Another way to think about this approximation is the Taylor series for a function \(f\left(x\right)\) near its minimum at \(x = a\).

$$f\left(x\right)\approx f\left(a\right)+f^{\prime}\left(a\right) \left(x-a\right)+\frac{1}{2}f^{\prime\prime}\left(a\right)\left(x-a\right)^2$$

At the minimum, the derivative is zero, so \(f^{\prime}\left(a\right)=0\). Therefore, the first order term drops out; the zeroth order term, \(f\left( a\right)\) and the second order term \(\frac{1}{2}f^{\prime \prime}\left(a\right)\left(x-a\right)^2\) remain, i.e.,

$$f\left(x\right)\approx f\left(a\right)+\frac{1}{2}f^{\prime\prime}\left(a\right)\left(x-a\right)^2$$

If the function \(f\) is your potential, then you now have an oscillator potential for this small neighborhood of \(x=a\). We know the solutions for \(x\left(t\right)\) in this neighborhood, as we discussed in remote session 26.

Many physical systems can be studied profitably from this standpoint, the Taylor series expansion of the potential near one of its minima.

In the electromagnetic field of electric field \(\vec{E}\) and magnetic field \(\vec{B}\), the energy density of the field, \(u\), is also proportional to the squares of the fields: $$u=\frac{1}{2}\epsilon_0E^2+\frac{1}{2}\frac{1}{\mu_0}B^2$$

The constants \(\epsilon_0\) and \(\mu_0\) are related to the speed of light \(c\) :

$$c^2=\frac{1}{\epsilon_0 \mu_0}$$

after a ton of algebra

molecule broken

the famous case of "small" displacements.

Entree to current research areas!

good

Because, when \(x=\pm A\), the kinetic energy is zero, no matter what your initial conditions were, and all the energy \(E\) is potential.

The heart of the matter

"small"

$$y_1$$

with only the mass of the spring

Cf., simulations of hanging oscillator

Uh oh. The Math Dept. would not approve!

I.e., use a differential equation based on \(F=m\ddot{x}\).

...as usual with equations of motion!

Alternate model uses the generic solution but with the phase \(\phi\) converted into a sine term. E.g.,

$$x\left(t\right)=a\, \cos\left(\omega t\right)+b\, \sin\left(\omega t\right)$$

The constants \(a\) and \(b\) look like components of a vector whose magnitude is \(A\), viz. $$a=A\cos\left(\varphi\right) \ b=-A\sin\left(\varphi\right)$$

Initial conditions easiest to use: $$x\left(0\right)=0.02\, m \ \dot{x}\left(0\right)=0.0\frac{m}{s}$$

Result: \(\varphi = 0\). Nice.

We know this already

Good.

The velocity is leftward, derivative is negative, $$\dot{x}<0$$ during the first half of the motion, then rightward, positive derivative, $$\dot{x}>0$$ in the second half of the cycle... from THESE initial conditions $$x\left(0\right)=A\ \dot{x}\left(0\right)=0$$

This figure is a nice synopsis

That is, when \(t=T,\,2T,\,3T\dots\)

so the work done is not in KE, it is in the potential energy.

✔︎

I think we demonstrated this with variation of the swing angle back when we used pendula in the classroom.

We already know this from earlier in the semester: $$\omega^2=\frac{k}{m}$$

Incorrect. Its acceleration is rightward as soon as it crosses equilibrium, \(x<0\), i.e., in (c).

also max KE

Initial conditions: $$\vec{F}\left(0\right)=kA\left(-\hat{i}\right)$$ $$x\left(0\right)=A$$ $$\vec{v}\left(0\right)=0$$

These words imply, as we have already seen, that the total energy \(E\) is proportional to the square of the displacement, \(q\), using a generic coordinate \(q\) and to the square of the coordinate velocity, \(\dot{q}\). Or, as the math department might say, the function \(E\) is "quadratic" in \(q\) and \(\dot{q}\).

$$\frac{\text{human}}{\text{ultra}}=\left(2\times 10^4\right)/ \left(2.5\times 10^6\right)\approx 0.01$$

So that is "small" in the sense of calculus shortcuts.

This also means the wavelengths are much smaller for ultrasound.

Basic terms, must be familiar with them and where they fit into the equations of motion.

Dissipates energy.

Cavendish, however, could measure quantities like this!

actually in Cavendish's day, it was $$G=6.754\times 10^{-11}N\,m^2/kg^2$$

"a few kilometers"

I.e., from the top of Mt. Everest (elevation \(\approx 9\, km\)) to the bottom of the Mariana Trench (depth \(\approx 11\, km\). Cf., NASA planetary fact sheet.

