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Unit 2: Waves & Sound

Unit 2: Waves & Sound. Simple Harmonic Motion. time to complete one cycle. period, T :. (s). frequency, f :. # of cycles per second (Hz). wall. equilibrium position. D x = 0. F elas = 0. a = 0. frictionless floor. At both extremes (i.e., at D x max )…. D x. v = 0.

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Unit 2: Waves & Sound

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  1. Unit 2: Waves & Sound

  2. Simple Harmonic Motion time to complete one cycle period, T: (s) frequency, f: # of cycles per second (Hz)

  3. wall equilibrium position Dx = 0 Felas = 0 a = 0 frictionless floor At both extremes (i.e., at Dxmax)… Dx v = 0 Felas = max a = max –Dx period of a mass-spring system: m = mass (kg) k = spring constant (N/m)

  4. A 5.5 kg cat is attached to a fixed horizontal spring of stiffness 22.8 N/m and is set in motion on a frictionless surface. Find the period of motion of… …the cat. = 3.1 s …a 120 g mouse, with the same spring and surface. = 0.46 s

  5. What stiffness must a spring have so that the period of the mouse’s motion is the same as that of the cat? = 0.49 N/m

  6. A 1645 kg car carries two passengers with masses 75 kg and 86 kg. The car has four shock absorbers, each with a spring constant of 1.7 x 104 N/m. Find the freq. of the vehicle’s motion after it hits a pothole. and so… = 0.98 Hz

  7. Frestore Dx restoring force: acts to move an object back to equilibrium simple harmonic motion (SHM): As Dx , F . As Dx , F . When Dx = 0, F = 0. For a mass-spring system, Hooke’s law applies: Frestore = Felas = k Dx

  8. energy: the ability to do work kinetic energy, KE: energy of mass m having velocity v KE = ½ m v2 m (kg); v (m/s) potential energy, PE: stored energy For a spring with spring constant k and stretch (or “squoosh”) Dx: PEelas = ½ k (Dx)2 k (N/m); Dx (m) For a mass m at a height h above a reference line: PEg = m g h m (kg); h (m); g = 9.81 m/s2

  9. KE PEelas X PEg KE X PEelas PEg

  10. A A Energy (J) PEg amplitude, A: maximum displacement from equilibrium frictionless Period T is NOT affected by amplitude. Energy of a Mass-Spring System total energy PEelas KE (arbitrary)

  11. length L amplitude Q mass m of bob Energy (J) PEelas The Pendulum For Q < 15o, a simple pendulum approximates SHM. Energy of a Simple Pendulum total energy PEg KE

  12. period of a simple pendulum: Period T is independent of… mass and amplitude. The period of a pendulum is 5.2 s. Find… A. …its length = 6.7 m

  13. B. …the mass of the bob NOT ENOUGH INFORMATION

  14. vibrations moving through space and time Waves Waves transmit energy, not matter. medium: the matter through which the energy of mechanical waves moves Harman Kardon soundsticks

  15. transverse waves: particles of medium move to direction of wave travel Energy (Amplitude)2 crest amplitude A trough wavelengthl For a transverse wave:

  16. “Dr. Drake…. I forgot: What are the units for frequency?” That hertz. I mean, that REALLY hertz. That hurts. I mean, that REALLY hurts.

  17. particles of medium move // to direction of wave travel longitudinal (compressional) wave: Each particle experiences a jostling, back-and-forth motion. compression rarefaction disturbance velocity (i.e., the wave) l each particle pulse wave: a single vibration periodic wave: rhythmic, repeated vibrations

  18. A A v v v v Wave Reflection fixed boundary free boundary waves are reflected and inverted waves are reflected and upright

  19. A1 + A2 A1 – A2 A1 A1 A2 A2 Wave Interference Two waves (unlike two objects) can occupy the same place at the same time. This condition is called interference. constructive interference: destructive interference: displacements have a canceling effect displacements are additive A1 A2 A1 A2

  20. Wave Velocity Equation: v (m/s); f (Hz); l (m) A water wave of wavelength 8.5 m washes past a boat at anchor every 4.75 seconds. Find the wave’s velocity. and so… = 1.8 m/s

  21. The velocity of any mechanical wave depends only on the properties of the medium through which it travels. e.g., string tension, water depth, air temperature, material density, type of material The speed of sound in air is related to the air temperature: Ta = air temp. in oC vsound = 331 + 0.6Ta

  22. Standing Waves Incident and reflected waves interfere to produce an unchanging wave pattern. Antinodes have a max. amplitude, while nodes have zero amplitude. Standing waves are most easily visualized on a string, where nodes remain motionless and antinodes go from max. (+) to max. (–) displacement.

