Section 4.6-4.7
• Stress, Strain, & Young’s Modulus
• Relating the micro to macro.
Wires as Masses on Identical Parallel and Series
Springs
The Micro-Macro Connection:
Stiffness for springs in
series and in Parallel
Length=Nser*dbond
Area=Npar*d2bond
F  ktot L
N par
A
 2
d
N ser
L

d
 A

F 
k1 L
Q4.6.d: Wires
d L 
You hang a 10 kg mass from a What happens?
copper wire, and the wire
a) The second wire stretches
stretches by 8 mm.
4 mm
Now you hang the same mass b) The second wire stretches
from a second copper wire,
8 mm
whose cross-sectional area is c) The second wire stretches
half as large (but whose
16 mm
length is the same).
Q4.6.e: Wires
 A

F 
k1 L
d L 
You hang a 10 kg mass from a
copper wire, and the wire
stretches by 8 mm.
Now you hang the same mass
from a second copper wire,
which is twice as long, but
has the same diameter.
What happens?
a) The second wire stretches
4 mm
b) The second wire stretches
8 mm
c) The second wire stretches
16 mm
Young’s Modulus
Two wires with equal lengths are
made of pure copper. The diameter
of wire A is twice the diameter of
wire B.
You make careful measurements
and compute Young's modulus
for both wires. What do you
find?
When 6 kg masses are hung on the
wires, wire B stretches more than
wire A.
Y = (F/A)/(DL/L) = k/d
1) YA > YB
2) YA = YB
3) YA < YB
Example: You hang a heavy ball with a mass of 14 kg from a silver rod
2.6m long by 1.5 mm by 3.1mm. You measure a stretch of the rod, and
find that the rod stretched 0.002898 m. Using these experimental data,
what value of Young’s modulus do you get?
The density of silver is 10.5 g/cm3 and you can look up its atomic mass.
What’s the inter-atomic spring stiffness?
Normal Force
Speed of Sound in a Solid: the logic
n
n-1
n+1
d
n-1
d
n
n+1
x
xn-1
xn
xn+1
stretchleft stretchright
((xn – xn-1 ) – d) ((xn+1 - xn ) – d)
Fn , Left  k s (( xn  xn 1 )  d ))
Fn , Right  k s (( xn 1  xn )  d ))
Fn ,net  Fn , Left  Fn , Right  k s (( xn  xn 1 )  d ))  k s (( xn 1  xn )  d ))
Fn ,net  k s ( xn 1  xn 1  2 xn )
Copmutational simulation
Speed of Sound in a Solid
n
n-1
m
n-1
xn-1
k
d m
n
xn
n+1
d
k
m
n+1
x
xn+1
Fn ,net  k s ( xn 1  xn 1  2 xn  xn 1 )
Informed Guess at v’s dependence
Stiffer, for a given atomic displacement, greater
force pulling it so greater velocity achieved.
ks
v
d
m
More distance between atoms means further the distortion
can propagate just through the light weight spring /bond
without encountering the resistance of massive atoms.
More massive, more inertial resistance to
applied force, less velocity achieved.
Example: The spring constant of aluminum is about 16 N/m. The
typical separation of Al atoms was 2.6×10-10m. Recall also that the
atomic mass of aluminum is 27 g/mole. So what is the speed of sound
in Aluminum?
Microscopic to Macroscopic Springs
Solid
Molecule
F=-katomic(x-d)
Lo
F
d
x

  
A

F  kkspring
atmicL  L
Ld 

Case Study in Three Modes of Exploration with
Varying Force: Mass on Spring
Experimentation / Observation
Case Study in Three Modes of Exploration with
Varying Force: Mass on Spring


 p
r f  ri  t
m
Experimentation / Observation
Observations to Understand
•
•
•
•
force, velocity, and position vary sinusoidally
force and position vary in synch
velocity varies out-of-synch
Period’s dependence
• Mass
- greater mass, slower
• Stiffness - greater stiffness, faster
• Amplitude – no effect !
Observations to Understand
• Changing gravity only changes center of
oscillation
Computation / Simulation

