Physics 220

1
Physics 220

• Dr. Martin Partlan

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Chapter 11 - Waves

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

3
Chapter 11 - Waves

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

4
Chapter 11 - Waves

• Recommended Problems
• 34-41, 51-60

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Chapter 11 - Waves

• Solutions to Problems 11- 38
• 11- 56
• 11- 61
• Follow

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Problem 11-38

• Because the modulus does not change, the speed
  depends on the density
• v µ (1/r)1/2.
• Thus we see that the speed will be greater in the
  less dense rod.
• For the ratio of speeds we have
• v1/v2 (r2/r1)1/2 (2)1/2 1.41.

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Problem 11-56

• All harmonics are present in a vibrating string
  fn nf1 , n 1, 2, 3, 4 . The difference in
  frequencies for two successive overtones is
• fn1 fn (n 1)f1 nf1 f1 ,
• so we have f1 350 Hz 280 Hz 70 Hz.

8
Problem 11-61

• The hanging weight creates the tension in the
  string FT mg. The speed of the wave depends
  on the tension and the mass density
• v (FT/µ)1/2 (mg/µ)1/2
• and thus is constant. The frequency is fixed by
  the vibrator, so the constant wavelength is
• l v/f (1/f)(mg/µ)1/2 (1/60 Hz)(0.080
  kg)(9.80 m/s2)/(5.6 104 kg/m)1/2 0.624 m.
• The different standing waves correspond to
  different integral numbers of loops, starting at
  one loop. With a node at each end, each loop
  corresponds to l/2. The lengths of the string
  for the possible standing wavelengths are
• Ln nl/2 n(0.624 m)/2 n(0.312 m), n 1,
  2, 3, ¼ , or
• Ln 0.312 m, 0.624 m, 0.924 m, 1.248 m, 1.560
  m, ¼ .
• Thus we see that there are 4 standing
  waves for lengths between 0.10 m and 1.5 m.

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Problem 11-60

• The hanging weight creates the tension in the
  string FT mg. The speed of the wave depends
  on the tension and the mass density
• v (FT/µ)1/2 (mg/µ)1/2.
• The frequency is fixed by the vibrator, so the
  wavelength is l v/f (1/f)(mg/µ)1/2. With a
  node at each end, each loop corresponds to l /2.
  
• (a) For one loop, we have l 1/2 L, or 2L
  v1/f (1/f)(m1g/µ)1/2
• 2(1.50 m) (1/60 Hz)m1(9.80 m/s2)/(4.3 x 104
  kg/m)1/2, which gives m1 1.4 kg.
• (b) For two loops, we have l 2/2 L/2, or L
  v2/f (1/f)(m2g/µ)1/2
• 1.50 m (1/60 Hz)m2(9.80 m/s2)/(4.3 x 104
  kg/m)1/2, which gives m2 0.36 kg.
• (c) For five loops, we have l 5/2 L/5, or 2L/5
  v5/f (1/f)(m5g/µ)1/2
• 2(1.50 m)/5 (1/60 Hz)m5(9.80 m/s2)/(4.3 x104
  kg/m)1/2, which gives m5 0.057 kg.
• The amplitude of the standing wave can be much
  greater than the vibrator amplitude because of
  the resonance built up from the reflected waves
  at the two ends of the string.

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Problem 11-62

• (a) The wavelength of the fundamental for a
  string is 2L, so the fundamental frequency is
• f (1/2L)(FT/µ)1/2.
• When the tension is changed, the change in
  frequency is
• Df f f (1/2L)(FT/µ)1/2 (FT/µ)1/2
• (1/2L)(FT/µ)1/2(FT/FT)1/2 1 f(FT
  DFT)/FT1/2 1
• f1 (DFT/FT)1/2 1.
• If DFT/FT is small, we have
• 1 (DFT/FT)1/2 1 !(DFT/FT), so we get
• Df f1 1/2(DFT/FT) 1 1/2(DFT/FT)f.
• (b) With the given data, we get Df
  !(DFT/FT)f
• 442 Hz 438 Hz 1/2(DFT/FT)(438 Hz), which
  gives DFT/FT 0.018 1.8 (increase).
• (c) For each overtone there will be a new
  wavelength, but the wavelength does not change
  when the tension changes, so the formula will
  apply to the overtones.

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Problem 11-74

• The frequency of the sound will be the frequency
  of the needle passing over the ripples. The
  speed of the needle relative to the ripples is v
  rw, so the frequency is
• f v/l rw/l
• (0.128 m)(33 rev/min)(2prad/rev)/(60
  s/min)(1.7x103 m) 260 Hz.

