Download Lecture 19 - Purdue Physics

Midterm Exam 2
March 29, 7:30-8:30 pm, STEW 183
will cover Chapters 6-10.
Lecture 18
Purdue University, Physics 220
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PHYSICS 220
Lecture 18
Elasticity and Oscillations
Lecture 18
Purdue University, Physics 220
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Contact Force: Spring
Force exerted by a spring is
directly proportional to the
amount by which it is
stretched or compressed.
Fspring = - k x
always trying to restore its original length
Lecture 4
Purdue University, Physics 220
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Hooke’s Law Revisited
F = - k DL; where k depends on
size of the bar and material
Hooke’s Law
F/A = Y *DL/L0
k = Y A/L0
• Y (Young’s Modules)
independent of L or A
Has the unit of (N/m2) pressure (i.e.,
pascal)
Inherent stiffness of a material
•
For many materials, Y is the same for
compression and tension, elastic constant
• Stress = Y*Strain
Define
Strain = DL / L0
Stress = F/A
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Purdue University, Physics 220
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Beyond Hook’s Law
Deformed beyond elastic
limit material will no longer
return to its original length
even when F is completely
removed.
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Purdue University, Physics 220
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Simple Harmonic Oscillation
The force exerted by a
spring is proportional
to the distance the
spring is stretched or
compressed from its
relaxed position.
FX = -k x
x is the displacement
from the relaxed
position and k is the
constant of
proportionality.
Lecture 18
Purdue University, Physics 220
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Potential Energy in Spring
• Force of spring is conservative
F = -k x
DPE = -W
W = -½ k x2
DPE= ½ k x2
Force
work
x
Work done only depends on initial and final position
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Purdue University, Physics 220
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Energy Conservation
• A mass is attached to a spring and set to
motion. The maximum displacement is x=A
SWnc = DKE + DPE
0 = DKE + DPE or Energy PE+KE is constant!
Energy = ½ k x2 + ½ m v2
PE
At maximum displacement x=A, v = 0
S
Energy = ½ k A2 + 0
At zero displacement x = 0
Energy = 0 + ½ mvm2
0
Since Total Energy is same
½ k A2 = ½ m vm2
m
vm = sqrt(k/m) A
x=0
Lecture 18
Purdue University, Physics 220
x
x
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Vertical Mass and Spring
• If we include gravity, there are two forces acting on
mass. With mass, new equilibrium position has spring
stretched d
At equilibrium: F=kd–mg=0
d = mg/k
Let this point be y=0
At a displacement, F=k(d-y)–mg=-k y
Total energy:
E=1/2k(d-y)2+mgy+1/2mv2
=1/2kd2+1/2ky2+1/2mv2
Same as the horizontal case,
except for a new equilibrium position
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Purdue University, Physics 220
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What is harmonic motion?
Lecture 18
Purdue University, Physics 220
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iClicker
A mass on a spring oscillates back & forth with simple harmonic
motion of amplitude A. A plot of displacement (y) versus time (t) is
shown below. At what points during its oscillation is the total energy
of the mass and spring a maximum? (ignore gravity).
A) When y = +A or -A (i.e. maximum displacement)
B) When y = 0 (i.e. zero displacement)
C) The energy of the system is constant
“When the kinetic energy is at a
minimum, the potential energy is at a
maximum and vice-versa.”
y
+A
t
Lecture 18
-A
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Exercise
A mass on a spring oscillates back & forth with simple harmonic
motion of amplitude A. A plot of displacement (y) versus time (t) is
shown below. At what points during its oscillation does the speed of
the block reach maximum?
A) When y = +A or -A (i.e. maximum displacement)
B) When y = 0 (i.e. zero displacement)
C) The speed of the mass is constant
y
+A
t
-A
Lecture 18
Purdue University, Physics 220
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Question
• A spring oscillates back and forth on a frictionless horizontal surface. A
camera takes pictures of the position every 1/10th of a second. Which
plot best shows the positions of the mass.
A)
EndPoint
Equilibrium
EndPoint
B)
EndPoint
CORRECT
Equilibrium
EndPoint
Equilibrium
EndPoint
C)
EndPoint
Lecture 18
Purdue University, Physics 220
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Analogy to Circular Motion
What does moving in a circle have to do with moving back &
forth in a straight line?
x = A cos  = A cos (wt)
since  = w t
x
x
1
1
2
A
3
A
8
2
8

