Page Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 1Review for the third testPhysics 207: Lecture 19, Pg 2Chapter 14, SHMkm-AA0(≡Xeq)T=2π(m/k)½T=2π(L/g)½for pendulum The general solution is: x(t) = A cos ( ωt + φ)with ω= (k/m)½and ω= 2πf = 2π/TT = 2π/ωA-APage Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 3Energy of the Spring-Mass SystemPotential energy of the spring is U = ½ k x2= ½ k A2cos2(ωt + φ)The Kinetic energy is K = ½ mv2=½ k A2sin2(ωt+φ)x(t) = A cos( ωt + φ )v(t) = -ωA sin( ωt + φ )a(t) = -ω2A cos( ωt + φ )U~cos2K~sin2E = ½ kA2Physics 207: Lecture 19, Pg 4Damped oscillationsx(t) = A exp(-bt/2m) cos (ωt + φ)-1-0.8-0.6-0.4-0.200.20.40.60.811.2ωωωωtAx(t) t lFor small drag (under-damped) one gets:Page Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 5Exercisex (m) t (s) 0.1 -0.1 lWhat are the amplitude, frequency, and phase of the oscillation?x(t) = A cos ( ωt + φ)1 2 3 4 A=0.1 m T=2 sec f=1/T=0.5 Hz ω= 2πf = πrad/sφ=πradPhysics 207: Lecture 19, Pg 6Chapter 15, fluidsArea=AyPressure=P0F=P0A+MgP=P0+ρgyP0APage Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 7Pressure vs. DepthlIn a connected liquid, the pressure is the same at all points through a horizontal line.pPhysics 207: Lecture 19, Pg 8Buoyancyy1y2F2F1F2=P2AreaF1=P1AreaF2-F1=(P2-P1) Area=ρg(y2-y1) Area=ρ g Vobject=weight of the fluiddisplaced by the objectPage Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 9Pascal’s PrincipleAny 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.FF12d2d1AA21P1= P2 F1/ A1 = F2/ A2A2/ A1 = F2/ F1Hydraulics, a force amplifierPhysics 207: Lecture 19, Pg 10Continuity equationA1A2v1v2A1v1: units of m2m/s = volume/sA2v2: units of m2m/s = volume/sA1v1=A2v2Page Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 11Energy conservation: Bernoulli’s eqnP1+1/2 ρ v12+ρgy1=P2+1/2 ρv22+ρgy2P+1/2 ρ v2+ρgy= constantPhysics 207: Lecture 19, Pg 12Chapter 16, Macroscopic description lIdeal gas lawP V = n R TP V= N kBTn: # of molesN: # of particlesPage Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 13PV diagrams: Important processeslIsochoric process: V = const (aka isovolumetric)lIsobaric process: p = constlIsothermal process: T = constVolumePressure2211 TpTp=12IsochoricVolumePressure2211 TVTV=12IsobaricVolumePressure2211VpVp=12IsothermalPhysics 207: Lecture 19, Pg 14Work done on a gasW= - the area under the P-V curveVolumePressurePiPfViVfW = − PdVinitialfinal∫W = −nRTVViVf∫dVW = −nRT1VViVf∫dVW = −nRT lnVfVi Page Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 15Chapter 17, first law of ThermodynamicsΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+QlFor systems where there is no change in mechanical energy:ΔEthermal =Wexternal+QPhysics 207: Lecture 19, Pg 16ExerciseVolumePressureifVi3ViPilWhat is the final temperature of the gas?lHow much work is done on the gas?Page Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 17Thermal Properties of MatterPhysics 207: Lecture 19, Pg 18Heat of transformation, specific heatlLatent heat of transformation L is the energy required for 1 kg of substance to undergo a phase change. (J / kg)Q = ±MLlSpecific 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 ΔTPage Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 19Specific heat for gaseslFor gases we typically use molar specific heat (Units: J / K mol )Q = n C ΔTlFor gases there is an additional complication. Since we can alsochange the temperature by doing work, the specific heat dependson the path. Q = n CVΔT (temperature change at constant V)Q = n CPΔT (temperature change at constant
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