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Waves & Sound

Waves & Sound. Objectives. FCAT Periodicity of waves Movement of particles in transverse vs longitudinal wave. Sunshine State Standards.

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Waves & Sound

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  1. Waves& Sound

  2. Objectives • FCAT • Periodicity of waves • Movement of particles in transverse vs longitudinal wave

  3. Sunshine State Standards • SC.H.3.4.2 Students know that technological problems often create a demand for new scientific knowledge and that new technologies make it possible for scientists to extend their research in a way that advances science. • SC.H.3.4.6 Students know that scientific knowledge is used by those who engage in design and technology to solve practical problems, taking human values and limitations into account

  4. Chapters 14 & 15 • Waves are periodic disturbances that propagate through a medium or space • a medium does not travel with the wave • Mechanical waves require a medium • Electromagnetic waves do not require a medium

  5. Key Formulae Periodic Motion F = -kx PE = ½ kx2 T = 2p L/g Waves f = 1/T v = f l I = P/A b = 10 log I fd = fs v + vd Io v + vs VT inoC = 331.5 m/s + (0.6m/soC)(DT)

  6. Assignments P&P14 Waves 32: 14/1-3,6,7,9,10 33: 14/15,16,79 P&P15 Sound 34: 15/6-8,14

  7. Periodic Motion • is repetitive motion • Simple harmonic motion is the result of a restoring force on an object being directly proportional to the object’s displacement from equilibrium. That type of force obeys Hooke’s Law: F = -kx • F, force, n k, spring constant, n/m • X, distance, m • Ignore the “–” (means restoring force working against applied force)

  8. Sample Problem Page 378. 1. How much force is necessary to stretch a spring 0.25 m when the spring constant is 95 N/m? F = kx F = (95 N/m)(0.25m) F = 24 N

  9. Pendulums T, period, sec L, length of pendulum, m g, gravity, 9.8 m/s2 on Earth What is the period on Earth of a pendulum with a length of 1.0 m? = 2 p sr(1.0m / 9.8 m/s2) = 2.0s

  10. Measuring Waves l, (lambda), wavelength, m v, speed, m/s f, frequency, 1/s, hertz f = 1/T f, frequency, 1/s T, period, s

  11. p. 386 15. A sound wave produced by a clock chime is heard 515 m away 1.50 s later. • What is the speed of sound of the clock’s chime in air? • V = d/t = 515 m / 1.50 s = 343 m/s • The sound wave has a frequency of 436 Hz. What is the period of the wave? • t = 1/f = 1/436 Hz = 2.29 x 10-3 s • What is the wave’s wavelength? • g = V / f = 343 m/2 / 436 Hz = 0.787 m

  12. More on waves: • Longitudinal waves – particles of the medium move parallel to the direction of the wave Sound is an example of this type of wave. Sound also requires a medium, categorizing it as a mechanical wave.

  13. Transverse Waves • Transverse waves – displacement of the particles of the medium are perpendicular to the direction of propagation of the wave.

  14. Phase of a wave physics.mtsu.edu

  15. Frequency

  16. Properties of Waves • 1. Rectilinear propagation – advancement of a wave is perpendicular to the wave front • 2. Reflection – waves bounce off barriers and rebound in opposite direction • Law of reflection: • Incident angle = reflected angle (i = r) • 3. Refraction – bending (changing direction) of a wave as it travels from one medium into another

  17. More properties… • 4. Diffraction – spreading of a wave as it passes beyond the edge of a barrier • http://www.physicsclassroom.com/Class/waves/U10L3b.html • 5. Interference – result of 2 or more waves passing through the same medium at the same time http://www.physicsclassroom.com/Class/waves/U10L3c.html

  18. Refraction – bending after passing from one substance into another Source: chemicalparadigms.wikispaces.com Source: matter.org.uk

  19. Diffraction – spreading after passing by a barrier Source: newgeology.us geographyfieldwork.com

  20. Interference Constructive – in phase, increase of amplitude, sounds get louder Destructive – out of phase, cancel one another out Source: discoverhover.org

  21. Math associated with waves • Frequency = 1 / Period (f = 1/T) • Period = 1 / Frequency • i.e., f = 60cycles per second = 60 hz = 60/s • Also: v = f l • where velocity = frequency x wavelength

