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Physics 153

2007 by Don WittThermodynamics

Thermometric Property: is a physicalproperty that changes with temperature

Different ways of measure temperature1) Expansion of a fluid.

2) Electrical

3) Mechanical expansion of solid.

4) Expansion of a gas.There are many others not listed.

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Physics 153

2007 by Don WittThermodynamics

Thermal Contact

A B C

A B C

A B

C

2)

1)

3)

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Physics 153

2007 by Don WittThermodynamics

Thermal Contact

A B C

A B C

A B

C

2)

1)

3)

None are in thermo contact

A&B are in thermo contact not C

Which are in thermal contact?

A&C and C&B

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What can we say about the temperature of

the the three objects?

Thermal Equilibrium: when the cooling stops.

The zeroth law of thermodynamics:

If two objects are in thermal equilibrium with athird, then they are in thermal equilibrium with

each other.

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Two objects are defined to have the same

temperature if they are in thermal equilibrium

with each other.

How to measure temperature:

Simple start is to pick at thermometric property

which linear.

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Physics 153

2007 by Don WittMercury Thermometer example

Example: Mercury Thermometer (Celsius temperature scale)

L100

L0

L100 is the height at

boiling point of water

L0

is the height at

freezing point of water

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2007 by Don Witt(Mercury Thermometer contd)

tc =L100 L0

-

L t L0- X 100 C

Fahrenheit:

t for ice if 32 F and t for boiling is 212 F

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2007 by Don Witt

Constant volume gas

thermometer

Example: Constant volume gas thermometer

tc =P

100P

0

-

Pt P0- X 100 C

P

t

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Physics 153

2007 by Don WittKelvin scale

The thermometer will go to P = 0 at sometemperature which is -273.15C.

Experiment give us this number along with

PV

T

= constant

We can also define a new temperature so that0K (kelvin) is -273.15C and take 1K to be 1C.

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2007 by Don Witt

Triple point of water as a

standard

Still a problem with temperature how do we calibrate theThermometer? Namely, how do we decide when we at

Freezing or boiling?

The triple point of water is for pressure at 4.58mm Hgand temperature 0.01C.

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2007 by Don WittErrors in gas thermometer

T = 273.16K P

Ptriple

T

P

O

air

N

H

2

2

2

446C

444C

Graph of temperature as a function pressure for

constant volume gas thermometer with different gases.

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Physics 153

2007 by Don WittThermal Expansion

As the temperature changes the length of a material

changes.

L = L0T

is a constant for the material, L is the change in

length, L0

the length, and T is the change in

temperature.

Not all material expand with increase in temperature.

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Physics 153

2007 by Don WittLinear Expansion

The equation can be written in the following form

L = L0T

IfT is small L should be small and so L0 L, then

one can take the limit of our above equation, namely,

L = L0 LT

Thus, ifT 0, the expansion equation becomes

dL = LdT

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2007 by Don WittExpansion coefficients

The constant is determined by the material be considered,it can be positive or negative.

Examples of values are Al 2.4X10-5 K-1

Brass 2.0X10-5 K-1

Cu 1.7X10-5 K-1

Glass 0.4-0.9X10-5 K-1

Quartz 0.04X10-5 K-1

Invar 0.09X10-5 K-1

Steel 1.2X10-5 K-1

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Physics 153

2007 by Don WittArea and Volume Expansion

The expansion formula given thus far is for linearmaterials, however, the area and volume also change

and those changes can be obtained from the linear

equation.

Example: In order to see, this take a square of metal of

area A, at temperature T, and expansion coefficient .

L

W A=LW (initial area)

If the temperature changes by T, find the

change in area A.

A+ A = (L+ L T) (W+ W T)

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2007 by Don Witt

A+ A = (L+ L T) (W+ WT)

A+ A = LW+ 2 LW T + 2W L (T)2

IfT is small, thenA+ A = A+ 2 A T .

Finally, A = 2 A T.

For volume, V = 3 V T.

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Basic Expansion Formulae and

Stress

L = L T

A = 2 A T

V = 3 V T

Stress and strain:

(read Chap 11-4 & 11-5)

Hookes Law:

(Stress)/(Strain) = (Elastic modulus)

Linear Stress and Strain:

(F/ A ) = Y (L/L)

Y is the Youngs modulus,

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Bulk stress and strain:

(F/ A ) = p = -B (V/V)

where B is the bulk modulus.

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Physics 153

2007 by Don WittExample

You are told to design a steel bridge is 600m long. How

Much allowance must be made for the linear expansion in

A temperature range -40C to +40C?

L = L T Use steel = 1.05x10-5 K-1.

L = (1.05x10-5 K-1) (600m) (80K)

L = 50 cm

Standard expansion joints allow this much expansion.

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Physics 153

2007 by Don Witt Example

Design a gasoline tank to store gasoline in a temperature range

-20C to 40C. Must be able to fill it any time of year and not

lose more than 0.1% of gasoline to overflow.

Use gasoline = 9.5 x 10-4 K-1. If you never spill any gasoline

you need =1/3 gasoline

V

V

V

Vtank gas

= -0.001

The maximum is obtain by filling the tank at the high temperature.

3 maxT- gasT= -0.001

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2007 by Don WittSolve for max.

max = 3.2 X10-4 K-1

Similarly, min is found by filling the tank at low temperature.

