Download 28 Aug 2006 (First Class)

28 Aug 2006 (First Class)
Introductions. Josh, Eli, and Peter, too. Susie Hauger? Go through the class list.
Hand out and go through the course outline and syllabus and ask for questions.
Chapter 1 “Measurement”: Encourage them to read the chapter
• Emphasize SI units, but mention CGS and discourage British
• Precision and Significant Figures
• Dimensional analysis(!)
Example: What is the “radius” of a black hole with mass M? Have G with [G]=L3/T2M
and can also use c, with [c]=L/T. Then need L=MxGycz=Mx-yL3y+zT-2y-z so x=y, 3y+z=1,
and z=-2y. This gives y=1=x so z=-2, and “radius”=GM/c2. (Actually missing factor of 2.)
Laboratory Activity
Using the cart with the fan. Get used to using the motion detector, measure x(t) and
show that it looks parabolic. Derive the velocity graph and value for the acceleration.
Extension: “How do you know the acceleration is correct?” See if they will get the idea
of tilting the cart until the cart and fan are stationary. Measure the angle, and see if
a=g*sin(angle). Good questions about the uncertainty in measurement should result.
31 Aug 2006
Do (1) vectors, (2) 1D kinematics, and (3) F=ma during “class” time. Then break, and do
(4) discussion of analysis (including uncertainties) for first lab. Time to take more data.
Vectors
Refer students to Secs.2-1 and 2-3. “You should be familiar with this.” For now, think of
vectors as a shorthand for keeping track of orthogonal directions (“components”) which
behave independently of each other, i.e. “bookkeeping”. Allude to a “greater meaning.”
One Dimensional Kinematics
First stress x and then x(t). Draw a plot and show the slope v(t)=dx/dt. Introduce “x-dot.”
Next do a=dv/dt=d2x/dt2. Introduce “v-dot” and “x-dot-dot.”
Emphasize “motion under constant acceleration” (Section 2-5). Do some integrals:
dv
a=
dt
Z
dv =
Z
adt
v = at + v0
1
x = at 2 + v0t + x0
2
Give them another way to integrate the kinematics. (Call it the “energy method?”)
v̇ = a
vv̇ = va = ẋa
d
dt
!
"
d
1 2
v = xa
2
dt
v2 = v20 + 2da
Typical application: Freely falling bodies (Section 2-6)
Newton’s Second Law
Some philosophy: Kinematics vs dynamics and “Equation of Motion”
F=ma=mx-dot-dot. “Differential equation”
“Free fall” F=mg; Weight vs Mass; Tell them about the equivalence principle
Free body diagrams: Show them a mass hanging from one or two strings
Discussion of analysis for “acceleration” lab
Do a free body diagram for mass on track, flat and also inclined. Indicate how this is
used to check their answer.
Error analysis: (1) Reading the graph for acceleration and (2) error in vertical height of
triangle for tilting to check answer. For (2) what about the error in the length? Get into a
discussion of dominant uncertainties.
7 Sept 2006
Hand in homework! Check syllabus updated for next week’s assignment.
Read up on laboratory activity for Monday; some notes will follow.
Newton’s Laws in Three Dimensions
Make use of vectors for bookkeeping. Projectile motion is a good example. It is “free
fall” in y-direction, but motion with no force (in absence of drag!) in x-direction.
Drag Forces
Friction with a viscous medium. Force depends on velocity and acts in the opposite direction to motion. For many objects and media, good approximation is D=bv for drag
force, where the constant b will depend on the object’s size and shape.
Integrate equations for motion with a drag force. It’s a little tricky: Start with F=ma
⇒ −bẋ = mv̇
−b(x − x0 ) = m(v − v0 )
mv0 − b(x − x0 ) = mẋ This estab-
lishes that x=x0 when v=v0. At this point, we might as well set x0=0. Now integrate:
!
"
dx
m
b
dt =
t = − log v0 − x +C
v0 − bx/m and so
b
m
where C is an integration
constant. Want to have v=v0 when t=0. So choose C=(m/b)log(v0). Then
!
"
m
b
t = − log 1 −
x
b
mv0
x(t) =
"
mv0 !
1 − e−bt/m
b
. Check initial conditions. Also find x(t) and v(t):
v(t) = ẋ(t) = v e−bt/m
0
and
. Discuss the kind of
motion this equation describes. Plot x(t) and v(t). Want to make a big deal out of the fact
that this only makes sense if v0 is positive and nonzero.
Falling bodies with a drag force. This leads to “terminal velocity” which is simple to derive, namely mg=bv. This is a way to determine b. See table 4-1 in textbook. This is also
a way to “check your answer” for deriving a formula for a falling body y(t). F=ma gives
mg − bẏ = mv̇y
, where “positive” is “down.” The “constant” term makes the integration
dt =
dvy
g − bvy /m
even trickier, so we will take a short cut to the velocity:
(Eq.4-20 in
the textbook). Starting with zero initial velocity at time t=0, we have the same integration
"
mg !
−bt/m
vy (t) =
1−e
b
as for x(t) above. That is
, Eq.4-22 in the textbook.
Notice that as time increases to infinity, we get the correct terminal velocity!
Maybe take a break here? Want to finish up with enough time so that students can
make it to the picnic, and I can get over to the meeting with Sam Heffner!
Uniform Circular Motion
Emphasize velocity as a vector, which can change by changing its direction and not its
magnitude. Change in velocity is acceleration, so there has to be some force involved.
Figure 4-16 in the textbook shows it all. Draw two points on circle, each ϑ left or right
away from vertical, then tangent velocity vectors at that point. Difference in velocity vectors is 2vsin ϑ. Note that the direction of the difference is radially inward. Time difference
is path length s=2r ϑ divided by v. So, magnitude of average acceleration is
v2 sin θ
r θ
. Then take limit as ϑ goes to zero. Spend some time on Taylor series?
Lingo: Let them know that this is called “centripetal acceleration.”
The Conical Pendulum
Nice example of “tension” as a force, vectors, and circular motion. Figure 5-18 in the
textbook has all the details. The question to ask: “What is the period of this pendulum?”
Will get some demos from Scott to show them. One quarter length leads to half the period, doesn’t depend on mass, that sort of thing.
Draw the free body diagram (Fig.5-18b). Clearly Tcosϑ =mg. Also Tsinϑ =mv2/R. Would
rather work with the length L of the pendulum bob, so take sinϑ=R/L. Cancel T, and also
m, to get gsinϑ/cosϑ=v2/R. For period t this gives
t2 =
!
2πR
v
"2
= (2π)2 L sin θ
!
L cos θ
cos θ
t = 2π
= (2π)2
g sin θ
g
or
L cos θ
g
. This is
Eq.5-15 in the textbook.
Examine the limits. It is interesting for ϑ=0. (It is not zero, but is the same as for the ordinary pendulum, which we’ll study later.) For ϑ=90o it gives t=0. What does that mean?
11 Sept 2006
This class is dedicated to acquiring data for an experimental investigation motion involving frictional forces and two “free bodies.” It is modeled after Activity 04 for the regular
Physics I classes, and you should review that write-up and download the logger pro file
for that activity. You should also review the write-up of Activity 03 on motion with friction.
You will take data on the motion of a cart along a horizontal track, the same way you did
with the cart/fan activity with which you worked before. In this case, however, the cart
will move because it is attached to a string, which stretches over a pulley to a hanging
weight, which falls. The hanging weight and the cart are couple together by the string,
which exerts a tension force on both the cart and the weight. The hanging weight is also
acted on by its gravitational force, and the cart is acted on by a frictional force.