E.g., $$\text{Everest} \longrightarrow\frac{1}{6380^2}=\frac{1}{40704400}\approx 2.457\times 10^{-8}$$ $$\text{sea level}\longrightarrow\frac{1}{6371^2}=\frac{1}{40589641}\approx 2.464\times 10^{-8}$$ $$\text{Challenger Deep}\longrightarrow\frac{1}{6360^2}=\frac{1}{40449600}\approx 2.472\times 10^{-8}$$

Not true for the Earth-moon system, as discussed, but good for most terrestrial objects like the oceans and for human objects like stadia and great pyramids.

Earth-moon system:

Earth and the average depth of the ocean (blue line):

Spectrum stadium, a blip next to Earth itself:

due to the symmetry of the circle of radius \(R_i\)

i.e., the orbital angular momentum of a single pixel

pixellation

Not so important in this context, but a good exercise.

good example

Be sure to work this out with some careful sketching. You can even define orbital angular momentum \(\vec{l}\) for a straight line trajectory.

Orbital angular momentum as opposed to spin angular momentum. It is amazing to think that, on a complex gravitational trajectory like an ellipse, parabola or hyperbola, the vector product of \(\vec{r}\) and \(\vec{p}\) is conserved. The magnitudes of momentum and radial distance and their geometric relationship are controlled by gravity to lock the object into its elliptical, parabolic or hyperbolic orbit.

Bypass

Oscillator potential hides in this example!!

like a "grandfather clock"

an area integral, but not as hard as \(dx\, dy\)

We kind of did this integral in Session 21.

Session 21 intro

pixellation

We made a study of this several weeks ago

bypass

We will not dwell too much on angular acc.

not sure this is correct, but magnitude \(r\) is constant in uniform circular motion The real reason that \(\frac{d\vec{r}}{dt}=0\) is that in uniform circular motion, the position vector \(\vec{r}\) only changes in the \(\hat{\theta}\) or \(-\hat{\theta}\) direction, which are parallel or antiparallel to \(\vec{\theta}\). The cross product of parallel or antiparallel vectors is zero!

A good diagram

Vector product aka cross product, using the right hand rule.

...as in the motorcyclist problem.

Use linear mass densities 8.0g/m and 2.7 g/m.

Use this integral and its associated example to tackle the problem of the one meter rod, using linear mass densities \(\lambda_{iron}=8.0\frac{g}{cm}\) and \(\lambda_{al}=2.7\frac{g}{cm}\)

Try this with \(m_1=1.000\, kg\) and \(m_2=1.500\, kg\). A good study problem.

brain burner!

good study problem

unlike the motorcyclist making a jump

nice

I like this example

We discussed this

yup, this is it.

This is why an astronomer can focus on the trajectory of a star or planet or galaxy as if all of its mass is concentrated at a point located at the center of mass.

bypass this section

Good approximation for an ideal gas, for instance.

Like a set of boxcars in the freight yard hooking together and moving off at a new speed.

Skateboarder interaction

Newton's third law is best understood in this way. The "equal but opposite reaction" is encoded quantitatively by the changes of momentum, or exchanges of momenta.

So, for instance, the moon exchanges momentum with Earth as it travels around its orbit, but second by second, we can compute the amount of impulse, e.g.,

\(\vec{J}=\Delta \vec{p}=\int_{0.0\,s}^{1.0\,s} \left[ \frac{Gm_{\oplus}m_{moon}}{\left[r\left(t\right)\right]^2}\right]dt\)

good example

Sir Isaac Newton's invention, "quantity of motion."

turning points in the motion

Good strategy note

I like this Example!

Correctamente! Slight adjustment, indeed, plus E becomes a component of the total relativistic momentum in four dimensions, \(\vec{p}=\left(p_x,\, p_y,\, p_z\right)\longrightarrow \vec{p}=\left(E,\, p_x,\, p_y,\, p_z\right) \).

FALSE. Ed White made the first US spacewalk in 1965, on the Gemini 4 mission.

typo below?

This would not be allowed in the math department, however.

Very good description of the periodic motion of a string on a guitar. The frequency of oscillation is in the audible range, e.g., A above middle C is 440 Hz.

I hope to do this demonstration during Lecture 16, on Feb. 27.

Bypass

Bypass

Bypass

I can now see why a few of you, on HW 4, used this notation, \( F_{net\ x}=T_{2x} - T_{1x}=0\) rather than the way I usually write it, \(\sum F_x = T_{2x}-T_{1x}=0\).

Notice the subtle change of notation here, using the magnitude w as opposed to the vector notation in Fig. 6.3.

Spider-man problem

Bypass. This section requires Calculus 3 concepts to be done properly, although the weather diagrams, Fig. 6.28 are instructive to view.

