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Lecture PHYSICS 18
18
Elasticity and Oscillations
Textbook Sections 11.1 11.5
Lecture 18
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Overview
Last Lecture
Archimedes Principle: Buoyant force is weight of displaced fluid
F=gV
Mass flow rate: Av (kg/s)
Volume flow rate: Av (m3/s)
Continuity Principle
A1 v1 = A2 v2
Bernoullis Principle
Where the velocity is high, the pressure is low and where the velocity
is low the pressure is high
P1 + v12 + gh1 = P2 + v22 + gh2
Today
Simple Harmonic Motion
Springs Revisited
Pendulum
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Quiz
1) An artery with cross sectional area of 1 cm2 branches into
20 smaller arteries each with 0.5 cm2 cross sectional area.
If the velocity of blood in the thicker artery is what is the
If the velocity of blood in the thicker artery is v, what is the
velocity of the blood in the thinner arteries?
A)
B)
C)
D)
E)
0.1 v
0.2 v
0.5 v
v
2v
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Quiz
2) A flask of water rests on a scale. If you dip your finger into
the water, without touching the flask, the reading on the
scale.
scale.
A) Increases
B) Remains the same
C) Decreases
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Hookes Law Revisited
Spring
F = -k x
What happens to k if cut spring in half?
Increases
k is inversely proportional to length!
Define
Strain = L / L
Stress = F/A
Hookes Law
Stress = Y Strain
F/A = Y L/L
k = Y A/L from F = k L
Y (Youngs Modules) independent of L or A
Has the unit of pressure (i.e., Pascal)
Inherent stiffness of a material
stiffness of material
For many materials, Y is the same for compression and tension
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Beyond Hooks Law
Ultimate
Strength
Strength
Breaking
Point
Elastic limit
Proportional
limit
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Simple Harmonic Oscillation
relaxed position
position
The force exerted by a
spring is proportional
to the distance the
th di
th
spring is stretched or
compressed from its
relaxed position.
FX = -k x
x is the displacement
is the displacement
from the relaxed
position and k is the
constant of
constant of
proportionality.
Lecture 18
FX = 0
x=0
x
FX = - kx < 0
x
x>0
x=0
FX = -kx > 0
x<0
x=0
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x
7
Potential Energy in Spring
Force of spring is Conservative
F = -k x
W = - k x2
Force
work
x
Work done only depends on initial and final position
Define Potential Energy PEspring = k x2
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Energy Conservation
A mass is attached to a spring and set to
motion. The maximum displacement is x=A
Wnc = KE + PE
0 = KE + PE or Energy PE+KE is constant!
Energy = k x2 + m v2
PE
At maximum displacement x=A, v = 0
S
Energy = k A2 + 0
At zero displacement x = 0
Energy
Energy = 0 + mvm2
mv
0
Since Total Energy is same
k A2 = m vm2
m
vm = sqrt(k/m) A
x=0
Lecture 18
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x
x
9
Vertical Mass and Spring
If we include gravity, there are two forces acting on
mass With mass new equilibrium position has spring
mass. With mass, new equilibrium position has spring
stretched d
At equilibrium: F=kdmg=0
d = mg/k
Let this point be y=0
thi
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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ILQ
A mass on a spring oscillates back & forth with simple harmonic
motion of amplitude A. A plot of displacement (x) versus time (t) is
shown below. At what points during its oscillation is the total energy
(K+U) of the mass and spring a maximum? (ignore gravity).
A) When x = +A or -A (i.e. maximum displacement)
B) When x = 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.
x
+A
t
Lecture 18
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ILQ
A mass on a spring oscillates back & forth with simple harmonic
motion of amplitude A. A plot of displacement (x) versus time (t) is
shown below. At what points during its oscillation is the speed of the
block biggest?
A) When x = +A or -A (i.e. maximum displacement)
B) When x = 0 (i.e. zero displacement)
C) The speed of the mass is constant
Well it isnt constant, and it is zero at the
maximums, so the zero position is the only
other choice
x
+A
t
-A
Lecture 18
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ILQ
A mass on a spring oscillates back & forth with simple harmonic
motion of amplitude A. A plot of displacement (x) versus time (t)
is shown below At what points during its oscillation is the
is shown below. At what points during its oscillation is the
magnitude of the acceleration of the block biggest?
A) When x = +A or -A (i.e. maximum displacement)
B) When x = 0 (i.e. zero displacement)
C) The acceleration of the mass is constant
Acceleration has it's max where the force is
largest, at maximum displacements
a(max)
a(max)=A(k/m)
x
+A
t
Lecture 18
-A
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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
plot best shows the positions of the mass.
