Physics 270

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Physics 270

• Dr. Martin Partlan

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Chapter 15 - Waves

• Terms Period, Wavelength, Wave Speed, Amplitude,
  Transverse and Longitudinal
• Wave Velocity Transverse, Longitudinal
• Energy Transport - Amplitude, Frequency
• Terms Wave number, Phase, Phase Velocity
• Principal of superposition

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Chapter 15 - Waves

• Reflection Transmission
• Interference
• Standing Waves Nodes, Antinodes
• Over Tones Resonance

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Chapter 15 - Waves

• Important Problems 15-59, 62, 69, 73

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Problem 15-59

• The speed of the longitudinal wave in a solid is
  given by
• v (B/r)1/2.
• If we form the ratio for two rods with the same
  bulk modulus, we get
• v2/v1 (r1/r2)1/2 (1/2)1/2, or v1 v2v2
  .
• The speed will be greater in the less dense rod
  by a factor of v2.

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Problem 15-62

• We assume that the change in tension does not
  change the mass density, so the velocity
  variation depends only on the tension. Because
  the wavelength does not change, we have
• l v1/f1 v2/f2 , or FT2/FT1 (f2/f1)2.
• For the fractional change we have
• (FT2 FT1)/FT1 (FT2/FT1) 1 (f2/f1)2 1
  (200 Hz)/(205 Hz)2 1 0.048.
• Thus the tension should be decreased by
  4.8.

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Problem 15-69

• The fundamental standing wave will have a node at
  the
• fixed end and an antinode at the free end. The
  distance
• between an adjacent node-antinode pair is l/4, so
  the
• wavelength of the fundamental is 4L.
• The first overtone will have an additional
  node-antinode pair, so the wavelength is 4L/3.
• The wavelength of the next overtone, with another
  node-antinode pair, is 4L/5.
• Thus the general form for the wavelengths is
• ln 4L/(2n 1), n 1, 2, 3, .

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Quiz 1 Solutions

• 1) A string is 5.5m in length and fixed to a wall
  at one end. The string has a mass of 13g and is
  vibrated at a frequency of 12Hz. The tension in
  the string is 75N.
• What is the speed of the waves in the string?
• What is the wavelength of these waves?

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Quiz 1 Solutions (Continued)

• Standing waves of the second overtone are formed
  on a string 1.2m in length at a frequency of
  24Hz.
• What is the wavelength of this wave?
• The second overtone means L 3l/2. l 2(1.2)/3
  .8m

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Chapter 16 - Sound

• Speed of Sound, Loudness, Pitch
• Pressure Waves
• Intensity of Sound waves
• Air Columns, Strings Overtones and Harmonics
• Beats
• Doppler Effect
• Principle Problems 16 - 81, 85, 92

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Chapter 16 - Sound

• 16-81
• The tension and mass density of the string
  determines the velocity
• v (FT/m)1/2.
• Because the strings have the same length, the
  wavelengths are the same, so for the ratio of
  frequencies we have
• fn1/fn vn1/vn (mn/mn1)1/2 1.5, or
• mn1/mn (1/1.5)2 0.444.
• With respect to the lowest string, we have
• mn1/m1 (0.444)n.
• If we call the mass density of the lowest
  string 1, we have
• 1, 0.444, 0.198, 0.0878, 0.0389.

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Chapter 16 - Sound

• 16-85
• 85. We find the ratio of intensities from
• Db 10 log10(I2/I1)
• 10 dB 10 log10(I2/I1), which gives I2/I1
  0.10.
• If we assume uniform spreading of the sounds,
  the intensity is proportional to the power
  output, so we have
• P2/P1 I2/I1
• P2/(150 W) 0.10, which gives P2 15 W.

