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Principle of Linear Superposition and Wave Interference Phenomena

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Principle of Linear Superposition and Wave Interference Phenomena

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    1. Principle of Linear Superposition and Wave Interference Phenomena Chapter 17

    2. Linear Superposition Draw two casesDraw two cases

    3. Constructive Interference

    4. Destructive Interference

    5. Interference and Path Lengths So you can cause a changing interference by either moving the speakers or by moving your ear.So you can cause a changing interference by either moving the speakers or by moving your ear.

    6. Example Draw path lengths before Green-left = 2 1/2 green-right = 4 1/2 Blue-left = 3 blue-right = 3 red-left = 4 red-right = 2 1/2 A => constructive (diff = 2) B => constructive (diff = 0) C => destructive (diff = 2 1/2)Draw path lengths before Green-left = 2 1/2 green-right = 4 1/2 Blue-left = 3 blue-right = 3 red-left = 4 red-right = 2 1/2 A => constructive (diff = 2) B => constructive (diff = 0) C => destructive (diff = 2 1/2)

    7. Diffraction Fig 17-12Fig 17-12

    8. Dispersion Fig 17-13 note: disp is dependent on wavelength and size of slit. If small lambda (high freq) poor dispersion If big lambda (low freq) great dispersionFig 17-13 note: disp is dependent on wavelength and size of slit. If small lambda (high freq) poor dispersion If big lambda (low freq) great dispersion

    9. Dispersion tweeters!!!!tweeters!!!!

    10. Beats note absolute value signs!!! Fig 17-18, 17-19note absolute value signs!!! Fig 17-18, 17-19

    11. How to tune an instrument note absolute value signs!!! Fig 17-18, 17-19note absolute value signs!!! Fig 17-18, 17-19

    12. Example (a) you don’t know! The beat frequency is the same if she is 4 HZ higher than the tuning fork, and if she is 4 HZ lower than it! absolute values (b) since tightening it increases the number of beats, (and our goal is ZERO beats), she needs to loosen the string. decreasing T, decreasing v, and so decreasing f (she was low) since v=f*lambda and lambda (distance possible to vibrate) is constant (a) you don’t know! The beat frequency is the same if she is 4 HZ higher than the tuning fork, and if she is 4 HZ lower than it! absolute values (b) since tightening it increases the number of beats, (and our goal is ZERO beats), she needs to loosen the string. decreasing T, decreasing v, and so decreasing f (she was low) since v=f*lambda and lambda (distance possible to vibrate) is constant

    13. Group Problem Solving COLLECT (a) diffraction is based on wavelength -- bigger wavelength, more diffraction. so higher frequency means smaller wavelength so smaller diffraction high frequencies (girls) diffract less and can only be heard in front of the speakers, low frequencies (bass) refract more and so can be heard farther from the front of the speakers COLLECT (a) diffraction is based on wavelength -- bigger wavelength, more diffraction. so higher frequency means smaller wavelength so smaller diffraction high frequencies (girls) diffract less and can only be heard in front of the speakers, low frequencies (bass) refract more and so can be heard farther from the front of the speakers

    14. Transverse Standing Waves Fig 17-21Fig 17-21

    15. Transverse Standing Waves Draw: up-hump reflects into a down hump laterDraw: up-hump reflects into a down hump later

    16. Harmonic Frequencies (String) Draw: up-hump reflects into a down hump laterDraw: up-hump reflects into a down hump later

    17. Example first overtone (2nd harm) is twice the fun.freq so 200 Hz second overtone (3rd harm) is three times so 300 Hzfirst overtone (2nd harm) is twice the fun.freq so 200 Hz second overtone (3rd harm) is three times so 300 Hz

    18. Example (a) we want f1=v/2L solve for L=v/2f1 need v: v=sqrt(F/ m/L) = sqrt( 226 N / 5.28x10-3 kg/m) = 207 m/s then L = 207 m/s / 2*164.8hz = 0.628 m (b) now we want the fun. freq to be 329.6 Hz wave speed is same (no tension change) so L = 207 / 2 * 329.4 Hz = 0.314 m ==> half as long(a) we want f1=v/2L solve for L=v/2f1 need v: v=sqrt(F/ m/L) = sqrt( 226 N / 5.28x10-3 kg/m) = 207 m/s then L = 207 m/s / 2*164.8hz = 0.628 m (b) now we want the fun. freq to be 329.6 Hz wave speed is same (no tension change) so L = 207 / 2 * 329.4 Hz = 0.314 m ==> half as long

    19. Longitudinal Standing Waves (air) Fig 17-26 Fig 17.28 closed pipe can only fit every other frequency.Fig 17-26 Fig 17.28 closed pipe can only fit every other frequency.

    20. Example f=nv/2L solve for L: L=nv/2f for fundamental, n=1 v=343 m/s f=164.8 1/s so L=1.04 m (b) f=nv/4L so L=nv/4f=0.52 mf=nv/2L solve for L: L=nv/2f for fundamental, n=1 v=343 m/s f=164.8 1/s so L=1.04 m (b) f=nv/4L so L=nv/4f=0.52 m

    21. Group Problem Solving don’t collect (a) node at one end, anti node at other, no crossing in middle (1/4 of a wavelength will fit) (b) note at both ends, antinode in middle (1/2) (c) same as A (d) same as A (e) anti node at each end, one cross in middle (1/2)don’t collect (a) node at one end, anti node at other, no crossing in middle (1/4 of a wavelength will fit) (b) note at both ends, antinode in middle (1/2) (c) same as A (d) same as A (e) anti node at each end, one cross in middle (1/2)

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