Physics 207 – Lecture 19
Review for the third test
Physics 207: Lecture 19, Pg 1
Chapter 14, SHM
k
-A
0(≡Xeq
)
T=2π (m/k)½
T=2π (L/g)½ for pendulum
m
A
The general solution is: x(t) = A cos ( ωt + φ)
with ω = (k/m)½ and ω = 2π f = 2π /T
T = 2π/ω
A
-A
Physics 207: Lecture 19, Pg 2
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Physics 207 – Lecture 19
Energy of the Spring-Mass System
x(t) = A cos( ωt + φ )
v(t) = -ωA sin( ωt + φ )
a(t) = -ω2A cos( ωt + φ )
Potential energy of the spring is
U = ½ k x2 = ½ k A2 cos2(ωt + φ)
The Kinetic energy is
K = ½ mv2= ½ k A2 sin2(ωt+φ)
E = ½ kA2
U~cos2
K~sin2
Physics 207: Lecture 19, Pg 3
Damped oscillations
lFor small drag (under-damped) one gets:
x(t) = A exp(-bt/2m) cos (ωt + φ)
x(t)
1.2
1
0.8
0.6
0.4
A
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
ωt
t
Physics 207: Lecture 19, Pg 4
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Physics 207 – Lecture 19
Exercise
x (m)
0.1
2
1
4
3
t (s)
-0.1
l What
are the amplitude, frequency, and phase of the oscillation?
x(t) = A cos ( ωt + φ)
A=0.1 m
T=2 sec f=1/T=0.5 Hz
ω = 2π f = π rad/s
φ=π rad
Physics 207: Lecture 19, Pg 5
Chapter 15, fluids
Pressure=P0
F=P0A+Mg
P0A
P=P0+ρgy
y
Area=A
Physics 207: Lecture 19, Pg 6
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Physics 207 – Lecture 19
Pressure vs. Depth
l In
a connected liquid, the pressure is the same at
all points through a horizontal line.
p
Physics 207: Lecture 19, Pg 7
Buoyancy
y1
F2=P2 Area
F1=P1 Area
F2-F1=(P2-P1) Area
=ρg(y2-y1) Area
=ρ g Vobject
=weight of the fluid
displaced by the
object
F1 y2
F2
Physics 207: Lecture 19, Pg 8
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Physics 207 – Lecture 19
Pascal’s Principle
Any change in the pressure applied to an enclosed
fluid is transmitted to every portion of the fluid and to
the walls of the containing vessel.
Hydraulics, a force amplifier
F1
F2
P1 = P2
F1 / A1 = F2 / A2
d2
d1
A1
A2 / A1 = F2 / F1
A2
Physics 207: Lecture 19, Pg 9
Continuity equation
A
2
A
1
v
1
v
2
A1v1 : units of m2 m/s = volume/s
A2v2 : units of m2 m/s = volume/s
A1v1=A2v2
Physics 207: Lecture 19, Pg 10
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Physics 207 – Lecture 19
Energy conservation: Bernoulli’s eqn
P1+1/2 ρ v12+ρgy1=P2+1/2 ρ
v22+ρgy2
P+1/2 ρ v2+ρgy=
constant
Physics 207: Lecture 19, Pg 11
Chapter 16, Macroscopic description
l Ideal
gas law
PV=nRT
P V= N kB T
n: # of moles
N: # of particles
Physics 207: Lecture 19, Pg 12
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Physics 207 – Lecture 19
PV diagrams: Important processes
l Isochoric
process: V = const (aka isovolumetric)
l Isobaric process:
p = const
l Isothermal process: T = const
Isobaric
Isothermal
1
p1V1 = p2V2
1
V1 V2
=
T1 T2
Pressure
p1 p 2
=
T1 T2
Pressure
Pressure
Isochoric
2
1
2
2
Volume
Volume
Volume
Physics 207: Lecture 19, Pg 13
Work done on a gas
Pressure
W= - the area under the P-V curve
Pi
final
W = − ∫ PdV
initial
Vf nRT
W =−∫
Pf
Vi
Vi
V
dV
Vf
1
W = −nRT ∫ dV
Vi V
 Vf 
W = −nRT ln 
 Vi 
Vf
Volume
Physics 207: Lecture 19, Pg 14
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Physics 207 – Lecture 19
Chapter 17, first law of Thermodynamics
ΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+Q
l For
systems where there is no change in
mechanical energy:
ΔEthermal =Wexternal+Q
Physics 207: Lecture 19, Pg 15
Pressure
Exercise
Pi
i
f
Volume
Vi
3Vi
l What
is the final temperature of the gas?
l How much work is done on the gas?
Physics 207: Lecture 19, Pg 16
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Physics 207 – Lecture 19
Thermal Properties of Matter
Physics 207: Lecture 19, Pg 17
Heat of transformation, specific heat
l Latent
heat of transformation L is the energy required for 1 kg of
substance to undergo a phase change. (J / kg)
Q = ±ML
l Specific
heat c of a substance is the energy required to raise the
temperature of 1 kg by 1 K. (Units: J / K kg )
Q = M c ΔT
Physics 207: Lecture 19, Pg 18
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Physics 207 – Lecture 19
Specific heat for gases
l For
gases we typically use molar specific heat (Units: J / K mol )
Q = n C ΔT
l For
gases there is an additional complication. Since we can also
change the temperature by doing work, the specific heat depends
on the path.
Q = n CV ΔT (temperature change at constant V)
Q = n CP ΔT (temperature change at constant P)
CP=CV+R
Physics 207: Lecture 19, Pg 19
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