You can change the mass M of the cart by placing one or more of the large rectangular
bars into the tray on top. You may also choose one or more different hanging weights
(mass m), which you attach to the string by hand. The acceleration a of the cart will depend separately on both M and m. Note that the frictional force on the cart (which opposes the motion induced by the tension T in the string) depends on the coefficient of
kinetic friction μ, which you should be able to determine from your measurements.
Work out the equations which give you the acceleration a of the cart, in terms of M, m,
and μ. The tension T will disappear from the equation when you consider also the expression for the acceleration of the hanging mass, which you can assume is the same
as a. (What assumptions are you making when you use the same T and a for the cart
and for the hanging mass?)
You should use your logger pro data to determine a two ways. The first way is the same
as what you did for our first experiment. That is, use the program to follow the cart while
it is accelerating, and determine a from the second derivative of the displacement as a
function of time. The second way is to make use of our integrated expression which
eliminates time and combines velocity v, acceleration a, and total displacement d:
v2 = v20 + 2ad
This is possible because the hanging weight will hit the floor before the cart moves the
length of the track. As soon as this happens, the cart stops accelerating, so you use
logger pro to follow it afterwards to get the velocity v. The distance the cart travels during the time that it accelerates is the total displacement d. You will need to do some
work to combine your various uncertainties to check that the two values you get for a
are consistent with each other.
Once you decide which method give you the more precise value for a you can use your
data for different masses M and m to get the coefficient of kinetic friction μ. Note that
you can check this value using hints from the write-up of Activity 03.
14 Sept 2006
Homework: Hand it in!
Two topics for today: (1) Momentum, and (2) Center-of-Mass. Break in between. Will not
cover “Systems with variable mass” (i.e. we will skip Sec.7-6)
(Linear) Momentum
Vector!! p=mv; Newton said F=dp/dt; Generally take m constant, so F=mdv/dt=ma
J! ≡
Z
i
f
!Fdt =
Z
i
f
d!p = !p f −!pi = ∆!p
Collisions: Draw F12(t) vs t; Then prove “Conservation of Momentum” in collisions, i.e.
!F12 = −!F21 ⇒ ∆!p2 = −∆!p1 ⇒ ∆!p1 + ∆!p2 = ∆(!p1 +!p2 ) = 0
“Linear Momentum and Translation Symmetry”: Remind them that
!p = m!v = m
d!r
and !r →!r" =!r +!r0 ⇒ !p → !p" = !p
dt
“Linear Momentum and Boosts”
!v →!v" =!v −!v0
!r = xî + y jˆ + zk̂
“Symmetry!”
!p → !p" = m!v − m!v0
!p! = 0
You can boost to a “frame” where the total momentum
Good example can
be had in the collision of two bodies. For reasons to be seen (2nd period) this is called
the “center of mass” frame, but better here it should be called “center of momentum”:
!p = !p1 +!p2 − (m1 + m2 )!v0 = 0
!
!v0 =
so
m1!v1 + m2!v2
m1 + m2
“Collisions”: Explain Fig.6-17, redraw, emphasize this is shown in CM frame, and also
that different conditions all represent conserved momentum. (Energy? Later!!)
Center of Mass (aka Center of Momentum)
Consider a system of N particles. The “center of mass” is defined as the position
1 N
m1!r1 + m2!r2 + · · · + mN!rN
!rcm ≡
= ∑ mn!rn
m1 + m2 + · · · + mN
M n=1
Note that
!vcm =!v0 , i.e. the velocity needed to boost to the “center of mass” frame.
Now consider a solid object. Replace m with dm, and sum with integral sign. That is
Z
Z
!rcm =
1
M
!rdm =
1
M
!rρdV
Example 1: Linear rod with length L and constant density. Cross sectional area A.
dV=Adx
1
xcm =
M
M=ρAL
Z
0
L
1
L2 L
xρAdx = ρA =
M
2
2
Right!!
Example 2: Linear rod with length L and density increasing from zero by x2. This is like
having a (homogenous) cone, with the area increasing from zero to πR2 at the rod end.
M=
Z
L
dm = ρ
Z
L
π(xR/L)2 dx = ρπR2
0
0
First calculate the mass:
that the dimensionality of the result is correct!! Always keep this in mind!
L
3
Notice
Now calculate the center of mass. (Introduce the term “first moment”??)
1
xcm =
M
Z
0
L
1 R2 L4 3
xdm = ρπ 2 = L
M L 4
4
Note that this “halfway point” in terms of mass is 2L/3 (prove it!), which is farther to the
left than the center of mass. That’s okay. If you imagine the the CM as the place from
which you hang it and it doesn’t twist (we’ll talk about torque and rotations later), then
you expect the CM to be to the right of the halfway mass point, because the lever arm
can be shorter for the bigger mass.
18 Sept 2006
There’s a lot of stuff to do today! Want to get through it all, then gravity on Thursday,
with some time left over for a bit of review.
Work and Kinetic Energy
W=Fs=Fxcosϑ (for constant force); Figures; Work done “by” and “on”. Can be negative!
“Dot product”: See pg 233/234; Emphasize “projection” and use unit vectors to derive
!A · !B = Ax Bx + Ay By + Az Bz
W=
dW = !F · d!s
Variable force:
W = !F ·!s
so we write
Z
2
1
dW =
Z
x2
x1
F(x)dx
for 1D case.
Example: Constant force (obvious)
Example: Spring force F=-kx;
Talk about Hooke’s Law!
1
W = − kx2
2
Proof of the “Work Energy Theorem” in 1D. (Argue generality with parallel and perps.)
W=
Z
2
1
F(x)dx =
and we write
Z
2
1
K=mv2/2
dv
m dx =
dt
Z
1
2
dv dx
m
dt =
dt dt
Z
1
2
"
!
1
d 1 2
1
mv dt = mv22 − mv21
dt 2
2
2
so W=K2-K1=ΔK. This is called the “Work Energy Theorem.”
WE Theorem is independent of reference frame! Consider a boost to v′=v-v0:
∆K ! = ∆K − v0 (mv2 − mv1 ) and
W =
!
Z
Fdx =
!
Z
F(dx − v0 dt) = W − v0
Z
Fdt = W − v0 ∆p
In other words, if then W=ΔK then W′=ΔK′. Aka “Gallilean Invariance.”
Recall “elastic collisions”: Were defined as initial and final speeds same in CM frame.
Now define as kinetic energy before equals after. Independent of the reference frame!
Break Time
Potential Energy and Energy Conservation
Imagine there exists a function U(x) for which F(x)=-dU/dx. Then W=∫ Fdx=-ΔU. But
does this make sense? Consider moving from 1 to 2 along two different paths, a and b.
(Like Figure 12-4, but I changed notation to be consistent.) For W=-(U2-U1) to make
sense, we must have W(Path a)=W(Path b) regardless of the paths. In other words, it
only makes sense if the force F leads to “path independent” values of the work.
Let path b be traced backwards, going from 2 to 1. Then the value of W changes sign
but the magnitude is the same. So, coming back to 1 results in zero net work. We write
I
that “the integral of the work around a close path is zero”, i.e.
!F · d!s = 0
.
Examples: Gravity has U(y)=mgy and a spring has U(x)=½kx2.
Note: It doesn’t work for friction! Work around a closed path is obviously not zero!
“Total Mechanical Energy” E=K+U is a “conserved quantity”:
!