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• 16-92
• (a) Because both sources are moving toward the
  observer at the same speed, they will have the
  same Doppler shift, so the beat frequency will be
  0.
• (b) Because the wavelength in front of a moving
  source decreases, the wavelength from the
  approaching loudspeaker is
• l1 (v vcar)/f0 (343 m/s 10.0 m/s)/(200
  Hz) 1.665 m.
• This wavelength approaches the stationary
  listener at a relative speed of v, so the
  frequency heard by the listener is
• f1 v/l1 (343 m/s)/(1.665 m) 206 Hz.
• Because the wavelength behind a moving source
  decreases, the wavelength from the receding
  loudspeaker is
• l2 (v vcar)/f0 (343 m/s 10.0 m/s)/(200
  Hz) 1.765 m.
• This wavelength approaches the stationary
  listener at a relative speed of v, so the
  frequency heard by the listener is
• f2 v/l2 (343 m/s)/(1.765 m) 194 Hz.
• The beat frequency is fbeat Df 206 Hz 194
  Hz 12 Hz.
• Note that this frequency may be too high to be
  heard as beats.
• (c) Because both sources are moving away from
  the observer at the same speed, they will have
  the same Doppler shift, so the beat frequency
  will be 0.

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Problem 17-14

• (a) The expansion of the container causes the
  enclosed volume to increase as if it were made of
  the same material as the container. The volume
  of water that was lost is DV DVwater
  DVcontainer V0bwater DT V0bcontainer DT
  V0(bwater bcontainer) DT
• (0.35 g)/(0.98324 g/mL) (55.50 mL)210 106
  (C)1 bcontainer(60C 20C), which gives
  bcontainer 50 106 (C)1.(b) From
  Table 171, copper is the most
  likely material.

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Problem 17-18

• (a) We consider a fixed mass of the substance.
  The change in volume from the temperature change
  is DV bV0 DT.
• Because the density is mass/volume, for the
  fractional change in the density we have
• Dr/r (1/V) (1/V0)/(1/V0) (V0 V)/V (V0
  V)/V0 DV/V0 b DT, which we can write
  Dr br DT.
• (b) For the lead sphere we have
• Dr/r 87 106 (C)1( 40C 25C)
  0.0057 (0.57).

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Problem 17-20

• (a) The radius will increase as if it were a
  length r r(1 a DT).
• The new surface area will be
• A 4pr2 4pr2(1 a DT)2 A(1 a DT)2.
• Thus the change in area is
• DA A A A2a DT (aDT)2 2Aa DT (1 !a
  DT).
• Because a DT/2 1, we have DA 2Aa DT
  8pr2a DT.
• (b) For the iron sphere we have
• DA 8p(60.0 cm)212 106 (C)1(310C 20C)
  
• 3.1 102 cm2.

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Problem 17-24

• (a) The tensile strain must compensate for the
  thermal contraction. From the relation between
  stress and strain, we have Stress E(Strain)
  Ea DT
• F/A (200 109 N/m2)12 106 (C)1( 30C
  30C) 1.4 108 N/m2 (tensile).
• (b) No, because the ultimate strength
  of steel is 500 106 N/m2 5.0 108 N/m2.
• (c) For concrete we have
• F/A (20 109 N/m2)12 106 (C)1( 30C
  30C) 1.4 107 N/m2 (tensile).
• Because the ultimate tensile strength of concrete
  is 2 106 N/m2, it will fracture.

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Problem 17-

• (a) T1(K) T1(C) 273 4000C 273
  4273 K
• T2(K) T2(C) 273 15 106 C 273
  15 106 K.
• (b) The difference in each case is 273, so we
  have
• Earth (273)(100)/(4273) 6.4
• Sun (273)(100)/(15 106) 0.0018.

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Problem 17-34

• (a) For the ideal gas we have
• PV nRT
• (1.000 atm 0.350 atm)(1.013 105 Pa/atm)V
  (18.75 mol)(8.315 J/mol K)(283 K), which
  gives V 0.323 m3.
• (b) For the two states of the gas we can write
• P1V1 nRT1 and P2V2 nRT2 , which can be
  combined to give
• (P2/P1)(V2/V1) T2/T1
• (1.00 atm 1.00 atm)/(1.00 atm 0.350
  atm)(1/2) (T2/283 K),
• which gives T2 210 K 63C.

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Problem 17-42

• The pressure at the bottom of the lake is
• Pbottom Ptop rgh 1.013 105 Pa (1000
  kg/m3)(9.80 m/s2)(37.0 m) 4.64 105 Pa.
• For the two states of the gas we can write
• PbottomVbottom nRTbottom , and PtopVtop
  nRTtop , which can be combined to give
• (Pbottom/Ptop)(Vbottom/Vtop) Tbottom/Ttop
• (4.64 105 Pa/1.013 105 Pa)(1.00 cm3/Vtop)
  (278.7 K/294.2 K), which gives Vtop 4.83
  cm3.