V

V

V

Vgas tank

= 0.001

max

= -0.001+ gas

T = -0.001+9.5X10-4 K-1(-20-40)K3 (-20-40)K3 T

min = 0.001- gasT = 0.001+9.5X10-4 K-1(40-(-20))K

-3 (40-(-20))K-3 T

min= 3.1 X10-4 K-1 so 3.1 X10-4 K-1 < < 3.2 X10-4 K-1

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2007 by Don Witt

Pivot

point

Lsteel

LAl

L

Example:

The figure on the right below shows a simple model of a pendulum for

a grandfathers clock. It consists of a steel bar steel = 11 x 10-6 K-1of length Lsteel, an aluminum bar Al = 24 x 10

-6 K-1 of length LAl and the

pendulum bob. Both bars are of negligible weight and are connectedat

their bottom ends. The bob is attached to the top end of the aluminumbar.

Build a grandfathers clock using the pendulum in the sketchthat

with L = 24.82 cm such that the period does not change due to

temperature changes. In other words, pick Lsteel and Lalsuch that L doesnt change because the period depends

on L.

0 = L = Lsteel - LAl

0 = Lsteel - LAl = steel LsteelT - AlLAlT

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2007 by Don Witt0 = steel LsteelT - AlLAlT

So steel Lsteel= AlLAl this tells us that

Lsteel / LAl= Al /steel = 24/11

24.82 cm = L = Lsteel - LAl

(1)

(2)

Equation (1) gives Lsteel= 24/11 LAl . Plug this into equation

(2) yields 24.82 cm = Lsteel - LAl = 24/11 LAl - LAl .

Solving, 24.82 cm = (24/11 - 11/11)LAl =13/11 LAl,thus LAl = 21.00 cm and Lsteel = 24.82 cm + LAl = 45.82cm

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Physics 153

2007 by Don WittExample

A 300cm3 glass is filled with 100g of ice at -5C and 200g of water

at 25C.

What is the content of the glass at equilibrium and the temperature?

(Assume the system is isolated)Start

Ice and water

The equations needed are

Q = mc T and Q = m L

Qsystem = Qice + Qwater = 0

The total heat change for the system is zero.

The ice term is tricky. It includes the heat to raise the

ice from -5C to 0C plus the heat to melt the ice plus the

plus raise that water to its final temperature.

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Qice = Qice to melting point + Qmelt ice + Qwater from ice

Qice = mice ciceTice + mice Lice + mice cwaterTwater from ice

Thus the change in heat of the ice is

cice = 2100 J/kg K

Lice = 334 x 103

J/kgcwater = 4190 J/kg K

Qice = (0.100kg) (2100 J/kg K)(0C-(-5C))+

(0.100kg)(334 x 103 J/kg)+ (0.100kg) (4190 J/kg K)(T-0C))

Qice= 34,450 J + 419 J/K T

Qwater

= ((0.200kg) (4190 J/kg C)(T-25C)) = (838 J/C)(T-25C)

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Qice + Qwater = 0 so 34,450 J + 419 J/C T + (838 J/C)(T-25C) = 0

Solve for T: 34,450 J - 20,950 J + (1257 J/C)T = 0

T= -11C answer

This answer is just totally wrong note the temperature is lower

than the start temperature!

To understand this Qice= 34,450 J + 419 J/K T when T = 0Cgives 34,450 J which means you need, however,

Qwater = (838 J/C)(T-25C)gives -20,950 J at T = 0C.

So there is not enough energy to melt all of the ice.

The only explanation is the ice never melted!

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Try your self, find how much ice melts in last example.

There is not enough heat to freeze the water because tofreeze 200g of water it would take Q=(.200kg)(334 x 103 J/kg ) which

is 66800J.

Final temperature must be 0C and have an ice water mixture.

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Physics 153

2007 by Don WittHeat

Heat:When two objects are in contact and and the energy

transfer is due solely to temperature difference the

energy transferred is called heat. The transfer is called

heat flow. The letterQ is used to denote heat.

Units for heat

1 cal = 4.186 J

1 kcal = 1000 cal

1 Btu = 778 ft lb = 1055 J

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Specific Heat Capacity:

Q = mc T

Molar heat capacity:

Q = nC Tn is the number of moles.

Phase changes also involve heat:

Q = m L

M is the mass and L is the heat to change phase.

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Physics 153

2007 by Don WittHeat Transfer by Conduction

100C 0C

Steel Bar

H = dQ = A (TH - TC)

dt L

H = dQ = -A dT

dt dx

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Physics 153

2007 by Don WittThermal Resistance

H = dQ = A (TH - TC) = A T/R

dt L

V=IR for electric circuits. Recall I = dq/dt

where R = L/ . R is called the thermal resistance

Also called R value.

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Heat Transfer by

Thermal Radiation

Black body radiation

Power radiate = eA T4

T = temperature, A =surface area, e = 1 for black body,

and = 5.67 x 10-8 W/m2 K4.

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Physics 153

2007 by Don WittHeat Transfer by Convection

Read section in book.

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Physics 153

2007 by Don WittKinetic Theory

Gas in box at

temperature T.

The average velocity of the

gas molecules is 0, because

it is equally likely to move left

and right or up and down.

In order to relate energy and temperature let us take

the rms value of v denoted vrms, i.e. root mean square.

