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