Review for the third test 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thermal Properties of Matter

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

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