y
7
4
6

0


2
-A
5
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3
4
3
2
6
5
Purdue University, Physics 220
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Analogy to Circular Motion
Period = T (seconds per cycle)
Frequency = f = 1/T (cycles per second)
Angular frequency = w = 2f = 2/T
Lecture 18
Purdue University, Physics 220
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Analogy to Circular Motion
x(t) = Acos(wt)
x(t) = Asin(wt)
V(t) = -Awsin(wt)
V(t) = Awcos(wt)
a(t) = -Aw2cos(wt) =-w2 x(t)
a(t) = -Aw2sin(wt) =-w2 x(t)
For spring: F = ma = -kx, so a(t) = -(k/m)x(t).
By analogy, we have w2 = k/m
xmax = A
vmax = Aw
amax = Aw2
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Purdue University, Physics 220
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Lecture 18
Purdue University, Physics 220
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Example
A 3 kg mass is attached to a spring (k=24 N/m). It is stretched
5 cm. At time t=0 it is released and oscillates.
Which equation describes the position as a function of time
x(t) (in cm) =
A) 5 sin(wt)
B) 5 cos(wt)
C) 24 sin(wt)
D) 24 cos(wt)
E) -24 cos(wt)
We are told at t=0, x = +5 cm. x(t) = 5 cos(wt)
only one that works.
Lecture 18
Purdue University, Physics 220
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Example
A 3 kg mass is attached to a spring (k=24 N/m). It is stretched
5 cm. At time t=0 it is released and oscillates.
What is the total energy of the block spring system?
E = PE + KE
At t=0, x = 5 cm and v=0:
E = ½ k x2 + 0
= ½ (24 N/m) (5 cm)2
= 0.03 J
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Purdue University, Physics 220
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Example
A 3 kg mass is attached to a spring (k=24 N/m). It is stretched
5 cm. At time t=0 it is released and oscillates.
What is the maximum speed of the block?
E = PE + KE
When x = 0, maximum speed:
E = ½ m v2 + 0
(0.03 J) = ½ 3 (kg) v2
J=kg*m2/s2
v = .14 m/s
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Purdue University, Physics 220
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Example
A 3 kg mass is attached to a spring (k=24 N/m). It is stretched
5 cm. At time t=0 it is released and oscillates.
How long does it take for the block to return to x=+5cm?
w = sqrt(k/m)
= sqrt(24/3)
= 2.83 radians/sec
Returns to original position after 2  radians
T = 2 w = 6.28 / 2.83 = 2.2 seconds
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Purdue University, Physics 220
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Purdue University, Physics 220
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Pendulum Motion
• For small angles
T = mg
Tx = -mg (x/L) Note: F proportional to x!
Fx = m ax
-mg (x/L) = m ax
ax = -(g/L) x
Recall for SHO a = -w2 x
L
w = sqrt(g/L)
T = 2  sqrt(L/g)
T
Period does not depend on A, or m!
x
m
mg
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Purdue University, Physics 220
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Question
Suppose a grandfather clock (a simple pendulum) runs
slow. In order to make it run on time you should:
CORRECT
A) Make the pendulum shorter
B) Make the pendulum longer
w
g
L
2
L
T
 2
w
g
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Purdue University, Physics 220
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iClicker
In Case 1 a mass on a spring oscillates back and forth.
In Case 2 the mass is doubled but the spring and the
amplitude of the oscillation is the same as in Case 1.
Which case has the smallest maximum velocity?
A) Case 1
B) Case 2
C) Same
Same maximum Kinetic Energy
K = ½ m v2
Lecture 18
smaller mass requires larger v
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