Physics 207: Lecture 19, Pg 1 Lecture 20Goals: Wrap-up Chapter 14 (oscillatory motion) Wrap-up Chapter 14 (oscillatory motion) Start discussion of Chapter 15 (fluids) Start discussion of Chapter 15 (fluids) Assignment Assignment  HW-8 due Tuesday, Nov 15  Monday: Read through Chapter 15

Physics 207: Lecture 19, Pg 2 The general solution is: x(t) = A cos (  t +  ) where A = amplitude  = angular frequency  = phase constant SHM Solution... k m -A A 0(≡X eq )

Physics 207: Lecture 19, Pg 3 k m -A A0(≡X eq ) T = 1 s k -1.5A 1.5A 0(≡X eq ) T is: A)T > 1 s B)T < 1 s C) T=1 s m

Physics 207: Lecture 19, Pg 4 SHM Solution... The mechanical energy is conserved: U = ½ k x 2 = ½ k A 2 cos 2 (  t +  ) K = ½ m v 2 = ½ k A 2 sin 2 (  t+  ) U+K = ½ k A 2 U~cos 2 K~sin 2 E = ½ kA 2

Physics 207: Lecture 19, Pg 5 k m -A A 0 Which mass would have the largest kinetic energy while passing through equilibrium? A) 2k 2m -A A 0 B) k -2A 2A 0 m C)

Physics 207: Lecture 19, Pg 6 SHM So Far For SHM without friction l The frequency does not depend on the amplitude ! l The oscillation occurs around the equilibrium point where the force is zero! l Mechanical Energy is constant, it transfers between potential and kinetic energies.

Physics 207: Lecture 19, Pg 7 Energy in SHM l The total energy (K + U) of a system undergoing SHM will always be constant! -AA0 x U U K E U = ½ k x 2

Physics 207: Lecture 19, Pg 8 SHM and quadratic potentials l SHM will occur whenever the potential is quadratic. l For small oscillations this will be true: l For example, the potential between H atoms in an H 2 molecule looks something like this: -AA0 x U U K E U x

Physics 207: Lecture 19, Pg 9 What about Vertical Springs? k m k equilibrium new equilibrium mg=k ΔL ΔL k m y=0 ΔL y F net = -k (y+ ΔL)+mg=-kyF net =-ky=ma=m d 2 y/dt 2 Which has the solution y(t) = A cos(  t +  )

Physics 207: Lecture 19, Pg 10 The “Simple” Pendulum l A 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.  F y = ma y = T – mg cos(  )  F x = ma x = -mg sin(  ) If  small then x  L  and sin(  )   dx/dt = L d  /dt a x = d 2 x/dt 2 = L d 2  /dt 2 so a x = -g  = L d 2  / dt 2 and  =   cos(  t +  ) with  = (g/L) ½ T  = 2  (L/g) ½  L m mg z y x T

Physics 207: Lecture 19, Pg 11 What about friction? l One way to include friction into the model is to assume velocity dependent drag. F drag = -b drag v=-b drag dx/dt F net =-kx-b drag dx/dt = m d 2 x/dt 2 d 2 x/dt 2 =-(k/m)x –(b drag /m) dx/dt a new differential equation for x(t) !

Physics 207: Lecture 19, Pg 12 Damped oscillations x(t) = A exp(-bt/2m) cos (  t +  ) x(t) t l For small drag (under-damped) one gets:

Physics 207: Lecture 19, Pg 13 Driven oscillations, resonance l So far we have considered free oscillations. l Oscillations can also be driven by an external force.  ext     ext oscillation amplitude    natural frequency

Physics 207: Lecture 19, Pg 14 Chapter 15, Fluids l An actual photo of an iceberg

Physics 207: Lecture 19, Pg 15 l At 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 surface l What 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 16 Fluids l An intrinsic parameter of a fluid  Density (mass per unit volume) units : kg/m 3 = g/cm 3  (water) = x 10 3 kg/m 3 = g/cm 3  (ice) = x 10 3 kg/m 3 = g/cm 3  (air) = 1.29 kg/m 3 = 1.29 x g/cm 3  (Hg) = 13.6 x10 3 kg/m 3 = 13.6 g/cm 3 ρ=m/V

Physics 207: Lecture 19, Pg 17 Fluids l Another parameter  Pressure (force per unit area) P=F/A SI unit for pressure is 1 Pascal = 1 N/m 2 1 atm = x10 5 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in 2 (=PSI) l The atmospheric pressure at sea-level is

Physics 207: Lecture 19, Pg 18 F If the pressures were different, fluid would flow in the tube! Pressure vs. Depth l For a uniform fluid in an open container pressure same at a given depth independent of the container l Fluid level is the same everywhere in a connected container, assuming no surface forces l Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium? F Imagine a tube that would connect two regions at the same depth.