The kinetic energy is K=1/2 mv2 and the average

value of K with v = vrms

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2007 by Don WittPressure from Kinetic Theory

vxt

# of molecules hitting onewall of box = 1/2( N/V vxt)A

Momentum for one moleculeis px= 2m vx

Pressure = F/A = m vx

( N/V vx

t)A /A t

because F=dp/dt.

Total momentum transfer = 2m vx 1/2( N/V vxt)A

Pressure = F/A = (N/V) m(vx)2

average

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Pressure = P = (N/V) m(vx)2average one show that

1/2 m(vx2)av = 1/2 kT

So PV= NkT , the expression is true if weconsider the motion as 3-dimensional because

v2 = vx2 + vy

2 + vz2

so Kav = 3/2 kT

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Physics 153

2007 by Don WittIdeal Gas law

PV = NkTwhere k= 1.38 x 10-23J/K , N = number of molecules.

Using basic chemistry N = NAn where

NA = 6.0221367x1023 molecules/mole. So the ideal gas law canBe written as

PV = nRTwhere k = R/NA .

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Physics 153

2007 by Don Witt Phases of Matter

Triple point

Critical pointP

T

Sublimation

curve

Vaporization

curve

Fusion curve

Solid Liquid

Vapor

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Physics 153

2007 by Don Witt Work Energy theorem

Let Fx be the x-component of all the forces acting on anobject of mass m, then Fx= m ax

The work done the force Fx is W = Fx dx

Plug in F=ma to obtain

(ignore the limits of integration first)

W = Fx

dx = W = m ax

dx = m ax

vx

dt

= m vx (dvx /dt )dt = m vx dvx = (1/2)m(vx)2

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W = (1/2)m(vx)2

If moving in 3-dimensions, then

W = Fx dx + Fy dy + Fz dz = F dr

by definition.

W = (1/2)m(vx)2 + (1/2)m(vy)

2 + (1/2)m(vz)2 = (1/2)mv2

(Once limits of integration are put back in, one obtain

the changes in K.)

Work Energy Theorem: W = K

where K = (1/2)m(vfinal)2 - (1/2)m(vinitial)

2

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Physics 153

2007 by Don Witt The first law of thermodynamics

dW = Fdx = pA dx

dQ = dW + dUFirst Law:

dQ = heat added to the system

dW = work done by the system

dU = internal energy of the system

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2007 by Don Witt Work and Internal Energy

The work to go from state 1 to state 2 is the

W12 = P dV , this dependents on the

path!

The integral is the area under the curve.

V1

V2

The internal energy of the system is

U = (# of degrees of freedom)/2 N k T

or

U = (# of degrees of freedom)/2 nR T

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The change in internal energy never dependents on the

path on the initial and final temperature.

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2007 by Don Witt Types of thermal processes

Isothermal: temperature is constant during process,T = constant.

Adiabatic: no heat transfer during the process, Q=0.

Isochoric: constant volume process, V = constant.

Isobaric: constant pressure process, P = constant.

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2007 by Don Witt Heat Capacities

dQ = dW + dU if the volume is constant during a

process, then dW=0 for this process

dQ = dU = n CV dT so CV = f R/2

where f = # of degrees of freedom.

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2007 by Don Witt Heat capacity for ideal gas

dQ = dW + dU if the pressure is constant during

a process, then dW = pdV = nRdT

dQ = dW + dU = nRdT + n CV dT = n CP dT

thus R + CV = CPassuming ideal gas law.

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The adiabatic index is = Cp/ CV

= Cp/ CV = (R + CV )/Cv = 1+ R/ f R/2

= 1 + 2/f = (f+2)/f

where f = ( # of degrees of freedom)

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Physics 153

2007 by Don Witt Adiabatic Process

Recall an adiabatic process is when Q=0.

Two facts

dU = nCVdT and dW = pdV. Since there is no heat

gain,

nCVdT = - dW = pdV.

Next use the ideal gas law to obtain

nCVdT = -(nRT/V)dV

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Physics 153

2007 by Don WittAdiabatic contd

nCVdT = -(nRT/V)dV so dT/T + R/CV dV/V = 0

Recall = Cp/ CV so R/CV = -1

Thus dT/T + R/CV dV/V = dT/T + (-1)dV/V = 0

TV-1= constant

Or

pV = constant

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Physics 153

2007 by Don WittHeat Engines

Q=QH+ QC

where QH is the heat transferred from the hot reservoirand QC is the heat transferred from the cold reservoir.

Hot

Cold

Work

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2007 by Don Witt Efficiency

e=W/QH

and

W= QH +Qc = |QH | - |Qc|

for a cycle.

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Example of Stirling

Engine

P

V

1

2

3

4

Paths 2 and 4 are isothermal

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Physics 153

2007 by Don WittExample contd

First, the work done for one cycle is

W = W1+ W2+ W3+ W4

no work is done along 1 or 3 because volume is

constant.

Process 2 and 4 are isothermal so the work done

is given by

W= nRT ln(Vf/ Vi)

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P

V

1

2

3

4

TH

Tc

W=nR TH ln(Vf/ Vi) + nR Tc ln(Vi/ Vf)

W=nR (TH - Tc) ln(Vf/ Vi)

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Finally, find the heat with Q>0.

