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Physics 2101 Physics 2101 Section 3Section 3Section 3Section 3Apr 19Apr 19thth
Announcements:Announcements:•• SHW #11 has been postedSHW #11 has been posted
•• Midterm #4, April 28Midterm #4, April 28thth 6 pm6 pm
•• Final: May 11Final: May 11thth--7:30am7:30am
Class Website:
Final: May 11Final: May 11 7:30am7:30am•• Make up Final: May 15Make up Final: May 15thth--7:30am7:30am
Class Website:
http://www.phys.lsu.edu/classes/spring2010/phys2101http://www.phys.lsu.edu/classes/spring2010/phys2101‐‐3/3/
http://www.phys.lsu.edu/~jzhang/teaching.htmlhttp://www.phys.lsu.edu/~jzhang/teaching.html
Standing Waves: Resonance on string
H t i id l ith λHere, two sinusoidal waves with same wavelength travel in opposite directions
Interference produces Standing waves What happens if we fixed the length of the string
x = n λ2
p g= enforce that nodes are at the ends
LL = n λ
2Node at x =0 & L:
2n = 1 ⇒ 2L = λ1
n = 2 ⇒ L = 2 λ2
2= λ2
n = 3 ⇒ L = 3 λ3
2=
32
λ3
Standing Waves: Resonant frequenciesFrequencies at which standing waves are produced are the Resonant Frequencies
v = λ f
λn =2Ln
vwave = λn fn
fn =nvwave
2L
n = 1 λ1 = 2L f1 =vwave
2L
n = 2 λ2 = L f2 = 2 f1
n = 3 λ3 =23
L f3 = 3 f13
Resonant frequencies are given by n and properties of system (length, tension, and mass density)
A string that is stretched between fixed supports separated by 75 cm has resonant frequencies of 420 Hz and 315 Hz, with no intermediate
Problemλn =
2Ln
vwave = λn fn
f =nvwaveq ,
resonant frequencies. a) What is the lowest resonant frequency?
For a fixed length the resonant wavelengths are given by: kx = nπ 2π⎛ ⎜
⎞ ⎟ L nπ
fn =2L
For a fixed length, the resonant wavelengths are given by: kx = nπλ⎝
⎜ ⎠ ⎟ L = nπ
The resonant frequencies are then: 2πλ
⎛ ⎝ ⎜
⎞ ⎠ ⎟ L = nπ ⇒
2 fn
v⎛ ⎝ ⎜
⎞ ⎠ ⎟ L = n ⇒ fn =
nv2L
(n +1)v⎛ ⎞ Between adjacent resonant frequencies:
fn+1
fn
=
(n +1)v2L
⎛ ⎝ ⎜
⎞ ⎠ ⎟
nv2L
⎛ ⎝ ⎜
⎞ ⎠ ⎟
=n +1
n=
420⋅ Hz315⋅ Hz
⇒ n = 3
2L⎝ ⎠
With n = 1: f3
f1
=
3v2L
⎛ ⎝ ⎜
⎞ ⎠ ⎟
1v⎛ ⎜
⎞ ⎟
=315⋅ Hz
f1
⇒ f1 =315⋅ Hz
3=105⋅ Hz
b) What is the wave speed?
f1
2L⎝ ⎜
⎠ ⎟ f1 3
λ1 = 2L Lowest possible wavelength is and thus
vwave = λ1 f1 = 2(75 ⋅ cm)( ) 105 ⋅ Hz( )= 157.5 ⋅ m/s
Problem 16‐58: A string, tied to a sinusodial oscillator at P and running over a support Q, is stretched by a block of mass m. The length is L and the linear density is
and the frequency is fμ, and the frequency is f.a) what mass allows the to set up the fourth harmonic?
(f 4)nf τ (for 4)2
4 2
nf nL
mg
μ
τ
= =vwave = λn fn
fn =nvwave
2L 4 22
mgL L
τμ μ
= =2L
Wave Speed vs Particle SpeedTh d f i diff t th th d f th ti l th t iThe speed of a wave is different than the speed of the particles on the string.
v =F
speed of the wave
the particles on the string go up and down di
m /Lspeed of the wave
according to y = Acosωtv = −Aω sinωt
A 2 ta = −Aω 2 cosωt
so the speed of the particle has a maximum of Aωω = 2πfvparticle = 2πfA
Chapter 18: Temperature, Heat, and Thermodynamics
Definitions
“System”‐ particular object or set of objects
“Environment” ‐ everything else in the universe
What is “State” (or condition) of system?
i d i ti i t f d t t bl titi‐macroscopic description ‐ in terms of detectable quantities:
volume, pressure, mass, temperature(“State Variables”)
Study of thermal energy ‐‐> temperature
Temperature
How do we know about temperature? thermometers
Linear scale : need 2 points to define
Fahrenheit [° F] body temp and ~1/3 of body temp ~100 ° F ~33 ° F
Celsius [° C] “freezing point” and “boiling point” of water 0 ° C 100 ° C
Kelvin [K] Absolute zero and triple point of water[ ] p p0 K 273.16 K
Conversion factors
K→ ° C
T 9 T + 32 TC = TK − 273.15 (1 ΔK = 1 Δ C)
° C → ° F TF = 95 TC + 32
Temperature
Thermal Expansion
Most substances expand when heatedZrW2O8 is a ceramic with negative thermal expansion over a wide Most substances expand when heated
and contract when cooled
ptemperature range, 0‐1050 K
The change in length, ΔL ( = L ‐ L0 ), of almost all solids is ~ directly proportional to the change in temperature, ΔT ( = T ‐ T0 )to the change in temperature, ΔT ( T T0 )
ΔL L ΔTα = coefficient of thermal expansion
ΔL = αL0ΔTL = L0 1+ αΔT( )0( )What causes thermal expansion?
Example: Bimetal Strip
Common device to measure and control temperature
F = kx = kL0 1+ αΔT( )
Thermal Expansion of a Pendulum Clock
Problem 2: Pendulum Clock
A pendulum clock made of brass is designed to
Problem 2: Pendulum ClockT = 2π L
g
L L ΔL of brass is designed to keep accurate time at 20°C. If the clock operates at 0°C, does it If the original period was 1 second
L = L0 + ΔL= L0 + αbrassL0ΔT
p ,run fast or slow?
If so, how much?
If the original period was 1 second
L0 =1s2π
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
g = 24.824 cm
L = 24.824 cm 1+ (19 ×10−6 /°C)(−20°C)( )= 24.824 cm 0.9996( ) The new period is:
24 814= 24.814 T = 2π24.814
9.8= 0.9998 s
It runs slow (less time per tick) at 20°C# ti k 24 * 60*60 86400It runs slow (less time per tick) at 20°Cat 0°C: fewer ticks = 1.7hr/yr
# ticks = 24 * 60*60 = 86400# ticks = 86400 * 0.999 = 86383