0, and solve for $\theta$ to find the equilibrium point... $$\begineq \frac{1}{mg}\frac{dU}{d\theta}&=& -(r+b)\sin\theta +r\sin\theta +r\theta\,\cos\theta \\ &=&r\theta \cos\theta-b\sin\theta.\endeq$$ This is 0 when $\theta=0$.

Take another derivative to find the concavity of the function as a function of $\theta$: $$\frac{d^2U}{d\theta^2}=mg(r \cos\theta - r\theta \sin\theta-b\cos\theta).$$

At the equilibrium point, $\theta=0$, so... $$\left.\frac{d^2 U}{d \theta^2}\right|^{\theta=0} = mg(r-b)$$

As long as this is positive ($r > b$) there is a restoring force, one that gets larger as $\theta$ increases. The cube will rock back and forth like a weight on a spring.

But, if $b>r$, the concavity of the function changes sign, and the equilibrium point at $\theta=0$ will be an unstable one.

The general idea is...

• For any 'nice' (continuous, finite...) potential energy function, the force on a particle is $F= - \del U/\del r$.
• Look for equilibrium positions where the force vanishes: $\left.\del U/\del r \right|^{r_0} =0$.
• Taylor expand the potential energy about $r=r_0$.
• The second derivative at the equilibrium position $\left.\del^2 U/\del r^2\right|^{r_0}$="$K$" plays the role of the spring constant, and (together with the mass) determines the frequency of oscillations near equilibrium. (See below.)
• For small distances away from equilibrium, $U\approx C+\frac{1}{2}K(r-r_0)^2$.

Example: Problem 5.2

The Morse function: an approximate potential for some two-atom systems, $$U(r)=A[(e^{(R-r)/S} -1)^2 -1],$$

With $A$, $R$, $S$ all positive constants and $S \ll R$.

Now, consider first just the function in parentheses, $f(r)=e^{(R-r)/S}-1$. This function is zero when $r=R$ as shown below for $R=10$:

blue: $(e^{(10-r)/0.2}-1)$
green: $(e^{(10-r)/0.05}-1)$

So, when we square $f(r)$, the minimum is guaranteed to occur at at $r=R$:

green: $4[(e^{(10-r)/0.05}-1)^2]$
purple: $4[(e^{(10-r)/0.05}-1)^2 -1]$

Now, let's write $U$ in terms of $x$, the deviation from the equilibrium position, $R$. That is... $r=R+x$. We'll have: $$U(x)=A[(e^{x/S}-1)^2-1]$$. According to our considerations about Taylor expanding, close to equilibrium ($x=0$), we should be able to write $$U(x) \approx C + \frac 12kx^2$$ where $k=\left.U''(x)\right|^{x=0}$.

Taking the first and second derivatives w.r.t. $x$: $$U'(x)=A[2(e^{x/S}-1)\frac 1Se^{x/S}].$$ $$U''(x)=\frac {2A}{S}[\frac 1Se^{x/S}e^{x/S}+(e^{x/S}-1)\frac 1Se^{x/S}].$$ Evaluating the second derivative at $x=0$: $$k=\frac{2A}{S^2}[1*1+(1-1)]=\frac{2A}{S^2}.$$ So, the approximate potential close to the equilibrium position should be: $$U(x)=U(x=0)+\frac 12 kx^2$$ and subbing in $x=r-R$ $$U(r)=U(R)+\frac 12 \left(\frac{2A}{S^2}\right)(r-R)^2.$$ See www.desmos.com/calculator/iomcpvvdi3.

Simple harmonic motion

With a restoring force of $-kx$, the equation of motion of something like our cart on a spring is $ma=-kx$ which can be written as: $$\ddot{x} = -\frac{k}{m}x = -\omega^2x$$ The constants $k$ and $m$ are both positive and $$\omega \equiv \sqrt{k/m}.$$

This is a second order differential equation, $\Rightarrow$ any general solution will have two constants which need to be fixed by the boundary conditions.

There are several different ways of writing the solution:

Sine and Cosine functions

You should confirm that both $\sin (\omega t)$ and $\cos(\omega t)$ are solutions. A general solution is: $$x(t) = B_1 \cos (\omega t) + B_2 \sin (\omega t)$$

For example, if at $t=0$, I pull the cart a distance $x_0$ from equilibrium and release it with no initial velocity, the solution is ... $$x(t) = x_0 \cos (\omega t).$$ (That is, $B_2=0$.)

With a partner: What does the solution look like if instead, at $t=0$ you start the cart out at $x=0$, and then give it a swift kick such that it starts out with some initial speed $v_0$?

Phase-shifted Cosine (or Sine for that matter)

$$x(t) = A \cos (\omega t - \delta)$$

If this one works, another, just as general solution is $$x(t) = A' \sin (\omega t - \delta')$$

Exponential solutions

Returning to our original differential equation: $$\ddot{x} = -\omega^2x$$

It looks like $e^{i\omega t}$ satisfies this. And just as easily $e^{-i\omega t}$, so another general solution would be: $$x(t)= C_1 e^{i\omega t} + C_2 e^{-i\omega t}$$

This is the linear superposition principle in action:

• $d^n/dt^n$ is a linear operator, and our differential equation $\frac{d^2}{dt^2}x = kx$ is a linear combination of purely linear operators.
• $e^{i\omega t}$ and $e^{-i\omega t}$ are two linearly independent functions. ($\sin \theta $ and $4 \sin \theta $ are not, because one is a multiple of the other.)
• Any linear combination of independent solutions is also a solution.

Real part of a complex exponential

We know that the graph of $e^{i \omega t}$ is a point moving counterclockwise on the unit circle in the complex plane with angular speed (radians/second) $\omega$. See animation. The projection of this complex vector on the real axis is is written as $$Re(e^{i\omega t}) = \cos (\omega t)$$

A first step in generalizing this is to multiply it by a (real) constant $A$, the amplitude to allow for different, well, amplitudes of the motion. $$Re(Ae^{i\omega t}) = A\cos(\omega t)$$

And finally, to allow for different starting positions (phases) we can multiply this by $e^{-i\delta}$, a complex number with unit length, so that it doesn't change the amplitude. So our final (general) solution for simple harmonic oscillations can be written: $$x(t) = Re(Ae^{i\omega t}e^{-i\delta}) = Re(A e^{i(\omega t - \delta)})$$

Homework

Chapter 5: 6, 9, 10 (hint: Taylor expand the $\sinh$ function), 13