Q1= nCv(TH- Tc)

V

1

3

4

TH

Tc

p2

Q3= nCv(Tc- TH)

Q2=nR TH ln(Vf/ Vi)

Q4=nR T

cln(V

i/ V

f)

e= nR (TH - Tc) ln(Vf/ Vi)/ (nR TH ln(Vf/ Vi)+ nCv(TH- Tc))

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Physics 153

2007 by Don WittCarnot Cycle

P

V

1

2

3

4

1 and 3 are adiabatic

2 and 4 are isothermal

a

b

c

d

Ta= Tb

Td = Tc

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Physics 153

2007 by Don WittCarnot Engine

QH= nR Ta ln( Vf/ Vi )

W= W1 + W2+W3+W4Note that W

1=- W

3this is because

W1 = -nCV(Ta- Td) and W3 = -nCV(Tc- Tb)

Ta= Tb and Td = Tc

Thus W= W2+W4

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e= nR (Ta - Td) ln(Vf/ Vi)/ nR Ta ln( Vf/ Vi )

e = 1- Td/ Ta

Hot

Cold

WorkTd is the temperatureof the hot source and

Ta is the cold sourcetemperature.

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Physics 153

2007 by Don WittOtto Cycle

p

VCurves are adiabatic

Start & Finish

1

2

3

4

a

b

c

d

e

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Physics 153

2007 by Don WittPiston positions

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Physics 153

2007 by Don WittPiston position start and finish

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Physics 153

2007 by Don WittOtto Cycle

p

VCurves are adiabatic

Start & Finish

1

2

3

4

a

b

c

d

e

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Physics 153

2007 by Don WittOtto Engine

Q2= 0 and Q4= 0 because each process is adiabatic

Q1 = Qh and Q3 = Qc because the heat is

proportional to the change in temperature for these.Namely,

Q1 = nCV(Ta -Td) and Q3 = nCV(Tc -Tb)

W = W2 + W4 = -nCV(Tb -Ta) -nCV(Td -Tc)

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e = W/Qh = (-nCV(Tb -Ta) -nCV(Td -Tc))/ nCV(Ta -Td)

e = 1- (Tb - Tc )/ (Ta -Td)

e = (- (Tb -Ta) - (Td -Tc))/ (Ta -Td)

e = (- Tb +Ta - Td + Tc))/ (Ta -Td)

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e = 1- (Tb - Tc )/ (Ta -Td)

For an adiabatic process TV-1= constant .

So Ta Va-1 = Tb Vb

-1 and Tc Vc-1 = Td Vd

-1

From our cycle Va = Vd and Vb

= Vc

e = 1- (Tb - Tc )/ (Ta -Td) = 1- (Va/ Vb )-1 = 1 - 1/(Vb/ Va )

-1

e = 1 - 1/r-1 where r is the compression ratio.

Ta (Va/ Vb ) -1 = Tb , Tc = Td (Vd/ Vc )

-1 = Td (Va/ Vb )-1

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Physics 153

2007 by Don WittCar Example

A 6 cylinder car engine has a displacement of 3.00L.Operates at 4000rpm with a compression ratio of

r = 9.50. Fuel enters with atmospheric pressure at a

temperature of 27oC. During combustion the fuel

reaches a temperature of 1350oC. Use = 1.4 and

cV= 0.718kJ/ kg K and R = cp - cv.

Find the power delivered.

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Vb/ V

a= r = 9.5 The displacement is for all 6 cyclinders.

So Vb - Va = 3.00L/6 = 0.500 x 10-3 m3

Vb = 0.559 x 10-3 m3 and Va = 0.588 x 10

-4 m3

n = PcVc/RTc = 2.24 x 10-2 moles or

mass = 6.49 x 10-4 kg

Pc Vc = Pd Vd

so Pd = 2.34 x 103 kPa

( R=8.31451 J/ mol K or 0.287 kJ/kg K)

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Td = PdVd/nR

Td = 739 K using ideal gas law.

Ta = 1623 K by ideal gas law.

Pa = 5.14 x 103 kPa again ideal gas law.

e = 1 - 1/r-1 = 1- 1/(9.5)1.4 -1 so e = 0.59

W = e Qh= e mcV(Ta -Td) = 0.244 kJ

Power = 6(1/2 rev)(4000 rev/min)(1 min/60s)(0.244kJ)

Power = 49 kW = 66 hp

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Physics 153

2007 by Don WittHeat Pumps Refrigerators

Hot

Cold

Work

COP (heating mode) = coefficient of performance = Qh/W

COP (cooling mode) = Qc / W

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Physics 153

2007 by Don WittEntropy

The mathematical definition of entropy isdS = dQ/T

Second law of thermodynamics:

The total entropy of an isolated system never

decreases.

Thus, the change in entropy must never decrease it

and it must zero for a reversible process.

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Physics 153

2007 by Don WittPeriodic Motion

Periodic Motion is motion thatrepeats itself.

Periodis defined to be the time over which the

motion repeats. Frequencyis the number of

cycles per time.

Amplitude is the maximum displacement fromequilibrium.

Ph i 153

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Hookes Law

F = -kx

where k is the spring constant, x is how much

from equilibrium.

Ph i 153

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Physics 153

2007 by Don WittSimple Harmonic Motion

Consider the motion of mass with a spring force

actting on it

-kx = max

where F = ma was used.

First note this is not a constant acceleration problem

Because ax depends on x.

Ph i 153

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2007 by Don WittSHM contd

One also see the motion should repeat itself

because of the way ax changes.