"
d 1 2
dE
dU dx
=
mv +U = mvv̇ +
= v [mv̇ − F] = 0
dt
dt 2
dx dt
Discuss motion in a general potential well:
Conservation of Energy (in general): Postulate some form of “internal” energy, different
from mechanical. Two ways to realize this. One is “heat energy”, actually mechanical
motion of the atoms and electrons in matter, aka “thermal energy.” Another is concept of
“binding energy”, in fact a conversion of matter to energy through E=mc2. We then talk
about the work done by all external forces giving rise to W=ΔK+ΔU+ΔEint. It is also possible to transfer energy using “heat” instead of “work.” We will study this when we move
on to the phenomenology and “laws” of thermodynamics.
21 Sept 2006
As usual, homework to Josh right away!
Note: HW Due Monday! (Two exercise on “gravitation”.) Exam #1 in one week.
Today: Newton’s Law of Gravity. Spend some time talking about the dual role of mass,
equivalence principle, Einstein, and so forth. Say it in words first, i.e. “The force between two bodies is proportional to the product of their masses, inversely proportional to
to the square of their separation, and directed along the the line between them.” Then
!F = G M1 M2 r̂
r2
r̂ =
!r
r
give them
where
with a diagram.
Can emphasize in terms of “12” or “21” subscripts on F and unit vector for r.
Mention G as a “proportionality constant.” Also talk bout the unit vector in the r direction.
Mention that “inverse square” laws (like gravity, Coulomb) are now understood in terms
of quantum field theory and three spatial directions. “It had to work out that way.”
Gravitation near the Earth’s surface. Assume a mass m. Then F=GMEm/r2 where r is
really just the Earth’s radius RE. Therefore g=GME/RE2. Mention Cavendish experiment.
Variation with altitude? Sure. Remind them that (1+x)n=1+nx when x is very small. We
will use this in a little while (today) after we have the concept of gravitational potential.
Shell Theorems. Won’t prove these, but they are very useful. Not hard to buy the idea,
based on symmetry and vector forces and so on.
Shell Theorem #1: A uniformly dense spherical shell attracts an external particle as if all
the mass of the shell were concentrated at its center. See Section 14-5!
Shell Theorem #2: A uniformly dense spherical shell exerts no gravitational force on a
particle located anywhere inside it. See Section 14-5!
Gravitational Potential Energy. Spend some time talking about the path independence
of gravity. See Figure 14-11. Pick a path which goes straight between points 1 and 2, so
W1→2 =
Z
1
2
!F!r = −
Z
1
2
Fdr = −
Z
r2
r1
!
"
! "#r2
1
1 ##
1
GMm
dr = −GMm − # = +GMm
−
r2
r r1
r2 r1
Now use W=ΔK and ΔK=-ΔU so ΔU=-W. Agree that U=0 at r2=∞. Then we write
U(r) = −
GMm
r
Now imagine the change in potential energy near the Earth’s surface. Imagine that you
go from the surface up to a height h. Then
!
1
1
−
∆U = U(RE + h) −U(RE ) = −Gmm
RE + h RE
"
!
"
GME m
1
=
1−
RE
1 + h/RE
(Note that the book uses y instead of my h.) Now h is much smaller than RE, so we can
use 1/(1+x)=1-x to good approximation. So
!
"
h
GME m
1−1+
= mgh
∆U =
RE
RE
which we know is the right answer!
Interlude: Should we spend some time on Taylor expansions, and where these expressions for small things come from? Will do this on Monday for sure when we talk about
the Euler formula for exponentials, but not sure now.
Escape Velocity. Launch a rocket from the Earth’s surface, straight upward. Does it fall
back down or does it escape? Answer this with conservation of energy. At launch it has
energy ½mv2 and potential energy -GMEm/RE. For “escape velocity” it will go fast
enough so that when it is at “infinity” it is no longer moving with any velocity. Therefore it
has no energy. In other words, the total energy is zero. We have
GME m
1 2
mvescape −
=0
2
RE
vescape =
so
!
2GME
RE
.
Obviously you can do this for planets and other things.
Circular Orbits. Just use F=ma but here a=v2/r is the centripetal acceleration, and F is
GMm
v2
=m
r2
r
v=
the force due to gravity. So
and
This becomes Kepler’s laws, especially after doing it for elliptical orbits.
!
GM
r
Aside: Dark matter and the behavior of galactic rotation curves. (See next page.)
.
25 Sept 2006
Collect homework!
First exam is Thursday! Open Book, Open Notes, ...
Today we begin a study of “oscillations.” Expect some syllabus changes.
The Spring Force: Energy Considerations
Recall “Hooke’s Law”: F=-kx where x is the “extension of a spring.” Generalizable!! We
∆U = U(2) −U(1) = −
Z
x2
1
1
Fdx = kxx2 − kx12
2
2
x1
already have the potential energy:
Measure from x1=”no extension”=0, i.e U(1)=0 and x2=x so U(x)=½kx2.
Now consider a mass m attached to the spring. E=½mv2+½kx2=constant. Emphasize
that both x and v are squared. So v is max when x=0 and vice versa. “Oscillation!”
Draw the harmonic oscillator potential well and illustrate it that way.
The Equation of Motion
More information in the exact form of x(t). Want to solve equation of motion, i.e. F=ma,
2
ω2 ≡ k/m . The solution is the function x(t).
or −kx = mẍ or ẍ = −ω x where
Try to solve it in your head. Think of some function f(x) (different x!!) so that when you
take the derivative twice, you get the same function back with a minus sign. Then the
solution will be x(t)=f(ωt). Get them to realize there are two such functions, sin(x) and
cos(x). Of course! Second order equation, so need two “constants of integration.”
So x(t)=a cos(ωt) + b sin(ωt) Determine a and b from “initial conditions.” Start at t=0
with x(t=0)=x0 and v(t=0)=dx/dt=v0. So a=xo and bω=v0 and x(t)=x0cos(ωt)+(v0/ω)sin(ωt).
Different way of writing this same solution. Instead of a and b use A and φ where
a=Acos(φ) and b=Asin(φ). (Easy to write equations that relate a and b to A and φ.) Now
use the trig identity cos(X-Y)= cos(X)cos(Y)+ sin(X)sin(Y). This means we write
x(t)=Acos(ωt-φ).
Discuss motion in these terms. Call A the “amplitude” and φ the “phase.” Easy to see
the meaning of amplitude. Phase is a little harder. Draw two sine curves, one with zero
phase and one with non-zero and positive phase. Point out that they will learn more
about this in the “oscillation” laboratory work.
Break at this point. Will learn a fancier way (using Euler’s formula after talking about
Taylor expansions in general) after the lab next Monday, so starting following Thursday.
Use remaining class time for questions, prior to the midterm exam, if they want.
5 Oct 2006
Note updated syllabus with homework assignment due next week.
Quick Review of Exam 1
Don’t spend more than ten or fifteen minutes on this. Learn the stuff, and keep in mind
the places you tripped up. It will appear again on the final exam.
1. “Free Body Diagram” plus “Centripetal Acceleration” plus “Conservation of Energy”
plus “ΣF=ma.” Note that tension T does no work because it is always perpendicular
to the direction of motion. Answer is T=3mg. Learn how to get this answer!!
2. We did part (a) in class. Again need “ΣF=ma.” Then dE/dt=-bv2. Energy is lost!
3. Know this one, example straight from class and homework. “Mass per unit length!”
4. Simple use of stuff from class. Emphasize that (b) and (c) are the same question!
Math: Taylor Expansions and “Applications”
We will return to “oscillations” after break. For now, a math “review.” (Hopefully!)
Taylor expansion of y=f(x) about x=x0. Put y0=f(x0). Slope at (x0,y0) is (y-y0)/(x-x0)=f’(x0).