Not sure why this paragraph is here.

It also accelerates a tide of circulating blood into the pilot -- blackouts occur

"force of the air on the wing" ↔︎ Many scientists describe aerodynamic lift as the pressure differential \(\Delta p\) between the top and bottom of the wing, effecting every square inch of the wing.

Here is a mini-differential equation

MANY scientific jobs require this skill, numerical solution of differential equations.

about the same as a small 2" sphere

Amazing comparison.

Cf., also the NASA Langley Research Center's wind tunnel fact sheet.

The free body diagram on the right is a bit different than the sketches I used in Session 12.<br> The author's \(\vec{w}_y\) vector is attached at its tail to the mass point, whereas in my sketch, it is slid over and attaches at the tip of \(\vec{w}\). But both are acceptable and useful. As I have mentioned several times, how you diagram forces etc. can be different from the author's and from my way of diagramming, and still be fine.

A good study problem

Our demonstration in Session 10, with the two skateboarders out in the hallway, is good to remember here.

By "changes in motion," Newton meant \(\Delta \vec{v}\) and therefore \(\vec{a}=\frac{\Delta \vec{v}}{\Delta t}\)

<3

Or, once could redefine contact to mean an interaction "at a distance," as the author mentions here.

This is because force is a derived quantity, based on time, space and mass measurements.

Galileo's famous student!

Be careful checking your answers here, too.

It is not clear to me what this expression is supposed to be. Be careful when checking your answer for h. For example,

$$\frac{24\frac{m}{s}}{9.8 \frac{m}{s^2}}=2.49\, s$$

Another planet? No problem. A walk in the park. : )

also known as "bombs" "Accretionary lava ball comes to rest on the grass after rolling off the top of an ‘a‘a flow in Royal Gardens subdivision. Accretionary lava balls form as viscous lava is molded around a core of already solidified lava (photo by J.D. Griggs, 7/2/83, JG2562) (picture #012)."

Another problem that does not come with solution. : )

No solution offered for this problem, I see.

Hmmm. Like the tennis ball classwork problem.

This is a brain burner

An excellent study problem.

WIll be interesting to calculate.

"...give him the heater"

Newton was the first one to work out this idea of projectile becoming a spacecraft in orbit, depending on initial conditions \(\left(x\left( 0\right),\, y\left( 0 \right)\right)\) and \(\vec{v}\left( 0 \right)\).

Would make a nice multiple choice item.

I like this combo problem.

Similar to HW 3, but would be simpler to check.

I like this problem.

"analyze"

However, the author does not complete the square, as we did!

Hang time

I never use this equation. YUKK

This example is like driving a car off a cliff, like in a movie.

https://youtu.be/iUy_61hkGPM

Do not try this at home.

This angle, upon which the author concentrates, is not so important to me.

INCORRECT! The entire curve can be parametrized by \(x\) as we discussed in Session 3 and 4

They are not really independent except mathematically. They are actually coupled. For instance, the horizontal range, \(x_{max}\), is controlled by the size of \(v_y\left(0\right)\)

In the fourth equation here,

$$v_y^2=v^2_{0y}-2g\left(y-y_0\right),$$

if you have eyes to see, you can see the remnants of potential energy and kinetic energy terms.

Note that the author uses a slightly different notation than I use:

1. author, \(-g=-9.8\frac{m}{s^2}\)
2. Dr. Brueckner, \(g=-9.8\frac{m}{s^2}\)

Keep this in mind as you study.

standard assumption for basic physics like PHY2048

...plus the fact that there is no such thing as horizontal gravity!

We discussed this in session 7, concerning uniform circular motion. The direction of the acceleration vector \(\vec{a}\left( t\right)\) is the direction of the \(\Delta \vec{v}\left( t\right)\) vector.

I.e., draw the sketch then let the numbers do the talking

Simply the vector itself, divided by the magnitude of the vector. $$\hat{u}=\frac{\vec{U}}{U}$$

This will be a frequent task in PHY2048

We will discuss this problem on Thursday, among other things.

We will occasionally work in polar coordinates.

Important geometric tool

...except this is what a pilot would use, a distance and a heading.

easy to calculate components from each subdiagram

tail to tail

another important vocabulary term

Important vocabulary term.

Who does this? Measure distance to a fishing spot from your tent? Really!

I just use an arrow, \(\vec{A}\), or boldface B, but not both.

Ummm, no.

:)

This is why we do not emphasize formula sheets and memorization.

Cf., e.g, the index of refraction for BK-7 borosilicate crown glass at various wavelengths.