A good guess of a solution is

x = A cos(t +)

A is the amplitude that is maximum

displacement from equilibrium. is thephase

constantor phase shift. (All angles are measured inradians. )

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Physics 153

2007 by Don Witt

Now, lets check that the above expression is a

solution.

If x = A cos(t +), then

vx = -A sin(t +), and ax = -A 2cos(t +).

Plug into the force equation, next.

Ph i 153

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Physics 153

2007 by Don Witt

-kx = -k A cos(t +) = max-k A cos(t +) = max = m(-A

2cos(t +))

-k A cos(t +) = m(-A 2cos(t +))

Canceling like terms yields

x = A cos(t +) is solution if and only if2 = k/m

Ph sics 153 Definition of simple harmonic

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Physics 153

2007 by Don Witt

Definition of simple harmonic

motion

Motion is simple harmonic if and only if it satisfies thefollowing equation

2 x = axThe solution is x = A cos(t +).

In case of a spring 2 = k/m.

is the angular frequency measured in units of

radians/time.

Periodis defined to be the time over which the

motion repeats. Frequencyis the number of cycles

per time.

Physics 153

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Physics 153

2007 by Don WittExample Vertical Spring

M

k

The forces acting on the mass are

gravity and the spring forces.

F = ma

-ky -mg = may

Now, as written the equation doesnt looklike simple harmonic motion. Note y

measures how much the spring is

stretched.

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2007 by Don Witt

M

k

To make it easy to solve this problem,

find where the net force zero in other

words where the system is in

equilibrium

F = -ky -mg = 0.

This is where y= -mg/k .

Now,measure everything from this equilibriumposition. So pick a new coordinate, namely,

ynew = y + mg/k

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2007 by Don Witt

Plug ynew

= y + mg/k into

-ky -mg = may

yields -ky -mg = -k(ynew - mg/k) -mg = may

-kynew= may. Finally, since mg/k is constant

ay=aynew .

-kynew= maynew

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Physics 153

2007 by Don Witt

Finally, this means a vertical spring is simple

harmonic motion but the displacement must be

measure from the equilibrium position.

The angular frequency is given by 2 = k/m

Physics 153

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Physics 153

2007 by Don Witt Work Energy theorem

Let Fx be the x-component of all the forces acting on anobject of mass m, then Fx= m ax

The work done the force Fx is W = Fx dx

Plug in F=ma to obtain

(ignore the limits of integration first)

W = Fx dx = W = m ax dx = m ax vx dt

= m vx (dvx /dt )dt = m vx dvx = (1/2)m(vx)2

Physics 153

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Physics 153

2007 by Don Witt

W = (1/2)m(vx)2

If moving in 3-dimensions, then

W = Fx dx + Fy dy + Fz dz = F dr

by definition.

W = (1/2)m(vx)2 + (1/2)m(vy)2 + (1/2)m(vz)2 = (1/2)mv2

(Once limits of integration are put back in, one obtain

the changes in K.)

Work Energy Theorem: W = K

where K = (1/2)m(vfinal)2 - (1/2)m(vinitial)

2

Physics 153 C ti f E

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Physics 153

2007 by Don WittConservation of Energy

The work energy theorem states W = K. Forces can bedivided into conservative and non-conserved forces. A force is

conserved if the work done is zero when return to its initial

state.

Thus, the work can be divided into work done by

conserved and non-conservative. The work energy theorem

can be rewritten as

K = Wcon + Wnon

Finally, the above can be written as K + U = Wnonwhere U = -Wcon .

Physics 153 P t ti l E

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Physics 153

2007 by Don WittPotential Energy

K + U = Wnon

E = Wnon

where E = K + U is the total energy. U is the

potential energy and the change in U is defined byU = -Wcon

Physics 153 Common Potential Energies

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Physics 153

2007 by Don WittCommon Potential Energies

To common potential energies are for a spring and

gravity

For a spring Uspring =1/2 kx2

and

for gravity Ugravity = mgy .

The energy for a horizontal spring is

E=1/2 mv2 + 1/2 kx2

Physics 153 Comment on Vertical Springs

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Physics 153

2007 by Don WittComment on Vertical Springs

If a vertical spring is written in terms of a coordinate

measured from the equilibrium position, i.e. y=0 is where

the spring is has no force, then

-ky=mayand the energy is just

E=1/2 mv2 + 1/2 ky2 .

However, if y = 0 is were the spring is unstretched,then -ky -mg =may and the energy must include

Gravity, namely, E=1/2 mv2 + 1/2 ky2 + mgy.

Physics 153 Simple Pendulum

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Physics 153

2007 by Don WittSimple Pendulum

The pendulum is length l and the mass is

Is m. Drawing a free body diagram

W

T

Simple pendulum means wire is massless

And the mass is a point mass.

y

x

W

T

One write the forces to obtain

Physics 153 Simple Pendulum contd

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Physics 153

2007 by Don WittSimple Pendulum contd

-mg sin= max

Using radians the acceleration can be written as

ax = ld2/dt2.So the equation is

-mg sin= mld2/dt2The mass cancels and one is left with

-g/ l sin= d2/dt2

Physics 153 Small angle motion for a

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Physics 153

2007 by Don WittSmall angle motion for a

pendulum

So the motion of a pendulum is periodic but not simple

Harmonic in general. However, if the angle is small

sin

-g/ l d2/dt2

and the motion is simple harmonic for small

with 2 = g/ l .