Then y=f(x)=y0 +f’(x0)(x-x0)=f(x0)+f’(x0)(x-x0). “Linear Approximation for f(x).” Can extend
this procedure to higher powers. (Won’t do it, but check a math textbook if you want.)
f (x) = f (x0 ) + f ! (x0 )(x − x0 ) +
1 !!
1
f (x0 )(x − x0 )2 + f !!! (x0 )(x − x0 )3 + · · ·
2!
3!
Example: Expand f(x)=(1+x)a about x=0. Get f(x)=1+ax. Useful linear approximation!!
Make it clear that this doesn’t depend on the sign or magnitude of a.
More Examples: Do sin(x) and cos(x) and ex and draw the similarities.
Now introduce i2=-1.Then get to eix=cos(x)+i sin(x).
Time for a break
Back to the Physics of Oscillations
ω2 ≡ k/m . Recall solutions
2
Recall F=ma, or −kx = mẍ or ẍ = −ω x where
written as x(t)=x0 cos(ωt)+(v0/ω)sin(ωt) or x(t)=Acos(ωt-φ).
Now we discuss a different way to write the solution. Emphasize that the derivative of ex
is ex, so derivative of eax is aeax. Then second derivative of eiax is -a2eiax.
So, it is natural to write x(t)=a1eiωt+a2e-iωt. Work through initial conditions and show that
it turns into x(t)=x0 cos(ωt)+(v0/ω)sin(ωt) so it is all the same as before. This will lead to
the relations 2cos(ωt)=eiωt+e-iωt and so forth.
Probably good to mention “real part” is always assumed when you write the solution this
way.
Let them know that this will be a very useful technique, and it will be used a lot in future
course in physics and engineering.
Example: Damped harmonic motion from
ω2 ≡ k/m
mẍ = −kx − bv = −kx − bẋ Rearrange
β ≡ b/2m
ẍ + 2βẋ + ω2 x = 0 . Now try
0
with some definitions 0
and
so
an “ansatz” x(t)=Aeiωt. (Is ω=ω0? Keep going forward and we’ll find out.)
A(−ω2 + 2iβω + ω20 )eiωt = 0 so
!
ω = iβ ± ω20 − β2
. Continue the
discussion, including a few words about “real part”, “amplitude”, and “initial conditions”.
Point students to Eq.(17-39).
Maybe we’ll do some “forced oscillations” and “resonance” next Thursday. Let’s see.
Homework Assignment due Thursday 12 Oct 2006
1. Perform a Taylor expansion of the function f(x)=x2 about the point x0=1, writing down
all nonzero terms. Show that the result is is just a fancy way of writing f(x)=x2.
2. The “hyperbolic cosine” function is defined to be cosh(x)=(e+x+e-x)/2 and the “hyperbolic sine” function is sinh(x)=(e+x-e-x)/2. Show cosh(ix)=cos(x) and sinh(ix)= isin(x).
3. Chapter 17, Problem 12. (Recall the lab we did on Monday 2 October.)
4. Chapter 17, Exercise 44. (Damped harmonic motion, with numbers.)
Laboratory 10 October 2006
This paper introduces the laboratory exercise on “oscillations” which will form the largest
basis on which your lab books will be graded. We’ll start with a review of the laboratory
exercises which you’ve already seen, with some idea of what should be in your lab
books for these.
You are welcome, in fact encouraged, to do more work on the past labs some time prior
to handing in your lab books for a preliminary look (Oct.26th) and final grade (Nov.30th).
Remember that the materials I handed out then, and the summaries below, are just
meant to highlight guidelines. I am anxious to see what other things you discover that
you can investigate in these laboratories.
Cart/Fan Acceleration Measurements (28 Aug)
You first measured the acceleration of the cart on a track, with a fan attached to the cart
which provided a force. The acceleration with the fan, call it af, was determined from a
plot of position x(t) as a function of time taken with the sonic ranging device and your
computer. The program doubly differentiates the position to get a constant (?) value for
the acceleration.
You then also were asked to “check to see if you got the right answer” by raising the end
of the track just enough so that the force due to gravity cancels out the force from the
fan, and the cart stands still. The acceleration ag,=gsin(θ) should be the same as af,.
An important aspect is to use “uncertainties” when comparing af, and ag. The former has
uncertainties from the way you read the data from your computer plot, or otherwise get
it from the sonic ranging data. The latter has uncertainty from the precision with which
you can measure the tilt angle θ. The two should agree “to within the uncertainties.”
Cart/Hanging Mass Acceleration Measurements (11 Sept)
This exercise had you measuring the acceleration differently, by determining the velocity
after the cart accelerates a little while, and the distance over which it travels, before the
hanging mass hits the floor and acceleration ceases. With a zero initial velocity, there is
a simple formula that relates the final velocity to the acceleration and distance traveled.
I think I called this the “energy method” for reasons that should since have become
clear to you. (If not, ask about it!)
Again, you determined the acceleration two ways. Once using the double differentiation
of the x(t) data, and again using the “energy method.” Compare them, once more doing
what you can to estimate the uncertainties.
Mass Hanging from a Spring (2 Oct)
This was a “warm-up” exercise for what you’ll start today. Using the standard Physics I
setup with a mass hanging from a spring, you set the system into oscillation (“simple
harmonic motion”) and measured its motion with the ranger and computer. From this
you could determine the (angular) frequency ω of the oscillation.
You were also able to predict the correct answer for the frequency by measuring the
spring constant k. This is from observing how much the spring stretches for a certain
mass. (Some of you learned that it’s a good idea to check to see if the spring stretches
the same amount when you add 200g to the 50g hanger, and then another 200g to the
250g already there. Why?)
Do your measured and predicted values of ω agree to within uncertainties?
Oscillations of a Cart/Mass/Spring System (Today)
Now you will start the most ambitious of the laboratory exercises. You can work with it
today, do some analysis and think about it over the next week, and take more data with
the same setup. Again, here are some guidelines that you should consider while taking
and analyzing your data.
Your cart now has two springs attached, one on each side. A metal strip is magnetically
attached to the cart, extending to one side. This is what will intercept the beacon of the
sonic ranger. Some of you have spring with different spring constants, and there are
more of everything in the bins at the front of the classroom. You also have the long,
black metal weights which you can add to the cart.
This setup is fraught with sources of systematic error. Pay attention to your data and
think about what you should be checking up on.
Your goal, ultimately, is to determine both the mass of the cart and the spring constant,
just from a measurement of the motion of the oscillator. You can do this by observing
the frequency as a function of the number of black metal weights added to the cart. Try
to come up with a relationship between the frequency and the “added” mass, and use
this to figure out how to plot your data so that you can determine the mass of the cart
and also the spring constant.
Today is for exploration, for asking questions, and for starting to think about how to get
to your ultimate goal. Don’t forget about sources of uncertainty, but it is okay if you’d like
to see how things compare first. You’ll have a scale at your disposal next week to get
the mass of the cart and of the black weights, but for now you can do everything in
terms of “units of black weight” mass. That probably sounds strange to you, but think
about and ask questions.
12 Oct 2006
Homework in. Note new assignment posted, due next Thursday.
Today: Kinematics and Dynamics of “Rigid Bodies”
Kinematics: Draw a figure of an axis of some body with an axis. Emphasize “one axis”
for now. Now “position” variables x, v, and a become “angular” coordinates φ, ω, and α.
“Constant angular acceleration” just change the names! Note the new units, though.
Same ω as we used for harmonic motion? Do x=Rcos(φ)=Rcos(ωt).