A)
EndPoint
Equilibrium
EndPoint
B)
EndPoint
CORRECT
Equilibrium
EndPoint
Equilibrium
EndPoint
C)
EndPoint
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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 (t)
cos
cos
since = t
x
x
1
2
A
3
A
8
1
2
8
y
7
7
3
0
3
2
2
4
6
5
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4
5
15
Simple Harmonic Motion
Period = T (seconds per cycle)
Frequency = f = 1/T (cycles per second)
Angular frequency = = 2f = 2/T
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Springs and Simple Harmonic Motion
X=0
X=A; v=0; a=-amax
X=0; v=-vmax; a=0
X=-A; v=0; a=amax
X=0; v=vmax; a=0
X=A; v=0; a=-amax
X=-A
Lecture 18
X=A
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Simple Harmonic Motion
x(t) = [A]cos(t)
x(t) = [A]sin(t)
v(t) = -[A]sin(t)
a(t) = -[A2]cos(t)
OR
v(t) = [A]cos(t)
a(t) -[A2]sin(t)
xmax = = A
Period = T (seconds per cycle)
vmax = A
Frequency = f = 1/T (cycles per second)
amax = A2
Angular frequency =
= 2 f = 2 / T
For spring: 2 = k/m
since F = ma = -kx
ma
kx
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Period of a Spring
Simple Harmonic Oscillator
=2f =2/T
x(t) = [A] cos(t)
v(t) = -[A] sin(t)
[A sin(
a(t) = -[A2] cos(t)
For a Spring F = -kx
amax = (k/m) A
A2 = (k/m) A
= sqrt(k/m)
Lecture 18
=
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k
m
m
T = 2
k
19
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.
cm At time t=0 it is released and oscillates
Which equation describes the position as a function of time
x(t) =
A) 5 sin(t)
B) 5 cos(t)
C) 24 sin(t)
D) 24 cos(t)
E) -24 cos(t)
We are told at t=0, x = +5 cm. x(t) = 5 cos(t)
only one that works
only one that works.
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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.
cm At time
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 = k x2 + 0
= (24 N/m) (5 cm)2
= 0.03 J
0.03
Lecture 18
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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.
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 = m v2 + 0
.03 = 3 kg v2
v = .14 m/s
.14 m/s
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.
cm At time
it is released and oscillates
What is the maximum speed of the block?
A) .45 m/s
B) .23 m/s
C) .14 m/s
E=U+K
When x = 0, maximum speed:
E = m v2 + 0
.03 = 3 kg v2
kg
v = .14 m/s
Lecture 18
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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.
cm At time
it is released and oscillates
How long does it take for the block to return to x=+5cm?
= sqrt(k/m)
= sqrt(24/3)
= 2.83 radians/sec
Returns to original position after 2 radians
T = 2 / = 6.28 / 2.83 = 2.2 seconds
6.28 2.83 2.2 seconds
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Question
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.
In which case is the maximum kinetic energy of the mass the
biggest?
A) Case 1
B) Case 2
C) Same
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Answer
PE = 1/2kx2
KE = 0
x=-A
x=0
A)
same
for both
x=-A
x=0
x=+A
Case 1
B)
x=+A
PE = 0
KE = KEMAX
same
for both
Case 2
C) Same
Lecture 18
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Velocity ACT
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
amplitude of the oscillation is the same as in Case 1.
Which case has the largest maximum velocity?
A) Case 1
B) Case 2
C) Same
Same
Same maximum Kinetic Energy
K = m v2
Lecture 18
smaller mass requires larger v
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Potential Energy ACT
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.
In which case is the maximum potential energy of the mass and
spring the biggest?
A) Case 1
B) Case 2
C) Same
Look
Look at time of maximum displacement x = A
Energy = k A2 + 0 Same for both!
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Simple Pendulum
The restoring force is the component of gravity directed
along the body path:
along the bodys path:
Frestore = Fparallel = - m g sin
If the angle is small, sin
The angle is also related to the displacement by = y/L
Applying Newtons Second Law: Fparallel = mgy = ma
The force is proportional to the displacement
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L
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Simple Pendulum: Frequency
The frequency of a simple pendulum is given by
1g
=
2 L
The frequency is independent of the mass of the
pendulum bob
The frequency is independent of the amplitude of
the motion
the motion
However, small angles were assumed
As long as the angle is under ~30, this equation is valid
long as the angle is under 30
this equation is valid
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Pendulum Motion
For small angles
T = mg
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 = -2 x
L
= sqrt(g/L)
T = 2 sqrt(L/g)
T
Period does not depend on A, or m!
x
m
mg
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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
=
T=
Lecture 18
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2
g
L
= 2
L
g
32
Elevator ACT
A pendulum is hanging vertically from the ceiling of an elevator.
Initially the elevator is at rest and the period of the pendulum is T.
Now the pendulum accelerates upward. The period of the pendulum
will now be
A) greater than T
B) equal to T
CORRECT
C) less than T
=
T=
Effective g is larger when accelerating upward
2
g
L
= 2
L
g
(you feel heavier)
Lecture 18
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Elevator ACT
A pendulum is hanging vertically from the ceiling of an elevator.
Initially the elevator is at rest and the period of the pendulum is T.
Now the pendulum accelerates upward. If you are accelerating
upward your weight is the same as if g had
A) increased
B) same
C) decreased
Effective g is larger when accelerating upward
is
(you feel heavier)
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Summary of Concepts
Simple Harmonic Motion
Occurs when have linear restoring force F= -kx
x(t) = [A] cos(t)
v(t) = -[A] sin(t)
a(t) = -[A2] cos(t)
cos(
Springs
F = -kx
PE = k x2
= sqrt(k/m)
Pendulum (Small oscillations)
= sqrt(g/L)
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