Physics 153 Example

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Physics 153

2007 by Don WittExample

A massless spring with a spring constant k=19 N/m is

hung vertically. A mass of 0.20 kg is attached and then

released. Assume the spring was unstreched before the

mass was released. Find how far below the initial point

the mass descends. Find the frequency and amplitude ofthe resulting motion.

Solution: M

Initial position

Equilibrium

y

v=0

yf

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E= 1/2 mv2 + 1/2 ky2

measure y from the equilibrium position. Now,

E is conserved so E = 1/2 kA2 .

Plug in v=0 which means y = A. Also, v=0 at yf. So

yf= 2y.

yf= 2mg/k= 0.2063 m

and y = A = 0.1032 m.

2 = k/m and = 2f so f = (1/2)(k/m)1/2

f = 0.01633 Hz .

Physics 153 Example

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Physics 153

2007 by Don WittExample

A 3.0 kg block moving at a speed of 1.8 m/s strikes ahorizontal spring as shown in the figure. If the spring

constant is 100 N/m and the maximum compression of the

spring is 21 cm, what is the coefficient of friction between

the block and the surface?

M

k

Physics 153 Solution

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y

2007 by Don WittSolution

E = 1/2 mv2 + 1/2 kx2

E = Wnon

Wnon= F r = -fx= -Nx= -mgx

Efinal = 1/2 kx2 and Einitial = 1/2 mv2

So E = 1/2 kx2 -1/2 mv2 = -mgx = Wnon

= (1/2 kx2 +1/2 mv2 )/mgx = 0.4300

m = 3.0 kg , v= 1.8 m/s, k= 100 N/m, and x= 0.21 m

Physics 153 Example

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y

2007 by Don WittExample

A spring of negligible mass is hung vertically by

attaching its upper end to a fixed point. A small

basket with a 0.40-kg copper block inside is placed

on the lower end of the spring. The system is set into

vertical simple harmonic oscillation. The oscillationperiod is found to be 0.90 s. When the copper

block is removed, the oscillation period changes to

0.60 s.What is the mass of the basket? What is the

spring constant of the spring?

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y

2007 by Don WittSolution

Solution:M

Tinitial = 2/ = 2 ((mbasket+mCu)/k)1/2

Tfinal = 2/ = 2 ((mbasket)/k)1/2

Tinitial = 0.90 s

Tfinal = 0.60 s

mCu= 0.40-kg

(Tinitial )2 = 2/ = 42 (mbasket+mCu)/k

(Tfinal )2 = 2/ = 42 mbasket/k

Physics 153

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y

2007 by Don Witt

(Tinitial )2 = 2/ = 42 (mbasket+mCu)/k

(Tfinal )2 = 2/ = 42 mbasket/k

Combining the equations

(Tfinal )2/ mbasket = 4

2 /k = (Tinitial )2 / (mbasket+mCu)

Solve for mbasket which gives

mbasket = mCu / ((Tinitial )2/ (Tfinal )

2 -1)

mbasket = (0.40 kg)/((0.9s)2/ (0.6)2 -1) = 0.8 kg

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y

2007 by Don Witt

(Tfinal )2/ mbasket = 4

2 /k

So k = 42 mbasket /(Tfinal )2 = 42 (0.8 kg)/ (0.6s)2

k = 87.7 N/m

Physics 153 Damped Simple Harmonic Motion

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y

2007 by Don WittDamped Simple Harmonic Motion

A damping force is a non-conservative force which causesthe simple harmonic motion to loose energy. A good model of

the force is

Fdamp = -b v

where b is constant depending on the system.

If the above damping force is added to spring system

-b vx - kx = max

Physics 153 Small b motion

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2007 by Don WittSmall b motion

If the damping constant b is small the motion should

be approximately simple harmonic motion with loss

of energy.

The energy for the system is E=1/2 mv2 + 1/2 kx2

If the average is taken over one period

The average energy is Eaverage = mv2 .

P = dEaverage /dt = -bv2. (Using the fact thatP = F v )

dEaverage /dt = -bv2 = -b/m Eaverage

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dEaverage

/dt = -b/m Eaverage

Solvingfor Eaverage, E(t) = E0 e-t/ where

= m/b .

For small b, the energy is near constant

and E=1/2 k A2 . Hence, A(t) = A0 e-t/2

Thus, the solution is x = A0 e-t/2cos(t +)

Physics 153 General Solution

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2007 by Don WittGeneral Solution

One can check by plugging in that

x = A0 e-t/2cos(t +)

is general solution of the equation

-b vx - kx = max

if and only if = 0(1 - (b/2m 0 )2)1/2where0 = (k/m)

1/2 is the natural angular frequency.

Physics 153 Quality Factor and Types of

D i

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2007 by Don Witt Damping

Q = 2 E/||

|| is the energy loss in one period and E is the

total energy.

Critically Dampedis when bc= 2m 0and

Over dampingis b > bc .

Physics 153 Plot of damping

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2007 by Don WittPlot of damping

x

t

Critical damping

Over damping

Physics 153 Driven Periodic Motion

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2007 by Don WittDriven Periodic Motion

In order to keep an oscillating system running

one can adding energy by applying a periodic

driving force.

