1. Review for the third test1

2. T = 2/ Chapter 14, SHM T=2 (m/k)½ T=2 (L/g)½ for pendulum
0(≡Xeq) A
The general solution is: x(t) = A cos ( wt + f)
with w = (k/m)½ and w = 2p f = 2p /T
T = 2/ A -A

3. Energy of the Spring-Mass System
x(t) = A cos( t +  )
v(t) = -A sin( t +  )
a(t) = -2A cos( t +  )
Potential energy of the spring is
U = ½ k x2 = ½ k A2 cos2(t + )
The Kinetic energy is
K = ½ mv2= ½ k A2 sin2(t+f)
E = ½ kA2
U~cos2
K~sin2

4. Damped oscillations
For small drag (under-damped) one gets:
x(t) = A exp(-bt/2m) cos (wt + f)
x(t) t

5. Exercise x(t) = A cos ( wt + f) A=0.1 m T=2 sec f=1/T=0.5 Hz
x (m)
0.1
t (s) 1 2 3 4
-0.1
What are the amplitude, frequency, and phase of the oscillation?
x(t) = A cos ( wt + f)
A=0.1 m
T=2 sec
f=1/T=0.5 Hz
w = 2p f = p rad/s
f=p rad

6. Chapter 15, fluids
Pressure=P0
F=P0A+Mg
P0A
P=P0+ρgy y
Area=A

7. Pressure vs. Depth
In a connected liquid, the pressure is the same at all points through a horizontal line. p

8. F1 F2 Buoyancy 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 y1 F1 y2
1) It directly follows from this that, the density of the object basically determines if the object is going to float or not. If the density of the object is larger than the density of the liquid, it will sink. If it is smaller than the density of the liquid than it will float. F2

9. Hydraulics, a force amplifier
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
P1 = P2
F1 / A1 = F2 / A2
A2 / A1 = F2 / F1

10. A1v1=A2v2 A1v1 : units of m2 m/s = volume/s
Continuity equation A 1 2 v
A1v1 : units of m2 m/s = volume/s
A2v2 : units of m2 m/s = volume/s
A1v1=A2v2

11. Energy conservation: Bernoulli’s eqn
P1+1/2 ρ v12+ρgy1=P2+1/2 ρ v22+ρgy2
P+1/2 ρ v2+ρgy= constant

12. Chapter 16, Macroscopic description
Ideal gas law
P V = n R T
P V= N kB T
n: # of moles
N: # of particles

13. PV diagrams: Important processes
Isochoric process: V = const (aka isovolumetric)
Isobaric process: p = const
Isothermal process: T = const
Volume
Pressure 1 2
Isochoric
Volume
Pressure 1 2
Isothermal
Volume
Pressure 1 2
Isobaric

14. W= - the area under the P-V curve
Work done on a gas
W= - the area under the P-V curve Pi
Pressure Pf Vi Vf
Volume

15. Chapter 17, first law of Thermodynamics
ΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+Q
For systems where there is no change in mechanical energy:
ΔEthermal =Wexternal+Q

16. Exercise Pressure Pi Volume Vi 3Vif Pi
Volume Vi
3Vi
What is the final temperature of the gas?
How much work is done on the gas?

17. Thermal Properties of Matter

18. Heat of transformation, specific heat
Latent heat of transformation L is the energy required for 1 kg of substance to undergo a phase change. (J / kg)
Q = ±ML
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

19. Specific heat for gases
For gases we typically use molar specific heat (Units: J / K mol )
Q = n C ΔT
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