Tangible: Ball and Spring Model
BallSpring.mov
Show ball and spring model and movie and discuss
what “pushing” and “pulling” is!
Tangible: Activity - What’s in a name…block?
WID 1082002 cubes
JCST_SCALE-UP_article.pdf
(article in the Journal of College Science Teaching describing the activity)
Have them measure the total mass and count the atoms to determine mass of one atom.
Have them do a cube root to verify number of atoms on a side and their separation.
Find your name block’s density.
(They’ll need scale and ruler.)
Chapter 4
1
Vblock+slot = ( 3.78 cm ) (1.92 cm ) (1.92 cm ) = 13.9 cm 3
Vslot = ( 0.1 cm ) (1.1 cm ) (1.9 cm ) = 0.2 cm 3
Vblock-slot = 13.7 cm 3
m = 36.3 g
r=
(
m
36.3 g
g
g
=
= 2.6 cm 3 accepted value is 2.7 cm 3
3
V 13.7 cm
)
One mole of Aluminum has a mass of 27 g, so how many moles are in your block?
36.3 g
= 1.3 moles
g
27 mole
How many atoms are in your block?
23
1.3 moles 6 ´ 10 23 atoms
atoms
mole = 7.8 ´ 10
(
)
Your block is roughly two cubes, each with 3.9e23 atoms.
How many atoms per side of that cube?
3
3.9 ´ 1023 = 73 ´ 106 atoms on a side
So block is roughly 73 by 73 by 150 million atoms.
How far apart are the atoms? This is approximately the diameter of an atom.
0.0192 m
m
datom =
= 2.6 ´ 10 -10 atom
= 2.6 A
6
73 ´ 10 atoms
What is the mass of a single atom?
36.3 g
g
-23
-25 kg
matom =
=
4.7
´
10
=
0.47
´
10
7.8 ´ 10 23 atoms
atom
atom
There is a direct way to calculate that value without referring to a specimen:
g
27 mole
M
g
kg
matom =
=
= 4.5 ´ 10 -23
= 0.45 ´ 10 -25
23 atoms
6 ´ 10 mole N Avogadro
atom
atom
(dimensional analysis)
Chapter 4
2
Also a direct way to calculate distance between atoms:
rmicro =
datom =
matom matom N atom matom
= 3 =
= rmacro
3
Vatom
d atom
N atom d atom
3
matom
rmacro
M
=
3
rmacro
( 4.5 ´ 10
)
( 2.7 ´ 10 )
-26
datom =
3
N Avogadro
kg
atom
3 kg
m3
(
rmacro = 2.7 cm
g
3
)
æ 1 kg ö æ 100 cm ö
4
çè 1000 g ÷ø çè 1 m ÷ø = 2.7 ´ 10
3
kg
m3
datom = 2.6 ´ 10 -10 m
Odd tables, find mass and diameter of gold atoms.
Look up density on web. Atomic mass from book cover.
g
kg
r = 19.3 cmg 3 , M = 197 mole
matom = 3.3 ´ 10 -25 atom
d = 2.6 ´ 10 -10 m
Even tables, find mass and diameter of lead atoms:
g
kg
r = 11.4 cmg 3 , M = 207 mole
matom = 3.5 ´ 10 -25 atom
d = 3.1 ´ 10 -10 m
FYI:
Prototypical solid has atoms with radius of about 10-10 m or one angstrom, mass about
10-25 kg; nuclei are 10-14 m radius
Ponderable: Activity - Interatomic spring stiffness
WID 1081774 pull
Odd tables, find interatomic spring constant for gold
Chapter 4
3
Web gives Y = 11.3e6 psi.
Use Google to convert with “11.3e6 psi in N/m^2” gives 7.79e10 N/ m2
ks, i = Yd = 7.79 ´ 1010 mN2 2.6 ´ 10 -10 m = 20.2 mN
(
)(
)
Even tables, find interatomic spring constant for lead
Web gives Y = 16GPa. Converting with Google gives 1.6e10 N/m2
ks, i = Yd = 1.6 ´ 1010 mN2 3.1 ´ 10 -10 m = 5.0 mN
(
)(
)
Clickers: Chapter 4
Tangible: Activity – Springs
Initially:
• If you are in an "a" group, do Lab: Macroscopic Springs
• If you are in a "b" group, do Lab: Young's Modulus
• If you are in a "c" group, do VPython: Spring
Once you have completed your group's part, switch to another, then another.
Lab: Macroscopic Springs
WID 1109323 bounce
(Can also use QID 849846 as a long question to replace QIDs 855731, 855732, 855733)
(Uses PASCO’s “Equal Length Spring Set)
Measure spring stiffness for springs in series and parallel (make sure that they use springs
of the same color). Discuss the results.
Have them use stopwatch or sonic ranger with bouncing hanging weight (use weight
hanger so there is sonic reflection). Show that a bigger amplitude leads remarkably to
the same period. Contrast with a bouncing ball. Emphasize that the dissipation isn’t the
issue here but the timing, with larger amplitude having longer period. Evidently the larger
forces with larger stretches makes the momentum increase enough to compensate exactly
for the longer distance traveled.
Check that their measurements of the stiffness are consistent. Also check that they do see
that doubling the amplitude doesn’t significantly change the period, and ask for ideas on
how that can be, considering that the mass has to go much farther (answer: bigger forces
generate bigger momentum changes, and we get bigger speeds which apparently exactly
compensate for the longer distance). Also make sure that they see with double the mass a
period that is about 1.4 times as long.
