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Physics 101:  Lecture 21  Waves ,[object Object],Exam III
Preflight 5 ,[object Object],Standing waves was difficult to understand. What does it mean and can you give another explanation as to how to calculate it? Also, reflection was not explained enough. What is the difference between a fixed and a free boundary?  Everything…   Nothing… Overtones! harmonic waves.  Longitudinal waves.  there was alot of info Not paying attention
Waves Overview ,[object Object],[object Object],[object Object],[object Object],[object Object]
Types of Waves 8 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Slinky Preflight 3 ,[object Object],[object Object],[object Object],the coil won't move, 5m/s is the speed for wave propagation 12 correct 5m
Velocity of Waves Act 17 A spring and slinky are attached and stretched. Compare the speed of the wave pulse in the slinky with the speed of the wave pulse in the spring. A) v slinky  > v spring   B) v slinky  = v spring   C) v slinky  < v spring   Slinky stretches more, so it has a smaller mass/length  
Harmonic Waves Wavelength:  The distance    between identical points on the wave. Amplitude:  The maximum displacement  A  of a point on the wave. 20 y(x,t) = A cos(  t –kx) Angular Frequency  :     = 2    f x y Wave Number k:  k = 2    /     Recall:  f = v /     “ the weird/new symbols get me confused i never know what they represent “  Wavelength Amplitude  A A
Period and Velocity 22 ,[object Object],[object Object]
Harmonic Waves Exercise ,[object Object],[object Object],[object Object],[object Object],T = 2     25 t +2 -2    / 2  /4  /4
Preflight 1+2 ,[object Object],[object Object],[object Object],[object Object],v = λ/ T 26 correct
ACT ,[object Object],(a)  1 GHz  (b)  10 GHz  (c)  100 GHz 1 GHz = 10 9  cycles/sec The speed of light is  c = 3x10 8  m/s 29
ACT Solution ,[object Object],1 GHz = 10 9  cycles/sec The speed of light is  c = 3x10 8  m/s 30 H H O Makes water molecules wiggle
Absorption coefficient of water as a function of frequency. 31 f = 10 GHz Visible  “ water   hole”
Interference and Superposition ,[object Object],[object Object],[object Object],34
Reflection Act ,[object Object],[object Object],[object Object],[object Object],37
Standing Waves Fixed Endpoints ,[object Object],[object Object],[object Object],44
Standing Waves: 48 A guitar’s E-string has a length of 65 cm and is stretched to a tension of 82N. If it vibrates with a fundamental frequency of 329.63 Hz, what is the mass of the string? v =    f = 2 (0.65 m) (329.63 s -1 ) = 428.5 m/s v 2  = T /      = T / v 2 m= T L / v 2 = 82 (0.65) / (428.5) 2   = 2.9 x 10 -4  kg f 1   = fundamental frequency   (lowest possible) L   / 2 f = v /   tells us  v  if we know  f  ( frequency ) and   wavelength)
Summary ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],50

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Lecture21

  • 1.
  • 2.
  • 3.
  • 4.
  • 5.
  • 6. Velocity of Waves Act 17 A spring and slinky are attached and stretched. Compare the speed of the wave pulse in the slinky with the speed of the wave pulse in the spring. A) v slinky > v spring B) v slinky = v spring C) v slinky < v spring Slinky stretches more, so it has a smaller mass/length 
  • 7. Harmonic Waves Wavelength: The distance  between identical points on the wave. Amplitude: The maximum displacement A of a point on the wave. 20 y(x,t) = A cos(  t –kx) Angular Frequency  :  = 2  f x y Wave Number k: k = 2  /  Recall: f = v /  “ the weird/new symbols get me confused i never know what they represent “  Wavelength Amplitude A A
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13. Absorption coefficient of water as a function of frequency. 31 f = 10 GHz Visible “ water hole”
  • 14.
  • 15.
  • 16.
  • 17. Standing Waves: 48 A guitar’s E-string has a length of 65 cm and is stretched to a tension of 82N. If it vibrates with a fundamental frequency of 329.63 Hz, what is the mass of the string? v =  f = 2 (0.65 m) (329.63 s -1 ) = 428.5 m/s v 2 = T /   = T / v 2 m= T L / v 2 = 82 (0.65) / (428.5) 2 = 2.9 x 10 -4 kg f 1 = fundamental frequency (lowest possible) L  / 2 f = v /  tells us v if we know f ( frequency ) and  wavelength)
  • 18.

Notas do Editor

  1. 1
  2. Stadium wave here, for fun.
  3. Long Spring and slinky equal length (8 feet each) speeds up a LOT in slinky because mu is small and T is same.
  4. Add stuff here
  5. Students make plot at x=0, give them line, they mark tics
  6. Do this with springs
  7. Use large spring as demo here
  8. Two people with large spring, can produce all of these as demo.
  9. Calculate tension in guitar string given mass density and frequency. Ch 11 # 45