physics 207: lecture 19, pg 1 lecture 20goals: wrap-up chapter 14 (oscillatory motion) wrap-up...
TRANSCRIPT
Physics 207: Lecture 19, Pg 1
Lecture 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 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...
km
-A A0(≡Xeq)
Physics 207: Lecture 19, Pg 3
km
-A A0(≡Xeq)
T = 1 s
k
-1.5A 1.5A0(≡Xeq)
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 x2 = ½ k A2 cos2(t + )
K = ½ m v2 = ½ k A2 sin2(t+)
U+K = ½ k A2
U~cos2K~sin2
E = ½ kA2
Physics 207: Lecture 19, Pg 5
km
-A A0
Which mass would have the largest kinetic energy
while passing through equilibrium?
A)
2k2m
-A A0
B)
k
-2A 2A0
mC)
Physics 207: Lecture 19, Pg 6
SHM So Far
For 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 7
Energy in SHM
The total energy (K + U) of a system undergoing SHM will always be constant!
-A A0x
U
U
KE
U = ½ k x2
Physics 207: Lecture 19, Pg 8
SHM and quadratic potentials
SHM will occur whenever the potential is quadratic. For small oscillations this will be true: For example, the potential between
H atoms in an H2 molecule lookssomething like this:
-A A0x
U
U
KEU
x
Physics 207: Lecture 19, Pg 9
What about Vertical Springs?
k
m
k
equilibrium
new
equilibrium
mg=k ΔL
ΔL
k
m
y=0ΔL
y
Fnet= -k (y+ ΔL)+mg=-ky Fnet =-ky=ma=m d2y/dt2
Which has the solution y(t) = A cos( t + )
Physics 207: Lecture 19, Pg 10
The “Simple” Pendulum
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.
Fy = may = T – mg cos()
Fx = max = -mg sin()
If small then x L and sin() dx/dt = L d/dt
ax = d2x/dt2 = L d2/dt2
so ax = -g = L d2/ dt2
and = cos(t + )
with = (g/L)½
T= 2π(L/g)½
L
m
mg
z
y
x
T
Physics 207: Lecture 19, Pg 11
What about friction?
One way to include friction into the model is to assume velocity dependent drag.
Fdrag= -bdragv=-bdrag dx/dt
Fnet=-kx-bdrag dx/dt = m d2x/dt2
d2x/dt2=-(k/m)x –(bdrag/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 + )
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
t
A
x(t)
t
For small drag (under-damped) one gets:
Physics 207: Lecture 19, Pg 13
Driven oscillations, resonance
So far we have considered free oscillations.Oscillations can also be driven by an external force.
extext
osci
llatio
n am
plitu
de
natural frequency
Physics 207: Lecture 19, Pg 14
Chapter 15, Fluids An actual photo of an iceberg
Physics 207: Lecture 19, Pg 15
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
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
An 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/cm3
ρ=m/V
Physics 207: Lecture 19, Pg 17
Fluids
Another parameter Pressure (force per unit area)
P=F/A
SI 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 is
Physics 207: Lecture 19, Pg 18
If the pressures were different, fluid would flow in the tube!
Pressure vs. Depth For a uniform fluid in an open
container pressure same at a given depth independent of the container
p(y)
y
Fluid level is the same everywhere in a connected container, assuming no surface forces
Why 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 depth.