Fs  k s  L  Lo Lˆ
L_mag = mag(ball.pos)
L_hat = ball.pos/L_mag
F=-k*(L_mag-Lo)*L_hat



p f  pi  Fs t
ball.p = ball.p + F*deltat


 p
r f  ri  t
m
ball.pos = ball.pos +
(ball.p/ball.m)*deltat
Finite changes to infinitesimal changes: derivatives
Case Study in Three Modes of Exploration with
Varying Force: Mass on Spring
Theory / Analysis

F  Fx ,0,0 
p  p x ,0,0
k
m

v  v x ,0,0
x̂
x
xo
0
System: Ball
Guess from experiment and simulation
t

xt   X cos 2   xo
 T
2
Shorthand:  
T
xt   X cost   xo
Fx t   k * x(t )  xo 
dp x (t )
 k * x(t )  xo 
dt
d mv x t 
 k * x(t )  xo 
dt
d  dx(t ) 
m 
 k * xt   xo 

dt  dt 
d 2 x(t )
k
  * xt   xo 
2
dt
m
Case Study in Three Modes of Exploration with
Varying Force: Mass on Spring
Theory / Analysis
System:
 Ball
F  Fx ,0,0 
p  p x ,0,0
k
m v  vx ,0,0
x̂
x
xo
0
Fx t   k * x(t )  xo 
d 2 x(t )
k
  * xt   xo 
2
dt
m
Guess
xt   X cost   xo
2
where:  
T
Plug in and see if guessed solution works
d2
k
X cost   xo    * X cost   xo  xo 
2
dt
m
d2
k
X 2 cost    * X cost 
dt
m
d
k
  sin t    cost 
dt
m
k
2
  cost    cost 
m
k
2
 
m
Our guess works if

k
m
Case Study in Three Modes of Exploration with
Varying Force: Mass on Spring
Theory / Analysis
System:
 Ball
F  Fx ,0,0 
p  p x ,0,0
k
m v  vx ,0,0
x̂
x
xo
0
Fx t   k * x(t )  xo 
d 2 x(t )
k
  * xt   xo 
2
dt
m
Guess
Solution
xt   X cost   xo
where:

2

T
k
m
Concisely tells us…
xt   X cost   xo
• Sinusoidally oscillates
• About the equilibrium
• With a period that…
• Shortens with greater stiffness
• Lengthens with larger masses

2

T
k
m
 T  2
m
k
• Doesn’t care about amplitude
Period dependence on: mass
Suppose the period of a
spring-mass oscillator is 1 s.
What will be the period if we
double the mass?
a.
b.
c.
d.
e.
T = 0.5 s
T = 0.7 s
T = 1.0 s
T = 1.4 s
T = 2.0 s
Period dependence on Stiffness:
Suppose the period of a spring-mass
oscillator is 1 s. What will be the period if
we double the spring stiffness? (We could
use a stiffer spring, or we could attach the
mass to two springs.)
a.
b.
c.
d.
e.
T = 0.5 s
T = 0.7 s
T = 1.0 s
T = 1.4 s
T = 2.0 s
Period Dependence on Amplitude:
Suppose the period of a spring-mass oscillator
is 1 s with an amplitude of 5 cm. What will be
the period if we increase the amplitude to 10
cm, so that the total distance traveled in one
period is twice as large?
1) T = 0.5 s
2) T = 0.7 s
3) T = 1.0 s
4) T = 1.4 s
5) T = 2.0 s
Case Study in Three Modes of Exploration with
Varying Force: Mass on Spring
Theory / Analysis
How does gravitational interaction change behavior?
System: Ball
k
0
yo

Fs  0, Fs ,0
m
y
ŷ

v  0, v y ,0

p  0, p y ,0

FE  0, mg ,0
Note: I’ve defined down as +y direction
So Earth’s pull has + sign

 
Fnet  Fs  FE  0, Fs  FE ,0
Fnet . y t   k *  y (t )  yo   mg
k
Fnet . y t   k *  y (t )  yo   mg
k
 mg 
Fnet . y t   k *  y (t )  yo   k 

k


mg 

Fnet . y t   k *  y (t )  yo 

k



mg 

Fnet . y t   k *  y (t )   yo 

k 


Fnet . y t   k *  y (t )  yo 
mg
where yo  yo 
k
Case Study in Three Modes of Exploration with
Varying Force: Mass on Spring
Theory / Analysis
How does gravitational interaction change behavior?
System: Ball
k
m
0
yo