Vector quantities?! Do the “rotate the book” demo to show that rotations don’t commute,
so we can’t make φ a vector. However, small rotations look better. Insinuate that infinitesimal rotations do commute. Hence we could make a vector out of dφ, so make ω and
α vectors, with direction “along axis” via Right Hand Rule (sense is arbitrary, but axis is
the only non-biased direction you can use.) Use pictures.
Relationship between linear s=rφ and angular φ variables. Emphasize tangential speed
vT=rω and that “radial” speed is zero. Not for acceleration,i.e. aT=rα and aR=vT2/r=ω2r.
Dynamics: Start with F=ma on a point constrained to rotate about an axis, Fig.9-2. Want
to study α so focus on aT. No effect from constraining force. Tangential acceleration is
driven by FT=Fsin(ϑ)=maT=mrα. Define “torque” τ=rFsin(ϑ) so that τ=Iα where I=mr2 is
called the “rotational inertia” or (to my generation) “moment of inertia.”
Torque as a cross product. Define, show magnitude correct, and direction consistent.
Break Here
Rotational inertia of a collection of particles is I=Σmr2. Present and prove the “parallel
axis theorem” I=ICM+Mh2. Proof:
I = ∑ mi ri2 = ∑ mi [xi2 + y2i ] = ∑ mi [(xi! + xcm )2 + (y!i + xcm )2 ] = ICM + Mh2
i
i
i
after reminding students of the definitions of xCM and yCM.
Rotational inertia for solid objects: I=Σ(δm)r2 becomes I=∫ (dm)r2=∫ r2dm. Note: This is
the “second moment” of the mass, whereas CM is the “first moment.” If time, do this for
a rod with axis through center, through end, and show that parallel axis theorem works.
Important: Emphasize Fig.9-15 and that you will likely need these to do problems! I
suggest that you spend some time deriving one or two or more of these.
“Center of Gravity is the same as Center of Mass:” See Section 9-4. (Time to derive?)
The Pendulum: Figure 17-10 and text, including “small angle approximation.” Point out
correction term (Eq.17-25) and interesting exercise in laboratory.
Then do “Physical Pendulum”, Fig.17-11 by solving differential equation in angular coordinates. Result is T=2π(I/MGd)1/2. Point out that measuring period can give the rotational inertia about any axis.
Laboratory 16 Oct 2006
This is a continuation of the work you started in the laboratory on 10 October last week.
Based on the data you took last week, and the analysis of that data, you should have
the following results so far:
• For one pair of springs, measurements of the frequency for different “added” masses
• A plot of 1/ω2 as a function of “added mass” showing a straight line dependence
• Values for the “spring constant” k and the mass of the cart, derived from the slope and
intercept of the straight line above. Try to come up with values for the uncertainties on
these quantities, from your estimate of the uncertainty on the slope and intercept.
• An independent measurement of k from the extension of the spring from some hanging mass(es). Estimate the uncertainty in k from the extension measurements.
• An independent measurement of the cart mass, using one of the scales in the classroom. (Maybe you used both scales. The difference is a measure of the uncertainty.)
If you didn’t get all this done, use some of the time during this class to get this far.
After this is done, you should have discovered two things. First, the spring constants
you measured above should differ from each other by a factor of two. Why? You should
explain your answer by using the equation of motion for the cart, and not just say “it’s
because there are two springs.” Correcting for the factor of two, are the results the
same within your uncertainties? Did you measure the spring constant of each of the two
springs, or did you make the assumption that they were identical? If the latter, maybe
you want to check and see how good is that assumption.
Your second discovery should be that the mass of the cart you derive from the straight
line is more than that you got from the scale(s). Maybe 10% larger, maybe more, but in
any case, well outside your experimental uncertainties. Can you think of a reason for
that? Can you think of a way (or ways) to check your reasoning, at least to see if it is a
plausible explanation?
19 Oct 2006
Note: One problem assigned for next Monday! Prelim lab books due on Thursday.
This class: Basics of angular momentum, but then time to work on the lab again.
Angular Momentum
Very important concept in all of physics. Just scratching the surface here.
!" =!r ×!p
First, one particle. Need to have an “axis.” Use origin:
Magnitude is
l=rpsinϑ=r p=rp . Refer to Fig.10-1 and draw it. Next connect it to torque:
d!"
d!p
=!v ×!p +!r ×
=!r × ∑ !F =!τ
dt
dt
but emphasize order of derivatives and why the
first term is zero.
Nice emphasis that it is an “alternate description:” Do the freely falling particle using an
impact parameter b for the origin and show that it all works. (Sample Problem 10-1.)
Now system of particles. Use “total angular momentum” by adding them all up. So
!L =
N
∑ !"n
n=1
N
d!L
= ∑!τn = ∑!τext
dt n=1
and
since all the “internal” torques cancel. This
is analogous to Newton’s second law, an is indeed just an “alternate description” but it
gives counter-intuitive results because of “hidden” internal forces. Bike Wheel Demo!
Now rigid bodies. Focus on one piece of a rigid body (Fig.10-7) and the z-component of
angular momentum. Look “from the top.” Then lz=rp=rmv=rm(rω)=mr2ω=Iω. Adding up
all the pieces of the rigid body, ω is the same for all pieces and I becomes the moment
of inertia of the entire body. That is, L=Iω.
Conservation of Angular Momentum: No external torques, then angular momentum is
conserved. Analogous to linear momentum conservation. Read Sec.10-4. Do the demo
with the turntable and weights in the hands of some willing volunteer.
On to the laboratory...
23 October 2006
Turn in your homework! Preliminary lab books due on Thursday!!
Today: Waves, from a more or less formal point of view. (Basically following Chap.18.)
What is a Wave?
Spend some time getting them to think of waves with which they are familiar. Water
waves, sound waves, light waves in that order. Go from “bobbing up and down” to the
idea of “pressure as a wave” (including seismic waves, which could also be transverse!)
to the (rather bizarre, when you think about it) phenomenon of light as waving (but interconnected) electric and magnetic fields. We will only discuss “mechanical” waves now.
What do they have in common? Energy is transported from one place to another, but
not the “stuff” that is waving. (Read DaVinci quote on page 401.) Maybe they’ll have
some fun figuring out what the “stuff” is for light waves, but leave them hanging on it.
General Properties of Waves
1. Direction of particle motion, i.e. transverse or longitudinal.
2. Dimensionality. String/spring (1 trans/long); water (2); light/sound (3 trans/long)
3. Periodicity, i.e. is it periodic or not? A “pulse” is an extreme example of “not”
4. Shape of wavefront. Do 3D, i.e. planar or spherical. Introduce idea of “ray”
Good figures (18-1,2,3) in textbook.
Mathematics of a Traveling Wave
Use the string paradigm for this discussion. Shape of string is y(x) at any given time t,
but it will change with time. So, better to talk about shape as y(x,t). At this point, restrict
the discussion to an “ideal string” with no losses due to friction.
Suppose shape at t=0 is f(x)=y(x,0). Wave moves to the right with speed v. Then it will
have the same shape, but measured from x=vt instead of x=0. That is, the shape will be
f(x’) where x’=x-vt. Draw a picture to explain, i.e. textbook Fig.18-4.
So, y(x,t)=f(x-vt) is a rightward “traveling wave.” Obviously, y(x,t)=f(x+vt) is a leftward
traveling wave. Discuss y(0,t) a little, just like Fig.18-5, with an asymmetric shape.
Sinusoidal Waves
General language is based on this, but “Fourier Analysis” gets us back to general. Give
amplitude and wavelength with y(x,0)=f(x)=ymsin[2πx/λ], so y(x,t)=ymsin[2π(x-vt)/λ]. Now
have time “period” T for any fixed x point to make a full oscillation, i.e. T= λ/v. Generally
instead speak of “frequency” f (or ν) =1/T. That is, v=fλ.