Chapter 4
4
Lab: Young’s Modulus
WID 1109367 bounce (includes Excel sheet: Lab_Young’s_Modulus_Data.xls)
Need equipment. Instructions have vernier calipers, but can just give them this info:
Data: diameter of brass wire = 0.285e-3 m (0.285 millimeters)
Special apparatus for Young’s modulus. The intent is not to get an accurate measurement
(which is difficult because the wire necessarily has some kinks in it and therefore behaves
a bit like a coiled spring, which lowers the value obtained).
VPython: Spring
WID 594349 bounce
WID 599948 “VPython: Spring Extra Credit” (can be used if 3D oscillations are extra credit)
Spring3D.py shell program
Have them use VPython to model a ball and spring moving on y-axis, starting from a
shell on WebAssign.
Chapter 4
5
Discussion: Analytical solution for spring system
Now we’ll start using what we’ve learned about masses and springs and atoms and bonds
together… Covered in book, chapter 3.
Discuss analytical solution for an idealized spring system (no friction etc.)
Draw a cosine curve, define amplitude and period. Momentum principle says slope of p
vs t is F graph.
Guess the analytical result that x = Acos(wt), explain meaning of A and  = 2/T.
æ 2p ö
x = A cos (w t ) = A cos ç
t
è T ÷ø
For example, if T = 1.2 s, after one period back to maximum amplitude:
æ 2p ö
æ 2p
ö
x = A cos ç
t ÷ = A cos ç
1.2 s÷ = A cos ( 2p ) = A
è T ø
è 1.2 s
ø
w=
ks
m
Ponderable: Activity - Coiled wire
WID 1110037 rdf47 (4.P.55)
Coiled_wire_solution.pdf
Chapter 4
6
Chapter 4
7
Demo: Speed of sound in a solid
NEED ON TABLE: Ball-spring models
Imagine you have a long metal rod. You take a hammer and strike one end. What
happens?
Deforms, Vibrates, Makes a sound
Vibration: do all the atoms vibrate instantly, the moment you hit it?
Think about ball-spring model—look at model on the table!
No—ones on the end vibrate first—then what happens?
They compress bonds, exert forces on neighbors, and so on
Vibration/compression travels, or propagates
What is this?:
A sound wave!
What would influence the speed of sound in a solid?
Mass, Spring stiffness, Initial stretch?? (think about it)
How would you write a VPython program to model this and predict the speed of sound?
Very similar to your spring program, just more masses & springs!
VPython Demo: Speed of sound in a solid
04_speed_of_sound.py
Group A and B—look up speed of sound in Aluminum
Group C—look up speed of sound in Lead
Run program for Aluminum
Program models the propagation of sound through a metal.
Linear chain of (N+2) atoms, first and last atoms fixed in position.
Atoms connected by interatomic "springs" (16 N/m for Al, 5 N/m for Pb).
Four parts of program to model propagation of sound through a metal:
(1) Click an atom to displace it, then click repeatedly to see motion.
(2) Same as (1), but also shows graph of displacements.
(3) Click an atom to displace it, click anywhere when pulse reaches right end.
(4) Same as (3), but lead atoms (Pb) instead of aluminum (Al).
Move to next part or back by clicking an atom at the end of the chain.
Atoms at the end of the chain are fixed in position.
Why do you think lead has a lower speed of sound?
What’s different? Lead has lower interatomic ks and bigger atomic mass
Chapter 4
8
Let’s make connection between speed of sound and frequency of oscillating masses:
dp
= Fnet
dt
dpx
= Fnet, x = -ks x
dt
d
dvx
d2 x
m
=m 2 =m
dt
dt
(
2
( A cos (w t )) = m d ( -Aw sin (w t )) = m
dt
dt
2
)
m -Aw 2 cos (w t ) = -ks A cos (w t )
so
mw 2 = ks and w =
x
Also useful to define f º
( -Aw
2
)
cos (w t ) = -ks x
ks 2p
=
m
T
1
so w = 2p f
T
Lead (massive atoms, weak springs) 1200 m/s compared to aluminum (lighter atoms,
stiffer springs) 5000 m/s
Single atom—about how far does it move: d! In about how long? T! So v should be
proportional to d/T or d*omega. It turns out:
v = wd =
ks, i
ma
d
where k s, i is the stiffness of the interatomic bond,
d is the bond length, ma is the mass of one atom
Book does copper, gets 3640 m/s. on page 163
Do speed of sound in Aluminum
k
17.7 mN
v = w d = s, i d =
2.3 ´ 10 -10 m = 5160 ms
-26
ma
4.5 ´ 10 kg
(
)
Why is this important? Micro-macro connections
Tangible: It’s a race
Speed of Sound.xls
Fill in the table, given Y, M, and rho. A does Al, B does Au, C does Pb. Book does Cu.
Chapter 4
9
Aluminum
Young's Modulus, Y
Gold
Copper
units
1.60E+10
1.17E+11 N/m^2
197
207
64 g/mole
Density, rho
2.70E+03 1.93E+04
1.14E+04
8.94E+03 kg/m^3
Mass of an atom, ma
4.49E-26 3.27E-25
3.44E-25
1.06E-25
kg
=M/1000/6.02E+23
Diameter of atom,d
2.61E-10 2.63E-10
3.18E-10
2.33E-10
m
=(ma/rho)^0.333
Atomic Mass, M
6.90E+10 1.60E+10
Lead
27
Interatomic spring
constant, ksi
18.0
4.2
5.1
27.3
N/m
=Y*d
speed of sound
5226
941
1224
3740
m/s
=SQRT(ksi/ma)*d
Book does atomic
diameter of copper
on page 142,
interatomic bond
stiffness for copper
on page 145 (from
measurements),
speed of sound in
copper on page 163
http://www.engineer
ingtoolbox.com/youn
g-modulusd_417.html
2.7 g/cm^3 = 2.7e3
kg/m^3
Chapter 4
10