Fs  0, Fs ,0
yo
y
ŷ

v  0, v y ,0

p  0, p y ,0
Fnet . y t   k *  y (t )  yo   mg
.
.
.
Fnet . y t   k *  y (t )  yo 
where yo
mg
 yo 
k
• Exact same form as for horizontal mass-spring, but
shifted equilibrium
d2
m 2 y t   k *  y (t )  yo 
dt

FE  0, mg ,0
Note: I’ve defined down as +y direction
So Earth’s pull has + sign

 
Fnet  Fs  FE  0, Fs  FE ,0
Solution:
y t   Y cost   yo

2

T
k
m
 T  2
m
k
Period dependence on g:
Suppose the period of a spring-mass
oscillator is 1 s with an amplitude of
5 cm. What will be the period if we
take the oscillator to a massive
planet where g = 19.6 N/kg?
1) T = 0.5 s
2) T = 0.7 s
3) T = 1.0 s
4) T = 1.4 s
5) T = 2.0 s
Speed of Sound in a Solid: the logic
n
n-1
d
n-1
xn-1
n+1
d
n
n+1
xn
xn+1
x
2
xn k s
d
Fn ,net  k s ( xn 1  xn 1  2 xn )
 ( xn 1  xn 1  2 xn )
2
m
dt
dpn
 k s ( xn 1  xn 1  2 xn )
d 2 xn k s   xn 1  xn   xn  xn 1  
dt

 d

2
d
d
m 
dt

d mvn 
 k s ( xn 1  xn 1  2 xn )
dt
 dxn 1 dxn 



2
 dxn 
k s 2  dx
d n
dx 
d
d



d
m
dt 2
 dt   k s ( x  x  2 x )
n 1
n 1
n
dt
m
k s 2 d 2 xn 1
d 2 xn
 d
2
dx 2
m
dt
ks
Speed of Sound in a Solid:
v
d
the result
m
Speed of Sound in a Solid
n
n-1
d
n-1
xn-1
n+1
d
n
n+1
xn
xn+1
x
Stiffer, for a given atomic displacement, greater
force pulling it so greater velocity achieved.
ks
v
d
m
More distance between atoms means further the distortion
can propagate just through the light weight spring /bond
without encountering the resistance of massive atoms.
More massive, more inertial resistance to
applied force, less velocity achieved.
Period dependence on: mass
Suppose the period of a
spring-mass oscillator is 1 s.
What will be the period if we
double the mass?
a.
b.
c.
d.
e.
T = 0.5 s
T = 0.7 s
T = 1.0 s
T = 1.4 s
T = 2.0 s
Period dependence on Stiffness:
Suppose the period of a spring-mass
oscillator is 1 s. What will be the period if
we double the spring stiffness? (We could
use a stiffer spring, or we could attach the
mass to two springs.)
a.
b.
c.
d.
e.
T = 0.5 s
T = 0.7 s
T = 1.0 s
T = 1.4 s
T = 2.0 s
Period Dependence on Amplitude:
Suppose the period of a spring-mass oscillator
is 1 s with an amplitude of 5 cm. What will be
the period if we increase the amplitude to 10
cm, so that the total distance traveled in one
period is twice as large?
1) T = 0.5 s
2) T = 0.7 s
3) T = 1.0 s
4) T = 1.4 s
5) T = 2.0 s
Period dependence on g:
Suppose the period of a spring-mass
oscillator is 1 s with an amplitude of
5 cm. What will be the period if we
take the oscillator to a massive
planet where g = 19.6 N/kg?
1) T = 0.5 s
2) T = 0.7 s
3) T = 1.0 s
4) T = 1.4 s
5) T = 2.0 s
Wed. 4.6-.7, .9-.10 Stress, Strain, Young’s Modulus, Compression, Sound
InStove @ noon
Science Poster Session: Hedco7pm~9pm
Lab L4: Young’s Modulus & Speed of Sound (Read 4.11-.12)
Fri 4.11-.12; .14-.15 Sound in Solids, Analytical Solutions Quiz 3
Mon. 4.8, .13 Friction and Buoyancy & Suction
Tues.
.
.
Fri. Exam 1 (Ch 1-4)
RE 4.b
RE 4.c
RE 4.d
EP 4, HW4: Ch 4 Pr’s
46, 50, 81, 88 & CP