-b vx - kx + F0 cos t = max

The solution to this equation is solution of

damped plus

x =cos(t - ) where t is angularfrequency of the driving force.

Physics 153 Resonance

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2007 by Don Witt

A =F0 /(m2(022)2 +b22 )1/2

The amplitude is given by

and phase constant is given by

tan =b/m(022)

The amplitude is largest for0

Physics 153 Waves

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2007 by Don Witt

Waves are disturbances which transmit energy

and momentum but not matter.

Waves are described by an equation y = f(x,t).

Waves have two basic types transverse and

longitudinalwaves.

Physics 153 Types of waves

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2007 by Don Wittyp

Transverse waves are waves which the

disturbance is perpendicular to the direction

of propagation.

Longitudinal waves are waves which the

disturbance is parallel to the direction of

propagation.

Physics 153

2007Parameters that describe waves

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Amplitude - maximum displacement from equilibrium.

Velocity a) wave velocity

b) velocity medium displacement

Waveform - shape of wave

Physics 153

2007Wave function

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y = f( x, t) is called the wave function.

Traveling waves can be written as

y = f( x v t)

in this case the waveform is given by the function f.

y = f( x - v t) represents a wave moving left to right

andy = f( x + v t) represents a wave moving right to left.

Physics 153

2007Wave speed

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2007 by Don Wittp

R2Fsin (/2) = ma

F=ma = (s)v2/R

F=(R)v2/R

v2 = F/ (v = wave speed)

s

= mass per length

Physics 153

2007Harmonic Waves

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Waves which have a waveform which is a sinefunction are called harmonic waves.

Since the sine is periodic the wave will be

periodic in time and spatially.

The period used for simple harmonic motion appliesin addition the wavelength,, is defined to be the

distance over which the wave repeats spatially.

v = / T = f

Physics 153

2007 b D WittMost general harmonic wave

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y =A sin(kx t + )where k is the wave number k= 2/.

Again as before the + sign indicates right to left

motion and the - sign indicates a wave moving

left to right.

Physics 153

2007 b D WittEnergy

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Eaverage = 1/2 (m)22

=1/2 22x

= 1/2 22vt

Paverage = dE/dt = 1/2 22v

Physics 153

2007 b D WittSuperposition

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Two waves can be added to obtain a new wave

y = y1 + y2Example

y1 = A sin(kx - t)

and

y2 = A sin(kx - t + )

If=0, then y1+y2 = 2 A sin(kx - t)This is called constructive interference.

Physics 153

2007 b D WittInterference

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Example

y1 = A sin(kx - t)

and

y2 = A sin(kx - t + )

If= , then y1+y2 = 0

This is called destructive interference.

Physics 153

2007 b Don WittStanding Waves

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2007 by Don Witt

y1

= A sin(kx - t) and

y2 = A sin(kx + t)

sin(1) + sin(2) = 2cos((1-2)/2)sin((1+2)/2)

Let1= kx - t and 2= kx + t, then

1 -2 = kx - t - kx - t = - 2t

and 1 + 2 = 2kx

So y = 2cos(t)sin(kx)

Physics 153

2007 by Don WittHarmonics

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2007 by Don Witt

2nd harmonic

L =

In general, L = n /2 n =1, 2,3, . Using f= v

f = n (v/2L) = n f1

1st harmonic

L = /2

fundamental

frequency

Physics 153

2007 by Don WittExample of Singing Bar

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Take a long steel bar balancing in the middle

and hit (as shown) it so it generates a sound.

What is the frequency relation to length?

Both ends are the same, i.e. unfixed, so a standingwave looks like

Physics 153

2007 by Don WittExample of singing bar contd

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This is the fundamental frequency. L = /2

So just as in the string example

f = n (v/2L) = n f1 where n = 1, 2, 3, .

2nd

harmonic

Physics 153

2007 by Don Witt

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What happens when the bar is hit as shown?

f = n (v/2L) = n f1 where n = 1, 2, 3, .

Why does it produce different frequencies forthe two different hammer positions?

Physics 153

2007 by Don Witt

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2007 by Don Witt

Physics 153

2007 by Don Witt Transverse and Longitudinal

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2007 by Don Witt Transverse and Longitudinal

speed

v2 = F/ (transverse wave speed on string)

where F is the tension and the mass per

length.

v2 = B/ (logitudinal wave speed in fluid)where B is the bulk modulus and

is the

density.

Physics 153

2007 by Don Witt

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v2 = Y/ (logitudinal wave speed in solid rod)

where Y is the tension and the mass per

volume.

v2 = P/ (sound wave in an ideal gas)

where is the adiabatic index, the density,

and P the pressure.

Physics 153

2007 by Don WittIntensity

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The intensityof a wave is

I = Paverage / A

where Paverage is average power and A is

the area.

Physics 153

2007 by Don Witt

Intensity in terms of elastic

properties

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2007 by Don Witt properties

I = 1/2 (B)1/222

I = 1/2 (Y)1/222

Physics 153

2007 by Don WittAverage Human hearing

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The intensity range for average human

hearing is

10-12 W/m2 to 1 W/m2

Physics 153

2007 by Don WittIs the telephone ad truthful?

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Our phone lines are so clear you can hear

a pin drop!

Sprint Wholesale enable a Switched outbound service

allows your customers to use existing business and

residential lines for exceptional reliability and pin-drop

quality on all long-distance calls .