  23. L n = 1; 1st harmonic l1 = 2 L (fundamental) n = 2; 2nd harmonic l2 = L (1st overtone) n = 3; 3rd harmonic l3 = 2/3 L (2nd overtone) n = 4; 4th harmonic l4 = ½ L (3rd overtone) wavelength of the nth harmonic on a string: (n = 1,2,3,…)

  24. Waves travel along a 96.1 cm guitar string at 492 m/s. Find the fundamental frequency of the string. l1 = 2 L = 2 (0.961 m) = 1.922 m v = fnln = 256 Hz Find the frequency of the 5th harmonic. l5 = 2/5 L = 0.3844 m = 1280 Hz fn = n f1 frequency of the nth harmonic:

  25. Standing Waves in Open Tubes L wavelength of the nth harmonic of an open tube: (n = 1,2,3,…)

  26. Closed Tubes L wavelength of nth harmonic of a closed tube: (n = 1,3,5… ONLY i.e., the even harmonics are NOT present)

  27. 1.24 m Find the fundamental frequency for an open tube of length 1.24 m. Assume the air temperature to be 20.0oC. l1 = 2 L = 2 (1.24 m) = 2.48 m v = fnln 20 v = ? vsound = 331 + 0.6Ta = 343 m/s = 138 Hz

  28. 1.24 m (same as prev. prob.) 100 Hz 200 Hz 400 Hz 800 Hz doubling (or halving) of frequency = Find fundamental frequency for a closed tube of length 1.24 m. Air temp. is 20.0oC. l1 = 4 L = 4 (1.24 m) = 4.96 m = 69.0 Hz one octave (three octaves)

  29. 20 Hz 20,000 Hz Sound (a compressional/longitudinal wave) high pressure/density compression: low pressure/density rarefaction: audible frequencies (human hearing) infrasonic ultrasonic

  30. Fundamental frequency determines pitch. high pitch high f = = short l low pitch = long l low f =

  31. The # and intensity of an instrument’s harmonics give it its unique sound quality, or ______. timbre f1 f2 f3 f4 f1 f2 f3 f4 f1 f2 f3 f4 f1 f2 f3 f4

  32. The Doppler Effect Relative motion between wave source and observer causes a change in the ________ frequency. observed v = 0 femitted fobserved (higher) femitted femitted fobserved (lower)

  33. Sun R O Y G B V most stars Other examples of Doppler effect: race cars police radar (“red-shifted”) dolphins and bats (echolocation) expansion of universe

  34. wave barrier Traveling Very Fast vbug = 0 vbug < vwave bow wave vbug > vwave vbug = vwave

  35. supersonic: “faster than sound” (vs. subsonic) shock wave: a 3-D bow wave sonic boom: caused by high-pressure air, NOT roaring engine lion tamer’s whip cracking bullets The Matrix

  36. Sound Intensity If a piano’s power output is 0.302 W, find the sound intensity at a distance of… A. …1.0 m = 0.024 W/m2 B. …2.0 m = 0.0060 W/m2

  37. Intensity is related to volume (or relative intensity): -- how loud we perceive a sound to be -- measured in decibels (dB) A difference of 10 dB changes the sound intensity by a factor of 10 and the volume by a factor of 2. 50 dB vs. 40 dB 60 dB vs. 90 dB 8X louder half as loud 1/10 as intense 1000X more intense

  38. 1 second elapses 1 second elapses Beats  alternating loud-and-soft sounds resulting from interference between two slightly- different frequencies (“ooh-wAAHH- ooh-wAAHH…”) Equation: f1 = 16 Hz f1 = 16 Hz f2 = 17 Hz f2 = 18 Hz fbeat = 1 Hz fbeat = 2 Hz

  39. Forced Vibrations and Resonance natural frequency: the frequency at which an object most easily vibrates forced vibration: a vibration due to an applied force resonance: occurs when a force is repeatedly applied to an object AT the object’s natural frequency large amplitude -- result of resonance =

  40. Examples: swing shattering crystal wine glasses Tacoma Narrows Bridge (1940) British regiment (Manchester, 1831) aeolian harps “The wind in the wires made a tattletale sound, as a wave broke over the railing…”

  41. When the decibel level of traffic in the street goes from 46 dB to 66 dB, how much louder does the traffic seem to be? How much more intense is the sound? 4X louder; 100X more intense If two flutists play their instruments together at the same intensity, is the sound twice as loud as that of either flutist playing alone at that intensity? Explain. No; the sound will be twice as intense as either performer playing alone. In order to be twice as loud, we’d need 10X the intensity (i.e., ten flutists). Which of the following factors MUST change when… a) a sound gets louder? b) a pitch gets higher? changes ONLY if medium changes decibel level speed of the sound waves decibel level wavelength wavelength intensity frequency intensity frequency amplitude amplitude

  42. vsound = 331 + 0.6Ta (strings and open tubes) (closed tubes) Frestore = Felas = k Dx fn = n f1 KE = ½ m v2 PEelas = ½ k (Dx)2 PEg = m g h

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