Review for the third testChapter 14, SHMEnergy of the Spring-Mass SystemDamped oscillationsExerciseChapter 15, fluidsPressure vs. DepthBuoyancyPascal’s PrincipleContinuity equationEnergy conservation: Bernoulli’s eqnChapter 16, Macroscopic descriptionPV diagrams: Important processesWork done on a gasChapter 17, first law of ThermodynamicsSlide 16Thermal Properties of MatterHeat of transformation, specific heatSpecific heat for gasesPhysics 207: Lecture 19, Pg 1Review for the third testPhysics 207: Lecture 19, Pg 2Chapter 14, SHMkm-AA0(≡Xeq)T=2 (m/k)½T=2 (L/g)½ for pendulum The general solution is: x(t) = A cos ( t + )with = (k/m)½ and = 2 f = 2 /T T = 2 / A-APhysics 207: Lecture 19, Pg 3Energy of the Spring-Mass SystemPotential 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+)x(t) = A cos( t + ) v(t) = -A sin( t + )a(t) = -2A cos( t + ) U~cos2K~sin2E = ½ kA2Physics 207: Lecture 19, Pg 4Damped oscillationsx(t) = A exp(-bt/2m) cos (t + )-1-0.8-0.6-0.4-0.200.20.40.60.811.2tAx(t) t For small drag (under-damped) one gets:Physics 207: Lecture 19, Pg 5Exercisex (m) t (s) 0.1 -0.1 What are the amplitude, frequency, and phase of the oscillation?x(t) = A cos ( t + 1 2 3 4 A=0.1 m T=2 sec f=1/T=0.5 Hz = 2 f = rad/s radPhysics 207: Lecture 19, Pg 6Chapter 15, fluidsArea=AyPressure=P0 F=P0A+MgP=P0+ρgyP0APhysics 207: Lecture 19, Pg 7Pressure vs. DepthIn a connected liquid, the pressure is the same at all points through a horizontal line.pPhysics 207: Lecture 19, Pg 8Buoyancyy1y2 F2 F1 F2=P2 Area F1=P1 AreaF2-F1=(P2-P1) Area=ρg(y2-y1) Area=ρ g Vobject=weight of the fluiddisplaced by the objectPhysics 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.FF12d2d1AA21 P1 = P2 F1 / A1 = F2 / A2 A2 / A1 = F2 / F1 Hydraulics, a force amplifierPhysics 207: Lecture 19, Pg 10Continuity equationA1A2v1v2 A1v1 : units of m2 m/s = volume/sA2v2 : units of m2 m/s = volume/sA1v1=A2v2Physics 207: Lecture 19, Pg 11Energy conservation: Bernoulli’s eqnP1+1/2 ρ v12+ρgy1=P2+1/2 ρ v22+ρgy2P+1/2 ρ v2+ρgy= constantPhysics 207: Lecture 19, Pg 12Chapter 16, Macroscopic description Ideal gas lawP V = n R TP V= N kB Tn: # of molesN: # of particlesPhysics 207: Lecture 19, Pg 13PV diagrams: Important processesIsochoric process: V = const (aka isovolumetric)Isobaric process: p = constIsothermal process: T = constVolumePressure2211 TpTp12IsochoricVolumePressure2211 TVTV12IsobaricVolumePressure2211VpVp 12IsothermalPhysics 207: Lecture 19, Pg 14Work done on a gasW= - the area under the P-V curveVolumePressurePiPfViVf W =- PdVinitialfinalòW =-nRTVViVfòdVW =- nRT1VViVfòdVW =- nRT lnVfViæ è ç ö ø ÷Physics 207: Lecture 19, Pg 15Chapter 17, first law of ThermodynamicsΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+QFor systems where there is no change in mechanical energy:ΔEthermal =Wexternal+QPhysics 207: Lecture 19, Pg 16ExerciseVolumePressureifVi3ViPiWhat is the final temperature of the gas?How much work is done on the gas?Physics 207: Lecture 19, Pg 17Thermal Properties of MatterPhysics 207: Lecture 19, Pg 18Heat of transformation, specific heatLatent heat of transformation L is the energy required for 1 kg of substance to undergo a phase change. (J / kg)Q = ±MLSpecific 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 ΔTPhysics 207: Lecture 19, Pg 19Specific heat for gasesFor gases we typically use molar specific heat (Units: J / K mol )Q = n C ΔTFor gases there is an additional complication. Since we can alsochange the temperature by doing work, the specific heat dependson the path. Q = n CV ΔT (temperature change at constant V)Q = n CP ΔT (temperature change at constant
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