Final “wave speak” words: “Wave number” k= 2π/λ and “(angular) frequency” ω= 2πf.
Then we write a sinusoidal wave as y(x,t)=ymsin[kx-ωt]. You will see this forever more. If
there is time, talk about the “phase constant” ϕ, and relative phases of waves.
Break
Speed of Waves on a Stretched String
Now going for more “physics.” What is the “dynamics” of the mechanical wave set up on
a stretched spring? (Take some time with this.)
First, how do you describe the string? It is infinitely long, right? Consider a small length
of the string δl. It has a mass δm=μδl where we call μ the “linear mass density.” Also,
this small length of string is being “pulled” by a tension F to the left and to the right.
Do some dimensional analysis! [μ]=M/L and [F]=ML/T2. So, v=(const)×[F/μ]1/2
Now let the string deform with a “circular bulge” at the top. (Figure 18-8.) Imagine that
you are following the bulge, so the string is moving past you. The force in the y direction
is Fy=2Fsinϑ=2Fϑ=Fδl/R, and the string is moving through the top at the wave speed v.
It has a centripetal acceleration downward v2/R caused by this force. So, use Newton’s
Second law and get (Fδl/R)=(μδl)(v2/R) and so v=[F/μ]1/2. (The constant is unity.)
Good suggestion: Check out Sec.18-5 on “The Wave Equation.” Do you know how to
use partial derivatives? Good idea to get used to the idea. Important concept!
This is as far as we got. Will do more “waves” on Thursday.
Energy in Wave Motion
Work with sinusoidal waves. Consider a “piece” of the string. The kinetic energy is
1
1
dK = dmu2y = (µdx) [−ym ω cos(kx − ωt)]2
2
2
How much does the kinetic energy change in the time dt it takes the wave to move a horizontal distance dx=vdt? The
dK 1 2 2
= µω ym v cos2 (kx − ωt)
2
answer is dt
. For the potential energy, need to figure out
how much the string stretches due to the tension. This will give you the work done by F.
dU = F
!"
#
!"
#
dx2 + dy2 − dx = Fdx
1 + (∂y/∂x)2 − 1
Now we make a critical
1
dU = Fdx(∂y/∂x)2
2
.
assumption, namely that the “slope” is always small. Hence
With F= μv2 we find that dU/dt=dK/dt, so dE/dt=2dK/dt. From this get the average power
transmitted is Pav=(1/2)μω2ym2v.
This may be too much to get through in the second half.
Save the topics of Superposition, Interference, and Standing Waves for another day,
perhaps next Thursday and push “Fluids” and “Sound Waves” off for a while.
26 October 2006
Hand in lab books. Second exam next Thursday. Material through today will be on the
exam. One HW problem due on Monday (on waves).
Review from Monday
1. General form of a wave y(x,t)=f(x±vt)
2. Sinusoidal waves y(x,t)=ymsin[kx±ωt] with k= 2π/λ and ω= 2πf
3. Dynamics example: The stretched string, giving speed v=[F/μ]1/2
(My apologies for blowing you away with the “wave equation.” Save it for another day.)
Energy in Wave Motion
Work with sinusoidal waves. Consider a “piece” of the string. The kinetic energy is
1
1
dK = dmu2y = (µdx) [−ym ω cos(kx − ωt)]2
2
2
How much does the kinetic energy change in the time dt it takes the wave to move a horizontal distance dx=vdt? The
dK 1 2 2
= µω ym v cos2 (kx − ωt)
dt
2
answer is
. For the potential energy, need to figure out
how much the string stretches due to the tension. This will give you the work done by F.
dU = F
!"
dx2 + dy2 − dx
#
!"
#
2
= Fdx
1 + (∂y/∂x) − 1
Now we make a critical
1
dU = Fdx(∂y/∂x)2
2
.
assumption, namely that the “slope” is always small. Hence
With F= μv2 we find that dU/dt=dK/dt, so dE/dt=2dK/dt. From this get the average power
transmitted is Pav=(1/2)μω2ym2v.
Superposition and Interference
Give them some words about “linear” differential equations. Could show them using the
wave equation that if y1(x,t) and y2(x,t) are both solutions, then so is y(x,t)=y1(x,t)+y2(x,t).
Show them Figure 18-12. Mention “Fourier Analysis”, and that this is the reason why we
can get away with talking about waves as sinusoidal only.
Now interference: Point out the usefulness of discussing the “phase” of a wave. That is,
for two waves with same k and same ω but Δϕ=π, you get zero. Remind that power
goes like square of the amplitude, enhancing interference. This is a common theme!
Break
Standing Waves
Watch: y(x,t)=y1(x,t)+y2(x,t) with y1(x,t) =ymsin[kx±ωt] and y1(x,t) =ymsin[kx±ωt]. You end
up with y(x,t)=[2ymsin(kx)]cos(ωt). This does not travel! It is a “standing” wave. Note that
the amplitude varies as a function of position x. In fact, the amplitude is zero whenever
kx=nπ, n=0,1,2,..., or x=nλ/2.
Energy considerations. Show figure 18-8. Draw the analogy with “oscillations.”
Think about this “the other way around.” That is, take a string and fix one end at x=0,
and the other end at x=L. This will “support” standing waves with λ=2L/n. Frequency is
related to wavelength by fλ=v, where v is the speed of the wave. Therefore, frequency
of the standing wave is f=nv/2L.
Demo (Mechanical vibrations of a string): Start with some discussion of resonance. Can
show that “higher mode” frequencies go up with n. You can also increase the weights on
the end. Factor of four increase in tension should increase v (and so f) by factor of two.
30 October 2006
Turn in homework assignment. Exam #2 is on Thursday.
Today a lightning discussion of “fluids” before the break, and “sound waves” after.
Fluids: Statics and Dynamics
Actually our first discussion of “continuum mechanics”, aka “classical field theory.”
Solids can support tensile and shear forces. Fluids cannot. Liquids can support compressional forces, but gases not even that so well. “Fluids do not have any shape.”
Express all this with math. Density is ρ=Δm/ΔV or =m/V for “uniform density.” Material
response to pressure is “bulk modulus” B=-Δp/(ΔV/V). Bulk moduli for liquids and solids
are some 10,000 times as for gases. (See book, Sec.15-2.) “Incompressible.”
Fluid at rest. Thin horizontal slide, area A and thickness dy. Fig.15-2. Not moving, so
sum of vertical forces is zero. Find dp/dy=-ρg. This is how pressure changes vertically.
Nice example in atmosphere, assuming ideal gas and uniform temperature. Follow book
page 335 to find p=p0e-h/a where a=p0/gρ0=8.55 km. “Scale height” for the atmosphere.
Pascal’s Principle: p=pext+ρgh so Δp=Δpext for incompressible fluid, like water or oil.
Demonstrate using example of the hydraulic lever, Figure 15-8.
Archimede’s Principle: Figure 15-10: Weight of blob of water counteracted by pressure
of surrounding water. Replace blob with something else, so get “buoyant force.”
Now dynamics: Discuss “fluid flow.” Mass δm=ρδV crossing distance vδt is ρAvδt.
Mass conservation implies that ρAv is a constant. Called “mass flux.” Illustrate with a
simple example of faster moving fluid in a constricted pipe.
