Physics303L - DepartmentofPhysics E303:SoundWave:Resonant...

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MAPUA INSTITUTE OF TECHNOLOGY Department of Physics E303: Sound Wave: Resonant Frequency and Velocity XXXXXXXXXXXXX 2005XXXXXX XXXX, Group 1 PHY130L-XX
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XXXXXXX  SAMPLE COMPUTATIONS Table 1. Determination of the Resonant Frequencies of the Tube Resonant Frequencies for an Open Tube Given: Solution: Resonant Frequencies for a Closed Tube Given: Solution: Hz f Hz f 8 . 357 188 0 = = 90319 . 1 188 8 . 357 0 = = = resonant resonant resonant f Hz Hz f f f f Hz f Hz f 3 . 336 2 . 110 0 = = 0517 . 3 2 . 110 3 . 336 0 = = = resonant resonant resonant f Hz Hz f f f f
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GUIDE QUESTIONS 1. Which tube configuration gives a series of consecutive whole numbers? Open tube; 1, 2, 3…  2. Is the number series the same for both the open and closed tube? Support your  answer with a brief explanation. No, since the air can only freely move at the opened part of the tube, thus  the sound waves will be different since air is a medium on which sound travels. PROBLEM 1. The wavelength of a soundwave in an open tube of length 100cm is 15.5cm. If  the   surrounding   air   temperature   is   25   degrees   Celcius,   find   the   resonating  frequency of the air column inside the tube.
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Given: Solution: 2. A tube 1.2m long is open at both ends. A 5g wire of length 0.50m is stretched  and is placed near one of the ends of the tube. If the wire vibrates at its  fundamental mode and sets the air column in resonance at its fundamental  frequency, find (a) the tension in the wire. (b) its fundamental frequency. Given: Solution: ( 29 ( 29 s m V V t V air air air / 347 25 6 . 0 332 6 . 0 332 0 = + = + = C t cm 0 25 5 . 15 = = λ s f m s m f v f f v 71 . 2238 155 . 0 / 347 = = = = λ λ m L m l g m 2 . 1 5 . 0 5 = = =
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(a) (b) 3. Weak noises from an auditorium set up the fundamental standing wave in a  cardboard tube of length L = 60cm with two open ends. The temperature inside  the auditorium is 20 degrees Celsius. What is the frequency that you can hear  from the tube? Given: ( 29 g F m g m F L F F L f m g m g l m 434 . 0 5 . 2 2 . 1 2 1 2 1 2 1 5 . 2 5 . 0 5 2 2 1 = = = = = = = μ μ μ μ μ ( 29 ( 29 ( 29 ( 29 s f m m g g f L F n f L v n f / 417 . 0 5 . 0 2 / 5 . 2 434 . 0 1 2 2 1 1 1 1 = = = = μ
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Solution: ( 29 ( 29 ( 29 ( 29 s f m s m f L nv f / 67 . 286 60 . 0 2 / 344 1 2 = = = C t m cm L 0 20 60 . 0 60 = = = ( 29 ( 29 s m V V t V air air air / 344 20 6 . 0 332 6 . 0 332 0 = + = + =
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REMARKS AND CONCLUSION Sound is a disturbance of mechanical energy that propagates through matter as  a longitudinal wave. Sound is characterized by the properties of sound waves, which are  frequency, wavelength, period, amplitude, and speed. The latter is sometimes referred 
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