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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 12 - Sound

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

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Quiz 2 Solutions

• 1) 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
• 2) Shock waves are produced when an object moves
  faster than the speed of sound in that medium.
• 3) For the second overtone L 5l/4
• so l.456m and f 343/.456 752.2Hz

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Problem 12-26

• The wavelength of the fundamental frequency for a
  string is l 2L, so the speed of a wave on the
  string is
• v lf 2(0.32 m)(440 Hz) 282 m/s.
• We find the tension from
• v FT/(m/L)1/2
• 282 m/s FT/(0.35 10ñ3 kg)/(0.32 m)1/2 ,
  which gives FT 87 N.

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Problem 12-30

• For an open pipe the wavelength of the
  fundamental frequency is l 2L.
• We find the required lengths from
• v lf 2Lf
• 343 m/s 2Llowest(20 Hz), which gives Llowest
  8.6 m
• 343 m/s 2Lhighest(20,000 Hz), which gives
  Lhighest 8.6 10ñ3 m 8.6 mm.
• Thus the range of lengths is 8.6 mm 8.6 m

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Problem 12-38

• For an open pipe all harmonics are present, the
  difference in frequencies is the fundamental
• frequency, and all frequencies will be integral
  multiples of the difference. For a closed
• pipe only odd harmonics are present, the
  difference in frequencies is twice the
  fundamental frequency, and frequencies will not
  be integral multiples of the difference but odd
  multiples of half the difference. For this pipe
  we have
• ?f 280 Hz ñ 240 Hz 40 Hz.
• Because we have frequencies that are integral
  multiples of this, the pipe is open,
  with a fundamental frequency of 40 Hz.
• The wavelength of the fundamental frequency is l1
  2L. We find the length from
• v l1f 1 2Lf1
• 343 m/s 2L(40 Hz), which gives L 4.3 m.

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Problem 12-48

• For destructive interference, the path difference
  is an odd multiple of half the wavelength.
• (a) Because the path difference is fixed, the
  lowest frequency corresponds to the longest
• wavelength. We find this from DL l1/2, so we
  have
• f1 v/l1 (343 m/s)/2(3.5 m ñ 3.0 m)
  343 Hz.
• (b) We find the wavelength for the next frequency
  from DL 3l2/2, so we have
• f2 v/l2 (343 m/s)/2(3.5 m ñ 3.0 m)/3
  1030 Hz.
• We find the wavelength for the next frequency
  from DL 5l3/2, so we have
• f3 v/l3 (343 m/s)/2(3.5 m ñ 3.0 m)/5
  1715 Hz.

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Problem 12-52

• Because the bat is at rest, the wavelength
  traveling toward the object is
• l1 v/f0 (343 m/s)/(50,000 Hz) 6.86 10ñ3
  m.
• The wavelength approaches the object at a
  relative speed of v ñ vobject. The sound strikes
  and reflects from the object with a frequency
• f1 (v ñ vobject)/l1 (343 m/s ñ 25.0
  m/s)/(6.86 10ñ3 m) 46,360 Hz.
• This frequency can be considered emitted by the
  object, which is moving away from the bat.
  Because the wavelength behind a moving source
  increases, the wavelength approaching the bat is
• l2 (v vobject)/f1 (343 m/s 25.0
  m/s)/(46,360 Hz) 7.94 10ñ3 m.
• This wavelength approaches the bat at a relative
  speed of v, so the frequency received by the bat
  is
• f2 v/l2 (343 m/s)/(7.94 10ñ3 m)
  43,200 Hz.

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Problem 12-54

• Because the wavelength in front of a moving
  source decreases, the wavelength from the
  approaching automobile is
• l1 (v ñ v1)/f0 (343 m/s ñ 15 m/s)/f0.
• 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)/(343 m/s ñ 15 m/s)/f0
  (343 m/s)f0/(328 m/s).
• Because the frequency from the stationary
  automobile is unchanged, the beat frequency is
• fbeat ?f f1 ñ f0
• 5.5 Hz (343 m/s)f0/(328 m/s) ñ f0 , which
  gives f0 120 Hz.

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Problem 12-58

• Because the source is at rest, the wavelength
  traveling toward the heart is
• l1 v/f0 .
• If we assume that the heart is moving away, this
  wavelength approaches the heart at a relative
  speed of v ñ vheart. The ultrasound strikes and
  reflects from the heart with a frequency
• f1 (v - vheart)/l1 (v - vheart)/(v/f0) (v
  - vheart)f0/v.
• This frequency can be considered emitted by the
  heart, which is moving away from the source.
  Because the wavelength behind the moving heart
  increases, the wavelength approaching the source
  is l2 (v vheart)/f1 (v vheart)/(v -
  vheart)f0/v 1 (vheart/v)v/f01 -
  (vheart/v).
• This wavelength approaches the source at a
  relative speed of v, so the frequency received by
  the source is f2 v/l2 v/1
  (vheart/v)v/f01 - (vheart/v) 1 -
  (vheart/v)f0/1 (vheart/v).
• Because vheart v, we use 1/1 (vheart/v) 1
  - (vheart/v), so we have
• f2 f01 - (vheart/v)2 f01 - 2(vheart/v).
• The maximum beat frequency occurs for the maximum
  heart velocity, so we have
• fbeat f0 - f2 2(vheart/v)f0 600 Hz
  2vheart/(1.54 103 m/s)(2.25 106 Hz), which
  gives vheart 0.205 m/s.