This is news to me, but accurate. Note that this number of hydrogen atoms is much smaller than \(1.00\, mole\). Curious

This is a polite way of saying, "All science is either physics or stamp collecting," which Prof. Ernest Rutherford said way back in his days of discovering the nuclear structure of matter. (Ernest Rutherford)

This is how most scientists work:

• from easily remembered basic principles and examples,
• then extending to new concepts and previously unseen problems.

(Enrico Fermi)

This mental process will be helpful

1. during our exams, which, by definition, will have previously unseen problems,
2. during future courses deeper into your UCF curriculum, and
3. throughout all of your professional work as a professional user of physics knowledge.

...helpful IF you want to excel. (Albert Einstein)

Professor Galileo wrote, "Philosophy is written in this grand book — I mean the universe — which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language and interpret the characters in which it it written. It is written in the language of mathematics, and its characters are triangles, circles and other geometrical figures, without which it is humanly impossible to understand a single word of it; without these, one is wandering about in a dark labyrinth.” [(Assayer, VI)]

That is our challenge, the mountain we climb: to understand the grand book of nature.

(https://bibdig.museogalileo.it/Teca/Viewer?an=300984&lang=en)

Good example of what makes an image virtual. I always describe it this way: if you place a photographic film at the image location (behind the mirror, in this case) and no light rays actually hit the film, which is true in this case, then it is a virtual image.

Yuge

important

We observed this with the green laser pointer.

$$\frac{\lambda}{2},\:\frac{3\lambda}{2},\text{ etc.}$$

Green laser pointer

In our sketch, wavelet sources \(S_1\) and \(S_2\).

Sketched out in Session 37, https://youtu.be/BF5e89OmsQQ

We did this calculation during Session 37, at about 45:11. https://youtu.be/BF5e89OmsQQ

Yuge

Approximately the B one octave above middle C.

purple

i.e., plane waves

As we sketched plane waves in Session 37.

I.e., a real lens for which thin lens model is just an approximation

Also, adaptive optics, like the COSTAR optics placed in the Hubble Space Telescope to correct spherical aberration.

About aperture, pupil etc.

Yup, I have astigmatism, which I noticed during my first year of college. (Test diagram)

Interesting simulation of spherical aberration in a microscope system. Eyeglasses can have aspherical lenses to correct this aberation:

Here is a coma test panel, comparing stars imaged at the center of a Canon 24mm f/1.4L II lens, at intermediate distance and at the edge of the lens:

At the edge, you get a lot more coma abberation, but at f2.8, it seems minimized.

E,g., the Zeiss apochromat lens that focuses all colors nicely.

"Achieving this level of correction is not simple. It requires the use of exceptional quality optical materials: long-crown glass (fluor crown glass and calcium fluoride) and special short flint glass."

We have discussed chromatic aberration in class a few times.

Schwarzschild radius" of the Sun. We do not think it will ever get collapsed to this small of a size, because the Sun is not large enough to have a supernova detonation.

This is the classical way of thinking about black holes, and a century after Sir Isaac Newton wrote Principia, another scientist, John Michell, worked it out using non-relativity concepts.

Escape velocity:

1. from the surface of Earth, \(11,200\frac{m}{s}\)
2. from event horizon of a black hole, \(c=3\times 10^8\frac{m}{s}\)

Diameter of DNA helix is about 2 nanometers. So wavelengths on the order of a nanometer are good to use. Rosalind Franklin probably used \(\lambda = 0.15\:nm\) for her xray diffraction images.

Huge

E.g., our handheld diffraction gratings,

with 500 lines per millimeter. That is about one line every 2000 nm, so just a few times larger than the typical Roy G. Biv wavelengths.

ok

This is like the difference between Newtonian physics and quantum physics.

"only"

80 MHz is just below our FM radio band and 2 GHz is in the microwave region of the spectrum.

Used to be very highly classified technology.

Isaac Newton developed this concept.

Second flip = nice!

The best way to think of this idea is the concept of aperture: the pupil of your eye, a few millimeters, vs., e.g. the aperture of the Hubble Space Telescope (HST), about 2.4 meters. HST is about 300 times wider, about 90,000 times more area than the human eye's pupil.

The HST is about the size of a school bus.

(Hubble being transferred from the Vertical Assembly Test Area (VATA) to the High Bay at the Lockheed assembly plant in preparation for transport to the Kennedy Space Center (KSC) after final testing and verification.)

We will conclude Ch. 26.4 with this example, then bypass the remainder of Ch. 26.4

Notice that, in this diagram, the rays are almost parallel coming out of the eyepiece to the eyeball. If the rays are parallel, your eye does not have to strain to focus the image on your retina

Or just one in a simple microscope

Always a big proviso.