Lecture 20SHM Solution...Slide 3Slide 4Slide 5SHM So FarEnergy in SHMSHM and quadratic potentialsWhat about Vertical Springs?The “Simple” PendulumWhat about friction?Damped oscillationsDriven oscillations, resonanceChapter 15, FluidsSlide 15FluidsSlide 17Pressure vs. DepthPhysics 207: Lecture 19, Pg 1Lecture 20Goals:Goals:•Wrap-up Chapter 14 (oscillatory motion)Wrap-up Chapter 14 (oscillatory motion)•Start discussion of Chapter 15 (fluids)Start discussion of Chapter 15 (fluids)•AssignmentAssignment HW-8 due Tuesday, Nov 15 Monday: Read through Chapter 15Physics 207: Lecture 19, Pg 2The general solution is: x(t) = A cos (t + )where A = amplitude = angular frequency = phase constant SHM Solution...km-AA0(≡Xeq)Physics 207: Lecture 19, Pg 3km-AA0(≡Xeq)T = 1 sk-1.5A1.5A0(≡Xeq)T is:A)T > 1 sB)T < 1 sC) T=1 smPhysics 207: Lecture 19, Pg 4SHM Solution...The mechanical energy is conserved:U = ½ k x2 = ½ k A2 cos2(t + )K = ½ m v2 = ½ k A2 sin2(t+)U+K = ½ k A2 U~cos2K~sin2E = ½ kA2Physics 207: Lecture 19, Pg 5km-AA0Which mass would have the largest kinetic energy while passing through equilibrium?A)2k2m-AA0B) k-2A2A0mC)Physics 207: Lecture 19, Pg 6SHM So FarFor SHM without friction The frequency does not depend on the amplitude !The oscillation occurs around the equilibrium point where the force is zero! Mechanical Energy is constant, it transfers between potential and kinetic energies.Physics 207: Lecture 19, Pg 7Energy in SHMThe total energy (K + U) of a system undergoing SHM will always be constant!-A A0xUUKEU = ½ k x2Physics 207: Lecture 19, Pg 8SHM and quadratic potentialsSHM will occur whenever the potential is quadratic.For small oscillations this will be true:For example, the potential betweenH atoms in an H2 molecule lookssomething like this:-A A0xUUKEUxPhysics 207: Lecture 19, Pg 9What about Vertical Springs?kmkequilibriumnewequilibriummg=k ΔL ΔLkmy=0ΔLyFnet= -k (y+ ΔL)+mg=-ky Fnet =-ky=ma=m d2y/dt2Which has the solution y(t) = A cos( t + )Physics 207: Lecture 19, Pg 10The “Simple” PendulumA pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements. Fy = may = T – mg cos() Fx = max = -mg sin()If small then x L and sin() dx/dt = L d/dtax = d2x/dt2 = L d2/dt2so ax = -g = L d2/ dt2 and = cos(t + ) with = (g/L)½T= 2π(L/g)½LmmgzyxTPhysics 207: Lecture 19, Pg 11What about friction?One way to include friction into the model is to assume velocity dependent drag.Fdrag= -bdragv=-bdrag dx/dtFnet=-kx-bdrag dx/dt = m d2x/dt2 d2x/dt2=-(k/m)x –(bdrag/m) dx/dta new differential equation for x(t) !Physics 207: Lecture 19, Pg 12Damped 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 13Driven oscillations, resonanceSo far we have considered free oscillations.Oscillations can also be driven by an external force.extextoscillation amplitudenatural frequencyPhysics 207: Lecture 19, Pg 14Chapter 15, FluidsAn actual photo of an icebergPhysics 207: Lecture 19, Pg 15At ordinary temperature, matter exists in one of three states Solid - has a shape and forms a surface Liquid - has no shape but forms a surface Gas - has no shape and forms no surfaceWhat do we mean by “fluids”? Fluids are “substances that flow”…. “substances that take the shape of the container” Atoms and molecules are free to move.Physics 207: Lecture 19, Pg 16FluidsAn intrinsic parameter of a fluid Density (mass per unit volume) units :kg/m3 = 10-3 g/cm3 (water) = 1.000 x 103 kg/m3 = 1.000 g/cm3(ice) = 0.917 x 103 kg/m3 = 0.917 g/cm3(air) = 1.29 kg/m3 = 1.29 x 10-3 g/cm3(Hg) = 13.6 x103 kg/m3 = 13.6 g/cm3ρ=m/VPhysics 207: Lecture 19, Pg 17FluidsAnother parameter Pressure (force per unit area) P=F/ASI unit for pressure is 1 Pascal = 1 N/m2 1 atm = 1.013 x105 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in2 (=PSI)The atmospheric pressure at sea-level isPhysics 207: Lecture 19, Pg 18 If the pressures were different, fluid would flow in the tube!Pressure vs. DepthFor a uniform fluid in an open container pressure same at a given depth independent of the containerp(y)yFluid level is the same everywhere in a connected container, assuming no surface forcesWhy is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium? Imagine a tube that would connect two regions at the same
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