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Problem 17-48

• We find the number of moles in one breath from
• PV nRT
• (1.013 105 Pa)(2.0 103 m3) n(8.315 J/mol
  K)(300 K), which gives n 8.12 102 mol.
• For the number of molecules in one breath we
  have
• N nNA (8.12 102 mol)(6.02 1023
  molecules/mol) 4.9 1022 molecules.
• We assume that all of the molecules from the
  last breath that Galileo took are uniformly
  spread throughout the atmosphere, so the fraction
  that are in one breath is given by V/Vatmosphere
  . We find the number now in one breath from
• N /N V/Vatmosphere V/4pR2h
• N /(4.9 1022 molecules) (2.0 103
  m3)/4p(6.4 106 m)2(10 103 m), which gives N
  20 molecules.

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Problem 17-52

• The two temperatures of the gas are
• T1 (273.16 K)(P1/Ptp)
• (273.16 K)(218 torr/286 torr) 208.21 K
• T2 (273.16 K)(P2/Ptp)
• (273.16 K)(128 torr/163 torr) 214.51 K.
• For a constant-volume thermometer, the actual
  temperature is
• We do this limiting procedure by assuming a
  linear relation and extrapolating to Ptp 0
• (T T1)/(P1 0) (T T2)/(P2 0)
• (T - 208.21 K)/286 torr (T 214.51 K)/163
  torr, which gives T 222.9 K.

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Problems 56, 57, 58

• 56. The ideal gas law is
• PV nRT (m/M)RT,
• where m is the mass and M is the molecular
  weight. We write this as
• P (m/V)RT/M rRT/M.
• 57. We use the ideal gas law
• PV NkT
• (1.013 105 Pa)(8.0 m)(6.0 m)(4.2 m) N(1.38
  1023 J/K)(293 K),
• which gives N 5.1 1027 molecules.
• We find the number of moles from
• n N/NA (5.05 1027 molecules)/(6.02 1023
  molecules/mol) 8.4 103 mol.
• 58. We use the ideal gas law
• PV NkT, or
• N/V P/kT (1 1012 N/m2)/(1.38 1023
  J/K)(273 K)(106 cm3/m3) 3 102
  molecules/cm3.

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Problem 17-62

• (a) The ideal gas law is PV nRT.
• For a small change in volume at constant
  pressure we have
• P dV nR dT, or dV/V dT/T.
• The thermal expansion is dV/V b dT, so we see
  that b 1/T. At 293 K we have
• b 1/293 K 3.41 103 (C)1, which agrees
  with the value of 3400 106 (C)1 in Table
  171.
• (b) When the pressure and volume change at
  constant temperature, we can differentiate the
  ideal gas law
• P dV V dP 0 , or dV/V dP/P.
• From the definition of the bulk modulus, we have
• B dP/(dV/V) dP/( dP/P) P.

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Chapter 17 - Temperature

• Terms System, State, State Variables,
  Thermodynamics
• Atomic theory of matter AMU Phases of matter
• Temperature Centigrade Celsius Fahrenheit
  Kelvin
• Zeroth Law Thermal Equilibrium Definition of
  temperature
• Thermal expansion Atomic Theory Water
• Gas Laws Absolute Temperature, Ideal Gas Law,
  Avogadros Number
• Principle Problems 17 57, 66, 70

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Faster Than the Speed of Sound
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Quiz 2 Solutions

• An observer moves away from a fixed sound source.
  Is there a speed that will cause the observed
  frequency to be zero? If so, what is it?
• The speed of sound
• Shock waves are produced when an object moves
  faster than the speed of sound in that medium.
• For the second overtone L 5l/4
• so l.456m and f 343/.456 752.2Hz

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Chapter 18 Kinetic Theory

• Postulates of Kinetic Theory
• Temperature and Kinetic Theory
• Distribution of Speeds
• Real Gas and Change of Phase
• Vapor Pressure and Humidity
• Van der Walls Equation of State
• Principle Problems

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Chapter 19 Heat Thermodynamics

• Heat and Energy Transfer
• Internal Energy
• Specific Heat
• Latent Heat
• 1st Law of Thermodynamics
• Work
• Equipartition of Energy
• Heat Transfer
• Principle Problems