Bernoulli’s Equation: Conservation of energy for a fluid flowing in a pipe with vertical
motion against gravity. See Fig.16-6 and Eq.16-5. External work done on the fluid is
Wext = p1 A1 δx1 + (−p2 A2 δx2 )[ − δmg(y2 − y1 )] . Set that equal to the change in kinetic energy K=½δmv2. Find p+½ρv2+ρgy=constant.
Good reading in book: Examples, and also “fluid flow” as a prototype for learning about
vector fields. (Maybe Peter will do this next semester?)
Break
Sound Waves
Technically, these are mechanical waves in any three dimensional medium. Solids can
support transverse and longitudinal waves since they can support shear forces, as well
as compressional forces. Fluids can only have longitudinal waves since no shear forces.
Only present results here and not do the quantitative analysis. Some discussion in
Sec.19-2 and 19-3, but it i a bit difficult to follow. Better left for more advanced courses
in continuum mechanics.
Sound is a “pressure wave” or, alternatively, a “density wave.” These are related one
way or another based on the fluid properties. (Example is “ideal gas law.”) For a typical
sound wave, pressure and density waves are in phase with each other. (Fig.19-2.)
For “plane wave” of sound in pressure, write p=p0+Δp where Δp(x,t)=Asin(kx-ωt). The
speed of the wave is v=ω/k=[B/ρ0]½. See Table 19-1 for the speed of sound in different
materials. Faster in liquids and solids than gases since B is larger even though ρ is, too.
Sound and hearing. See Figure 19-5 and Table 19-2. Normal conversation is 10-6W/m2.
Equation 19-19 says this is (Δpm)2/2ρv. For air have density ρ=1.2kg/m3 (Table 15-2)
and speed v=340 m/s (Table 19-1). This gives (Δpm)2=816 (in whatever are the right SI
units, which happen to be Pa2=(N/m2)2.) One atmosphere is p0=105N/m2 (page 332).
Therefore the fluctuation in pressure that one hears in normal conversation is just a few
parts in 10,000 over normal atmospheric pressure. Your ears are very sensitive!
Standing sound waves, obvious application are pipe organs. One important difference
with mechanical waves on a string. A closed end of the pipe is an “antinode”, whereas
an open end of the pipe is a node. See Section 19-6.
Assuming time is left, open the floor to questions prior the exam. That’s it for now!
6 November 2006
Today we do “special relativity.” The in-class activity counts as homework. Note that
there is a homework assignment (fluids; special relativity) due on Thursday.
Objective for today: Gain an appreciation for what the subject is, and do the graphical
exercise to get a physical feeling. Lots more we won’t cover, like addition of velocities,
relativistic momentum and energy, and so forth, also relativistic four-vector notation and
the mathematics of boosts, but you will see that in later physics courses.
Problems with Classical Physics
Newton inconsistent with Maxwell. Something had to give, and it was Newton. Einstein
came up with the simplest way to state it, namely (1) physics is the same in all frames of
reference (“relativity”) and (2) the speed of light is the same to all observers (!).
∆t02 = (2L0 /c)2 but
Neat consequence: Bouncing light clock (Figures 20-4 and 20-5):
!
2 2
∆t 2 = (2L/c)2 = (2L0 /c)2 + (u∆t/c)2 or ∆t = γ∆t0 where γ = 1/ 1 − u /c . This is known
as “time dilation.” Also “length contraction.” See your textbook (and the exercise.)
The Lorentz Transformation
How does an observer in one reference frame see “events” in another reference frame?
The answer is the Lorentz Transformation. Not hard to derive, but we won’t bother to do
it here. Just quote the answer. (See Textbook, Eq.20-14 and 20-17.)
x! = γ(x − ut) and t ! = γ(t − ux/c2 ) and the other way is x = γ(x! + ut ! ) and t = γ(t ! + ux! /c2 )
Obviously, to go from one to the other, just change u to -u.
Natural Units
Why bother to carry around the c all the time? Think of it as a conversion factor between
space and time. Measure time in seconds, but measure distance in light-seconds. This
actually gives a suggestive form for the Lorentz Transformation:
x = γx! + γut ! and t = γt ! + γux! but γ2 − γ2 u2 = 1 so put γ = cosh α and γu = sinh α
!
!
!
!
In other words x = (cosh α)x + (sinh α)t and t = (sinh α)x + (cosh α)t which is sort of like
a “rotation” but using hyperbolic cosine and sine instead of circular. Hmm....
OK, so now move on to the worksheet, perhaps after a break.
Lecture 9 Nov 2006
Collect homework, including leftover “relativity” worksheets from Monday.
An apology: I never tried the “relativity” worksheet before, and there were obviously
some problems. For whatever it’s worth, David Statman got in touch with me yesterday
and told me that he has found it always takes longer and is more confusing than he
thought it would! Oh, well, next year.
Coupled Oscillations
Follow posted notes from today. Note homework assignment due next Thursday is at
the end of the notes. This will all be reinforced by the laboratory exercise we do next
Monday.
Laboratory 13 Nov 2006
This lab is on coupled oscillations, using the same setup that we had for demonstration
in class last Thursday. You will have your own two-mass and three-spring setups, and
you are encouraged to work in pairs in the same way as for previous labs. You will be
able to take data using the motion sensor, following just one of the two masses.
Refer to last Thursday’s class notes for the mathematics of coupled oscillations.
In this, your last laboratory for the term, I encourage you to be creative. Here are some
suggestions for data to take and results you can determine from your analysis.
1. Measure the normal mode frequencies and compare their ratio to expectations. It
can be tricky to get the masses into “exact” eigenmodes, but do the best you can.
You can take several trials, getting the period from the “zero” crossings, and use the
average and standard deviation (or just the range) for the value and uncertainty. For
identical masses and springs, you know what you should get. How do you calculate
the ratio if the coupling spring has a different value?
2. Predict absolute values of the eigenmode frequencies. This will be similar to earlier
labs. You’ll need the mass of the carts, and values for the spring constants.
3. Eigenmodes for different masses. How do you expect things to change if the two
masses have different values? Try this, and see if you can set the system in motion
in one or both of the two eigenmodes. Does this agree with what you expect from
the equations of motion?
4. General motion of the system. After you determine the eigenfrequencies, try setting
one of the masses in motion from rest but away from its equilibrium position. See if
its subsequent motion agrees with the predicted motion from the notes. It’s probably
easiest to try this first with equal masses and three identical springs, but you can of
course see what happens in a more general case.
Keep all this in your lab book. A good job on this lab will help your final lab report grade.
Class Notes 27 Nov 2006
Thanks to class: Prof Persans had a good experience! (Thanks to him, too!)
Today: (1) Review of Persans’ material on thermodynamics, (2) Some new stuff on
thermodynamics, (3) Hand out the IDEA forms
Announcements: See the new syllabus, posted online. Third exam is this Thursday! No
more homework is due. Final lab books are due one week from Thursday, which will be
the last class of the term. We’ll do a “course review” that day. Next Monday, we’ll see a
“glimpse of the future” by studying “The Principle of Least Action.”
Persans Review
Temperature, Heat, and the First Law of Thermodynamics. Kinetic model of gases and
the “ideal gas law.” Thermal expansion, heat capacity, and heat flow through slabs.
Temperature scales (F, C, and K) and units of heat (kCal and BTU, but also joule).
Ideal Gas Law: PV=NkBT=nRT
1
3
m!v2 " = kB T
2
Kinetic theory: Temperature as molecular motion, i.e. 2
Specific Heat Capacity: ΔQ=CΔT=mcΔT
See the tables for c in your textbook!
Latent heat of phase changes, i.e. transformation.
Heat flow thru cross sectional area A: dQ/dt=kA(dT/dx); k= specific thermal conductivity
Heat energy and work: Heat a gas andZ it can expand and do work against a piston.
