h6sol - HW6 solution 1 If the speed of sound in the air is...

Info icon This preview shows pages 1–6. Sign up to view the full content.

Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW6 solution February 25, 2008 1 If the speed of sound in the air is 340 m/sec, the wave length of the “chirps” in the air is, = § (1) = 0.0057m (2) = 5.7mm (3) From the lecture note “If we use waves to probe the shape of an object of size L, then the wavelength of the probe must be less than L” Thus, the bat can detect an object of size larger than 5 mm. 2 given conditions are, f0 = 400 Hz (4) f1 = 450 Hz: when the train is approaching (5) Rom the doppler effect equation, the speed of train is, f1 = 1 —1v/c f0 since the train is approaching (6) fo v/c — 1 — E (7) = 0.11 (8) v = 0.11c (9) if c = 350 m/sec or 340 m/sec (10) v = 38.89 m/secor37.78 m/sec (11) When the train is moving away from the observer, the frequency, f2, is, _ f0 f2 — 1 + v/c (12) = 360 Hz (13) We can plot this change on a graph, frequency vs. bastion of the source ‘N :I: v > u c u :1 a o .2 Damon or small) 6) The eq. 3.21 and eq 3.22 are, I ——=1—v/c o (14) I —=1+v/c f0 (15) The eq. 3.24 and eq 3.25 are, f ' 1 _ = 1 f0 1 — ”/0 ( 6) f ' 1 -— = 17 f0 1 + 'u/c ( ) When '0 = 0.010, compare (14) with (17). I —- = 1.01 from (14) (18) f0 I —- = 1.0101 from (17) (19) fo When 'u = 0.01c, compare (15) with (16). I —— = 0.99 from (15) (20) f0 I 7— = 0.99001 from (16) (21) 0 Thus, When 12 < c, the answers from the two cases are very close to eahc other. When u = 0.9c, compare (14) with (17). I _ = 1.9 from (14) (22) fo — = 10 from (17) (23) f0 When 1; = 0.9a, compare (15) With (16)- _’ = 0.1 from (15) (24) f0 _ = 0.526 from (16) (25) f0 Thus, When 1; z c, the answers from the two cases are very different. b) The principle of relativity does not hold for sound because moving source and moving observer are two different situation for sound wave. When a source is moving and an observer is stationary, the medium(e.g. Air) is stationary from the observer’s point of view, but when an observer is moving, he / she has velocity with respect to the medium. In other words, when an observer is stationary, the speed of sound is just c. However, when an observer is mov- ing the speed of sound is c — volumm because the midium is moving for the observer.Thus, the two sit- uation can not be indentical. On the other hand, the speed of light is just c on the both cases(you will learn about this in 60H) Thus, we can not distinguish two cases for the light wave. Thus, only relative velocity matters. vAt = 2L (the distance the sound travels is 2L) (26) L = $12615 (27) if c = 340 m/sec (28) L = 17m (29) 4.1 b) Since the echo frequency is higher than original fre- quency, the moth is moving toward the bat. So, we should use corresponding formulas Let’s define variables, f0 : original frequency (30) f1 : the frequency the moth would hear (31) f2 : echo frequency (32) vmou. : the speed of the moth (33) Then, the fl is, f1 = (1 + ”moth/0).