  22. Sound Waves

  23. Math and waves I = P/A Intensity = Power/Area Units: watt / cm2 Intensity relates to loudness

  24. More math… b = 10 log I Io 10-16 w/cm 2 b, Intensity level, decibels I, Intensity , w/cm2 Io , threshhold of hearing, w/cm2 We all don’t hear the same, so this is a comparative measurement in decibels

  25. The human ear can respond to minute pressure variations in the air if they are in the audible frequency range, roughly 20 Hz - 20 kHz donrathjr.com

  26. Loudness of Sounds Source: phl-caw.org

  27. Intensity Problem Sound energy is radiated uniformly in all directions from a small source at a rate of 1.2 watts. A) What is the intensity of the sound at a point 25 m from the source? Assume no energy is lost in transmission over a spherical area with radius of 25 m. P = 1.2 w r = 25 m

  28. Finding the intensity, I I = P = P A 4pr2 = 1.2 w 4p (2500 cm)2 = 1.5 x 10-8 w/cm2 Source: hyperphysics.phy-astr.gsu.edu

  29. Find intensity level, b b = 10 log I Io = 10 log 1.5 x 10-8 w/cm2 10-16 w/cm2 • = 10 log (1.5x108 ) • = 82 db

  30. Intensity Level to Intensity Let’s say that the intensity level of a sound is 25.3 dB. What is the intensity of the sound in w/cm2? b = 10 log I / Io 25.3 dB = 10 log (I/10-16 w/cm2) 2.53 = log I – log 10-16 2.53 + log 10-16 = log I 2.53 – 16 = log I

  31. 25.3 dB = 10 log (I/10-16 w/cm2) 2.53 = log I – log 10-16 2.53 + log 10-16 = log I  what power do you raise 10 to, to get 10-16? 2.53 – 16 = log I Adding on the left  -13.5 = log I Raise 10 to the -13.5 power by this sequence: 2nd 10x (-13.5)  3.16 x 10 -14 w/cm2 = I

  32. More to come… Doppler Effect is the apparent change in frequency as a result of relative motion between the source of a sound and a detector. f d = f s v v d vv s fd, frequency heard by detector fs, frequency of source v, velocity of sound in air vd, velocity of detector vs, velocity of source Source: onlinephys.com

  33. More on Doppler Stationary detector: Source moving away: v+vs on bottom Source moving toward: v-vs on bottom Stationary source: detector moving toward: v+vd on top detector moving away: v-vd on top

  34. Velocity of sound at various temps. * • VT inoC = 331.5 m/s + (0.6m/soC)(DT) • 331.5 m/s is speed of sound in air at 0oC • 0.6m/soC rate of change of speed peroC change What is the speed of sound in air at 10.00oC? At -5.20oC? #1  V10oC = 331.5m/s +(0.6m/soC)(10oC) V = 337.5 m/s #2  V-5.2oC = 331.5 m/s + (0.6m/soC)(-5.20oC) V = 331.5 - 3.12 = 328 m/s *Notice the 2 constants in red.

  35. Doppler Example A stationary civil defense siren has a frequency of 1000 Hz. What frequency will be heard by drivers of cars moving at 15 m/s? A) away from the siren? B) toward the siren? The velocity of sound in air is 344 m/s.

  36. Solution: Moving away from the siren fd = fsv - vd = v fd = (1000 Hz) (344 m/s - 15 m/s) 344 m/s fd= 956 Hz Apparent frequency heard by the detector decreases

  37. Solution: Moving toward the siren fd = fsv + vd = v f d = (1000 Hz) (344 m/s + 15 m/s) 344 m/s f d = 1044 Hz Apparent frequency increases

  38. One More… A police car with a 1000 Hz siren is moving at 15 m/s. What frequency is heard by a stationary listener when the police car is a) receding from the detector? b) approaching the detector?

  39. f d = f s v v + v s f d = (1000 Hz) 344 m/s = 958 Hz 344 m/s + 15 m/s P. car moving away f d = f s v v - v s f d = (1000 Hz) 344 m/s = 1046 Hz 344 m/s - 15 m/s P. car coming toward

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