Physics 153

2007 by Don WittdB level

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Sound level or sound level intensity level is

measured in dB (decibels)

= 10dB log10(I/I0)

where I is the intensity and I0 theintensity for threshold of hearing.

Physics 153

2007 by Don Witt

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Threshold of hearing 0dB threshold of

pain is 120dB

Physics 153

2007 by Don WittExample of pin drop

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y

When a pin of mass 0.1g is dropped from aheight of 1m , 0.05% of its energy is

converted into a sound pulse with duration of

0.1s. Estimate the range at which the pin can

be heard. First, assume sound is at threshold

of hearing. If you assume the intensity level

must be at least 40dB for the sound to be

heard, estimate the range at which the pin can

be heard.

Physics 153

2007 by Don WittSolution to Pin Example

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y

I = Paverage

/ 4r2

Paverage = (mgy/t) x 0.05%.

m=0.1g, y=1m, g=9.8m/s2, t=0.1s

In order to hear, must have I =10-12

W/m2

So r2 = Paverage / 4 I pluging in numbers yields

r = 624 m !

Physics 153

2007 by Don WittSolution at 40dB

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y

First, convert 40dB to intensity. 40dB=10dB log (I/I0)

where I0 =10-12 W/m2. Solving I= 10-8 W/m2.

Now, Paverage / 4= constant so I1r2

1= I2r2

2 .

(10-12 W/m2)(624m)2 = (10-8 W/m2) r22

solving r2= 6.24m.

Physics 153

2007 by Don WittBeats

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y

f1= 440Hz and f2= 443Hz added together

Physics 153

2007 by Don Witt

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y

fbeat = f1 - f2

Maximum difference you hear is about 10Hz.

After this tones merge.

Physics 153

2007 by Don WittInterference

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y

y1+ y

2= Asin(kx

1-t+

1) + Asin(kx

2-t+

2)

S1

S2

Listener

or detector

x1

x2

Physics 153

2007 by Don WittInterference

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Two special cases to add waves, added to

1) maximum value constructive interference

(kx1-t+1) - ( kx2-t+ 2) = 2n

where n=0, 1, 2, 3,

2) minimum value destructive interference

(kx1-t+1) - ( kx2-t+ 2) = nwhere n=1, 3, 5,

Note: can also have negative values.

Physics 153

2007 by Don WittDons Interference Equations

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(kx1-t+1) - ( kx2-t+ 2) = (kx1- kx2)+1 - 2= 2n

1) kx + = 2n

where n=0, 1, 2, 3,

2) kx + =n

where n= 1, 3, 5,

Physics 153

2007 by Don Witt

Interference in terms of

wavelength

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1) 2x/ + = 2n where n=0, 1, 2, 3,

2) 2x/+ =n where n= 1, 3, 5,

Rewriting the equations

1) x + /2=n where n=0, 1, 2, 3,

2) x + /2=nwhere n= 1, 3, 5,

Physics 153

2007 by Don WittStanding Waves in Pipes

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Open Pipes Half open pipes

The above waves are the displacement waves.

Closed Pipes

f = n (v/2L) = n f1 where n = 1, 2, 3, .

f = n (v/4L) = n f1where n = 1, 3, 5,

Physics 153

2007 by Don WittPressure Waves

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Open Pipes Half open pipes

Closed Pipes

f = n (v/2L) = n f1 where n = 1, 2, 3, .

f = n (v/4L) = n f1where n = 1, 3, 5,

Physics 153

2007 by Don WittMoving source

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The green dot is a sound source with

waves moving in all directions, speed

of source is 0.

The same source

moving at velocity u to

right.

In the moving case, the wavelength is shorter in thedirection of motion and longer as it moves away.

u

Physics 153

2007 by Don WittMoving source doppler

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For the stationary source the distance betweencrests is the wavelength 0. In the moving caseThe wavelength has changed and is . The new

wavelength is related the old by the following

= 0 - uT

where T is the period of the wave this the time

between crests. Next, divide both sides by the

wave speed, v.

/v= 0 /v - uT /v

1/f = 1/f0- u/f0v

Physics 153

2007 by Don WittMoving source formula

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1/f = 1/f0 - u/f0v

Finally, group like terms together

1/f = 1/f0 - u/f0v= (1/f0 )(1 - u/v )

f = f0 /(1 - u/v )

A similar calculation gives the formula for the

Moving detector.

Physics 153

2007 by Don WittDoppler Effect

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f = f0 (1 + ud/v)/(1 + us/v)

where v is the speed of the wave, f0 is the frequencyemitted, us is the speed of the source, ud is the speed

of the detector.

Rules: 1) As objects approach, f > f0 2) As objects recede, f < f0

Physics 153

2007 by Don WittExample

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A hungry bat, using ultrasonic beeps, finds a

doomed moth sitting on a plant. The sound

emitted by the bat is 40,000Hz. If the doomed

moth detects a frequency of 41,000Hz when thebat swoops at it, what is the speed of the bat?

What frequency does the bat detect reflected off

the doomed moth?

Physics 153

2007 by Don WittExample

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The string A on a street musician's violin is

too tight. Comparing it to 440Hz, 3 beats are

heard. What is the frequency of the violin

string? Passengers travelling on a bus along

the street hear the frequency at exactly 440Hz.

What is the velocity of the bus?