W=
V2
pdV
V1
Have “work done on the system”
and ∆Eint = Q +W where Q is the heat
added to the “working material.” This is the “First Law of Thermodynamics”.
Specific examples: Work done on an ideal gas under constant pressure Wp = p(V f −Vi )
or constant temperature WT = NkB T ln(V f /Vi ) .
Some discussion (?) of cycles and engines.
Entropy and the Second Law: New Stuff
Reversible, irreversible, and the “arrow of time.” This is profound stuff.
Z f
dQ
∆S ≡
T “along a reversible path.” See the book for a “proof” that
i
Entropy change is
S is a “state variable” for the case of an ideal gas. This is true in the general case.
Consider “free expansion”, Fig.23-22. Have dQ=-dW since no change in internal energy.
In order to calculate the change in entropy, one needs to calculate the integral over
some reversible path.
So, connect i and f with isothermal expansion so dW=-pdV=-NkBT(dV/V). (Note that the
sign is negative because W is the work done on the gas by the piston, so dW must be
Z f
Vf
dV
∆S = NkB
= NkB ln
Vi . This is a positive quantity
i V
negative if dV is positive.) Then
since the final volume is necessarily larger than the initial volume for free expansion.
This is an “isolated” system. There was no change in entropy outside the walls because
there was no heat flowing through the walls. If we actually did carry out an isothermal
expansion, though, then the heat flow would match what went in, and there would be an
equal but opposite change in entropy in the space outside the walls.
We have just seen an example of the “Second Law of Thermodynamics”, namely that
“When changes occur within a closed system, its entropy either increases (for irreversible processes) or remains the constant (for reversible processes). It never decreases.”
Time for questions prior to Thursday’s exam.
The test will not cover today’s material, but entropy may show up on the final exam.
Break and/or Hand out IDEA forms
Need a volunteer to pick them up and take them to the Physics office.
Class Notes 4 Dec 2006
Final class will be this Thursday, 7 Dec. Course review. Come with questions!
Final lab books are due on Thursday. Remember, this is 20% of your grade.
Final exam: Wednesday 13 Dec 6:30-9:30pm J-ROWL 2C30 (i.e. here). If you have
some sort of conflict let me know right away!
Physics and the “Action Principle”
John Cummings will introduce the subject. Want to get to the concept that “action” is the
integral over time of some quantity (i.e. “the Lagrangian”) which depends on the “path.”
A Specific Example: Motion under Constant Acceleration
Demonstrate that “the right path is the one that minimizes the action.”
Call x(t) a “path” through (one dimensional) space and time. A path has specific endpoints, i.e. (x1,t1) and (x2,t2) where x1=x(t1) and x2=x(t2).
Consider a particle moving under constant acceleration between (x1,t1)= (0,0) and
(x2,t2)= (1,1). We “know” that the right path is x=t2. (We measure position in meters, time
in seconds, and the acceleration is 2 in these units.)
S[(x(t)] =
Z t2
L(x, ẋ)
. The “physics” is in figuring out what L(x, ẋ) is supposed to be.
1
L(x, ẋ) = K −U = mẋ2 −U(x)
2
For mechanical motion, we now know that
. For U(x) take
-dU/dx=F=ma=2m so U(x)=-2mx.
"
Z 1!
Z 1!
"
1 2
4
2
S[x(t)] =
mẋ + 2mx dt
t 2 + t 2 dt = m
S[t ] = 2m
2
2
3
0
0
So
. Calculate for x(t)=t :
So
t1
Challenge to class: Calculate S[x(t)] for some other path x(t). The only restriction is that
x(0)=0 and x(1)=1. Does anybody get a smaller value than 10m/3? Try to get them to
work in pairs, and use different functions.
n
For me:
S[t ] = m
Z 1! 2
n
0
2
t
2n−2
+ 2t
n
"
#
2
n2
+
dt = m
2(2n − 1) n + 1
$
Another function would be x(t)=sin(πt/2). Use half-integral formula to integrate KE term.
Answer works out to (π/4)2+(4/π)=1.89>(4/3).
Class Notes 7 Dec 2006
Course review. Expect questions from students, but here is a list of topics that will be
covered on the final exam.
Kinematics and Dynamics of Particles
• Position, velocity, and acceleration as derivatives with respect to time
• Newton’s second law as a differential equation
• Uniform circular motion and centripetal acceleration
• Free body diagrams for solution to statics and dynamics problems
• Specific force laws for springs, friction, and drag
Momentum and Energy
• Momentum as a vector quantity; Translational symmetry
• Center of Momentum (aka Center of Mass)
• Work, kinetic energy, and the work-energy theorem; Galilean Invariance
• Potential energy and the conservation (or not) of total mechanical energy
The Law of Universal Gravitation
• Newton’s force law for gravitation; The dual role of mass
• Shell theorems
• Gravitational potential energy
• Circular orbits; Escape velocity
Mechanical Oscillations
• Simple harmonic motion as derived from a mass and spring system
• Oscillations as sines and cosines, or as exponentials of imaginary numbers
• Meaning of initial conditions for position and velocity, in terms of amplitude and phase
• Damped harmonic motion
• Coupled oscillations (See separate notes); Normal modes as “eigen” modes
Waves
• Waves as oscillations in time and space
• Wavelength and wave number; Period and (angular) frequency
• Traveling waves and standing waves; Normal modes of a standing wave
• The Wave Equation; Superposition and interference
• Energy in wave motion
Kinematics and Dynamics of Rigid Bodies
• Angular coordinates, velocity, and acceleration
• Tangential acceleration versus radial acceleration
• Rotational inertia (aka Moment of inertia); Determining for solid bodies
• Torque and Newton’s Second Law in angular variables; Center of gravity
• The Parallel axis theorem; The Pendulum and the Physical Pendulum
Fluid Statics and Dynamics
• Fluids vs Solids; Liquids (“incompressible”) vs Gases (“compressible”)
• Density, pressure, and “bulk modulus”
• Fluids at rest: Atmospheric “scale height”, Pascal’s Principle, Archimedes’ Principle
• Fluids in motion: Mass flux (also, a little about Bernoulli’s Equation)
• Sound waves: Longitudinal in fluids, but longitudinal and transverse in solids
Special Relativity and the Lorentz Transformation
• The inconsistency between Newton and Einstein: Make c the same for everyone
• Consequences: Time Dilation and Length Contraction
• The Lorentz Transformation: Equations and graphically; Natural units (c=1)
Thermodynamics
• Temperature and Fahrenheit, Centigrade (Celsius), and Kelvin scales
• The Ideal Gas Law: PV=NkBT
• Kinetic theory of ideal gases: Average molecular kinetic energy=(3/2)kBT
• Specific heat capacity; Heats of transformation; Heat flow; Thermal expansion
• Work done on an ideal gas
• Heat and work: The First Law of Thermodynamics
• Entropy and the Second Law of Thermodynamics
Principle of Least Action
• Unifying principle for all of (classical) physics
• The Action S[x(t)] as a “functional” of the path x(t)
• Action as the integral between (time) “endpoints” of the Lagrangian L=K-U
• Calculating the action for a given path
Topics in Mathematics
• Basic differentiation and integration formulas
• Integration as the addition of many small things; Integrating over solid bodies
• Taylor series and approximation formulas
• Euler’s formula for exponentials of imaginary numbers
• Hyperbolic sines and cosines
Laboratory Techniques and Data Analysis
• The role of redundancy in data taking
• Making use of diagrams and graphs
• Determining uncertainties