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Problem 12-62

• In a time t the shock wave moves a distance
  vsoundt perpendicular to the wavefront. In the
  same time the object moves vobjectt.
• We see from the diagram that sin q
  (vsoundt)/(vobjectt) vsound/vobject .

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Problem 13-12

• We can treat the change in diameter as a simple
  change in length, so we have
• ?L aL0 ?T
• 1.869 cm - 1.871 cm 12 10-6 (C8)-1(1.871
  cm)(T - 208C), which gives T -698C.

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Problem 13-

• 16. (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
• ?V ?Vwater- ?Vcontainer V0bwater ?T -
  V0bcontainer ?T V0(bwater - bcontainer)?T
• (0.35 g)/(0.98324 g/mL) (65.50 mL)210 10-6
  (C8)-1 - bcontainer(608C - 208C), which gives
• bcontainer 74 10-6 (C8)-1.
• (b) From Table 13-1, aluminum is the
  most likely material.

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Problem 13-20

• (a) We consider a fixed mass of the substance.
  The change in volume from the temperature change
  is
• ?V bV0 ?T.
• Because the density is mass/volume, for the
  fractional change in the density we have
• ?r/r (1/V) - (1/V0)/(1/V0) (V0 - V)/V (V0
  - V)/V0 - ?V/V0 - b ?T,
• which we can write
• ?r - br ?T.
• (b) For the lead sphere we have
• ?r/r - 87 10-6 (C8)-1(408C - 258C)
  - 0.0057 (0.57).

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Problem 12-30

• (a) T1(K) T1(8C) 273 40008C 273
  4273 K
• T2(K) T2(8C) 273 15 106 8C 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 13-34

• (a) For the ideal gas we have
• PV nRT (m/M)RT
• (1.013 105 Pa)V (18.5 kg)(103 g/kg)/(28
  g/mol)(8.314 J/mol ? K)(273 K),
• which gives V 14.8 m3.
• (b) With the additional mass in the same volume,
  we have
• PV nRT (m/M)RT
• P(14.8 m3) (18.5 kg 15.0 kg)(103 g/kg)/(28
  g/mol)(8.314 J/mol ? K)(273 K),
• which gives P 1.83 105 Pa 1.81 atm.

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Problem 13-40

• 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
• (0.70 atm/1.00 atm)(V2/V1) (278.2 K/293.2 K),
  which gives V2/V1 1.4.

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Problem 13-44

• (a)We find the number of moles from
• n rV/M r4pR2d/M
• (1000 kg/m3)3p(6.4 106 m)2(3 103 m)(103
  g/kg)/(18 g/mol) 6 1022 mol.
• (b) For the number of molecules we have
• N nNA (6 1022 mol)(6.02 1023
  molecules/mol) 4 1046 molecules.

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Problem 13-46

• We find the number of moles in one breath from
• PV nRT
• (1.013 105 Pa)(2.0 10ñ3 m3) n(8.315 J/mol ?
  K)(300 K), which gives n 8.02 10ñ2 mol.
• For the number of molecules in one breath we have
• N nNA (8.02 10ñ2 mol)(6.02 1023
  molecules/mol) 4.83 1022 molecules.
• We assume that all of the molecules from the last
  breath that Einstein 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/(4.83 1022 molecules) (2.0 10ñ3
  m3)/4p(6.4 106 m)2(10 103 m),
• which gives N 20 molecules.

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Chapter XX - 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

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Answers to Exam 1

• 1a) False, 1b) (ii) pass through each other,
  1c) True, 1d)False, 1e) True,
• 1f) Beats occur when two sound waves interfere
  in time. The two waves have different
  frequencies and so periodically, they
  constructively and destructively interfere
  causing periodic maxima and minima of sound.
• 2) D T 100 deg C
• 3) T-180 deg C
• 4) f35.78Hz
• 5a) I 10-7W/m2, 5b) 40.45dB
• 6) F 588Hz
• 7) d1 .66m, d2 1.44m, d32.22m

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Chapter xx 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 xx Heat Thermodynamics

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