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Physics 207: Lecture 26, Pg 1 Review for the third testGoals: Review Chapters 14-17 Review Chapters 14-17 Third test on Thursday at 7:15 pm.
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Physics 207: Lecture 26, Pg 2 Chapter 14, SHM k m -A A0(≡X eq ) T=2 (m/k) ½ T=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 -A
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Physics 207: Lecture 26, Pg 3 Energy of the Spring-Mass System Potential energy of the spring is U = ½ k x 2 = ½ k A 2 cos 2 ( t + ) The Kinetic energy is K = ½ mv 2= ½ k A 2 sin 2 ( t+ ) x(t) = A cos( t + ) v(t) = - A sin( t + ) a(t) = - 2 A cos( t + ) U~cos 2 K~sin 2 E = ½ kA 2
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Physics 207: Lecture 26, Pg 4 Damped oscillations x(t) = A exp(-bt/2m) cos ( t + ) x(t) t l For small drag (under-damped) one gets:
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Physics 207: Lecture 26, Pg 5 Exercise x (m) t (s) 0.2 -0.2 l What are the amplitude, frequency, and phase of the oscillation? x(t) = A cos ( t + 1 23 4 A=0.2 m T=2 secf=1/T=0.5 Hz = 2 f = rad/s rad
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Physics 207: Lecture 26, Pg 6 Chapter 15, fluids Area=A y Pressure=P 0 F=P 0 A+Mg P=P 0 +ρgy P0AP0A
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Physics 207: Lecture 26, Pg 7 Pressure vs. Depth l In a connected liquid, the pressure is the same at all points through a horizontal line. p
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Physics 207: Lecture 26, Pg 8 Buoyancy y1y1 y2y2 F 2 F 1 F 2 =P 2 Area F 1 =P 1 Area F 2 -F 1 =(P 2 -P 1 ) Area =ρg(y 2 -y 1 ) Area =ρ g V object =weight of the fluid displaced by the object
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Physics 207: Lecture 26, Pg 9 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. P 1 = P 2 F 1 / A 1 = F 2 / A 2 A 2 / A 1 = F 2 / F 1 Hydraulics, a force amplifier
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Physics 207: Lecture 26, Pg 10 Exercise m 1 =10 kg r 1 =0.1 m m 2 =? kg r 2 =1 m P 1 = P 2 m 1 g / A 1 = m 2 g / A 2 m 2 =m 1 A 2 /A 1 =1000 kg
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Physics 207: Lecture 26, Pg 11 m 1 =10 kg r 1 =0.1 m m 2 =? kg r 2 =1 m h A P A =1 atm +m 1 g/ A 1 +ρgh l What is the pressure at the bottom of the container?
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Physics 207: Lecture 26, Pg 12 Continuity equation A 1 A 2 v 1 v 2 A 1 v 1 : units of m 2 m/s = volume/s A 2 v 2 : units of m 2 m/s = volume/s A 1 v 1 =A 2 v 2
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Physics 207: Lecture 26, Pg 13 Energy conservation: Bernoulli’s eqn P 1 +1/2 ρ v 1 2 +ρgy 1 =P 2 +1/2 ρ v 2 2 +ρgy 2 P+1/2 ρ v 2 +ρgy= constant
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Physics 207: Lecture 26, Pg 14 Chapter 16, Macroscopic description l Ideal gas law P V = n R T P V= N k B T n: # of moles N: # of particles
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Physics 207: Lecture 26, Pg 15 PV diagrams: Important processes l Isochoric process: V = const (aka isovolumetric) l Isobaric process: p = const l Isothermal process: T = const Volume Pressure 1 2 Isochoric Volume Pressure 1 2 Isobaric Volume Pressure 1 2 Isothermal
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Physics 207: Lecture 26, Pg 16 Work done on a gas W= - the area under the P-V curve Volume Pressure PiPi PfPf ViVi VfVf
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Physics 207: Lecture 26, Pg 17 Chapter 17, first law of Thermodynamics ΔU+ΔK+ΔE thermal =ΔE system =W external +Q l For systems where there is no change in mechanical energy: ΔE thermal =W external +Q
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Physics 207: Lecture 26, Pg 18 Exercise Volume Pressure i f ViVi 3V i PiPi l What is the final temperature of the gas? l How much work is done on the gas?
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Physics 207: Lecture 26, Pg 19 Thermal Properties of Matter
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Physics 207: Lecture 26, Pg 20 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
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Physics 207: Lecture 26, Pg 21 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 C V ΔT (temperature change at constant V) Q = n C P ΔT (temperature change at constant P) C P =C V +R
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Physics 207: Lecture 26, Pg 22 Exercise Volume Pressure i f ViVi 3V i PiPi l How much heat energy is transferred to or from the gas?
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