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# Physics of Sound 029 044

UI

GPA 3.76

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This 4 page Class Notes was uploaded by Mr. Jalon Paucek on Friday October 23, 2015. The Class Notes belongs to 029 044 at University of Iowa taught by Frederick Skiff in Fall. Since its upload, it has received 26 views. For similar materials see /class/228014/029-044-university-of-iowa in Physics and Astronomy at University of Iowa.

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Date Created: 10/23/15

Impedance Matching Friday 5 We have been saying that if the impedance is matched between two media then there will be no re ection of waves as energy passes from one medium to another If a wave pulse passes from medium number 1 with impedance Z 1 and strikes a medium number 2 that has impedance Z 2 the ratio of the amplitude of a re ected wave pulse A to the incident wave pulse A is given by A 1 ZZZl A 1ZZZl Impedance Matching Relative Amplitude of Reflected Wave 5 Impedance Ratio Plot of the ratio AAi given the ratio Z Z 1 Fig 1 Notice that if a wave re ects from a low impedance Z 2 less than Z 1 then the re ected pulse has the same sign and nearly the same amplitude as the incident pulse If it re ects from a high impedance Z 2 greater than Z 1 then the sign is inverted This is what we saw with re ections on the wave machine in lecture If the impedance ratio is 1 this means that the impedances are the same and then the amplitude of the re ected wave is zero This is called quotimpedance matchingquot The energy in the wave is always a positive number and it is proportional to the square of the amplitude If we look at the power in the re ected wave we see that it goes to zero when the impedances are matched This is shown in the next gure Fraction of Reflected Power n l l l l l l l l 5 Impedance Ratio Fraction of power that is re ected as a function of the impedance ratio Fig2 Sound in a tube An important example of sound propagation is sound traveling inside of a tube Because sound propagation in the tube is con ned and guided by the tube the situation is one dimensional In open air you use the speci c impedance 251839 Inside of a tube the impedance has different units Z Z A where A is the diameter of the tube When sound that is inside of a tube comes to the end of a tube there is a missmatch of the impedance and the situation is compleX It turns out that the impedance of the end of the pipe depends on the ratio of the radius of the tube r to the sound wavelength A This is because sound diffracts out and around the end of the tube and diffraction depends on the size of the wavelength compared to the size of the opening Figure 3 shows both the impedance ratio 2221 and the ratio of the re ected power to the incident power for sound as a function of the ratio between the tube radius and the wavelength You can also think of the horizontal aXis as a frequency aXis once you specify the radius of the tube If you know the speed of sound S the tube radius r and the frequency f then you can determine the ratio of tube radius to wavelength using r A 3f Sound reflection at the end of an open tube l l l l l 09 7 08 7 07 7 0396 7 7 Impedance Ratio 2221 7 Ratio of Reflected to Incident power 05 7 04 7 03 7 02 7 01 7 n l l l l l l l l l 0 001 002 003 004 006 007 008 009 01 Ratio of tube radius to wavelength Figure 3 Sound re ection from the end of an open tube Examples 1 Sound traveling through concrete hits a layer of brick The sound impedance of brick is about 40000000 Kgmzs and the impedance of concrete is about 8000000 Kgmzs Using the graphs given how much wave power will be re ected rather than transmitted into the brick The ratio Z Z 1 is 5 so from gure 2 we see that about 45 or almost half of the energy is re ected 2 What is the amplitude of the re ected wave compared to the amplitude of the incident wave for the situation in question 1 This we read off of gure or from the formula that the ratio is about 07 So the wave re ects with a change of sign This is a shift of phase of 7 or 180 degrees 3 The tube of a ute has an inside diameter of 19 cm What is the fraction of power re ected when concert A 440 Hz hits the end of the tube Take the speed of sound to be 340ms 340 The wavelength is S 77m 440 L s The ratio of the tube radius r192095 cm0095m t0 the wavelength is W 0012 From the graph above you can see that the re ection is very high 995 which means that 12 of one percent escapes out of the tube

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