“ (34) We can write f2 in terms of vmom and f1, 5 = —— 35 f2 1_ 'Umoth/cfl ( ) .a) = Mitt/.2 (36) The wall undergoes simple harmonic motion. Thus, 1 “ ”moth/C f2/f0(1 " ”moth/C) = 1 + ”math/0 (37) z = A sin(wt) (x: position of the ventricular wall) — 1 ”moth/C = fz—HL— (38) . (48) f 2/ f0 + 1 dx from the given coditions (39) v = E = Aw (5030“) (49) f0 = 12 MHZ and f2 = 12.15 MHZ (40) ”max = Ag) (50) ”moth = 000620 (41) from the given conditions, (51) The v is a positve value. This means that the moth A = 0.0018 In and w ___ 21', x 115 (52) is indeed moving toward to the bat. If the moth ls 603cc moving away from the bat, the v will be a negative = 0.0217 m/sec (53) value. The minimum occurs when 11 = um.” and the ventric- c) ular wall is moving way from the sound source. When the wall is moving toward the source with v = um”, Let’s define new variables, the frequency is maximum. f3 the frequency the moth would hear I whenthebatismoving fmal = (1 + ”mm/9V0 (54) f4 : the frequency of echo = 2.000029 KHz (55) for the stationary observer fylnm = (1 ' ”mam/CW0 (55) vmoth : the speed of the moth = 1.999971 KHz (57) um : the speed of the bat The f3 is, b) _ 1 + ”moth/C f3 — —f0 (42) , . 1 — vbat/c The echo frequency f ’ 16, f4 can be witten in terms of f3, 1 II = fémz 58 f4 = T—TTf (43) fr” —1 — awn/c ‘ ) plug (42) into the eq. = M f0 (59) 1 -— vmam/c = 1 + ’Umoth/c 1 f (44) 1 — ”moth/C 1 — 'Ubat/C ° = 2.000058 KHz (60) 1 + vmoth/c ,, .’...~.. = . = —— 61 remember f2 1 _ vmom/cfo (45) "“n 1 + 'UmaI/C ( ) 1 1 — vmu/c f2 1 " “bot/c ( ) 1 + ”max/0ft) ( ) f4 = 12.29 MHz (47) = 1.999942 KHz (63) c) If we measure um“, and the oscillation frequency of the ventricular wall, we can calculate the aplitude of . the oscillation. So, we need to find a way to measure um” and oscillation. 11m” can be calculated from the equations in part b) and the measurements of f,’,’"-n and film, values from the piezo crystal. We can find the oscillation frequency by counting how many times fm“ (or fmin) occurs over certain time interval. Then, the oscillation frequency is fwa" = O "“118 ' ' __ U Lfimeimflwl. The amplitude is A — —m‘—2flm“ . 6 picture above, the speed of the source is, __ —21.7m — (—30)m v _ 10.5 sec (66) Tune: 10.3~ 9”,." M = 0.79 m/ sec (67) the speed of sound is,c = 11/08 (68) z 0.988 m/sec (69) From the picture above, the frequency is, 135.3%.- “V ~ X: '21.5 Y=5.8 f0 = 10/10.6 sec z 0.94 Hz 1 x:—19.e Y=0 b) You can use just one period of time, but I used 10 period of time for more accurate result. From the from the picture above, [\fmm = 9 crests = 0.211771. A ' _ (—21.5 — (—29.8)) behind — 5 crests = 1.66m A _ 5.8 — 0 ”a” — 9 crests = 0.644m From the theory, Afro'nttheo = C/ffront C = _.£o_ l—u/c = 0.210m Abehindtheo = 0/ f behind c 1+u7c = 1.89m (71) z (78) and (73) z (81). d) mm: .1 .5 Doppler (—19.6 — (—21.5)) (70) (71) (72) (73) (74) (75) (76) (77) (78) (79) (80) (81) from the picture above, (6.5 — 3.2 m = 0.33m (26.5 + 3.2) 10 crests = 2.97m 9.9—0 10 crests = 0.99m Afront = Abehind = Amart = From the theory, Afronttheo = c/ff'ront C 1 (c f0" 1::/c = 0.35m )‘behindtheo = C/fbehind c l-v c f0 1+vc = 3.19m (83) z (90) and (85) z (93). 7 (82) (83) (84) (85) (86) (87) (88) (89) (90) (91) (92) (93) (94) If the size of' hole is large, the image becomes blurred. As you can see from the picture, there will be three virtual images. 9 You need draw a parabola and find a focal point from the picture. You should draw at least three light rays to find the focal point. Then, you will find that -, where P is the position of the focal point. Thus, if you were Archimedes and wanted to attack warships about 1 Km away from the mirror, the In value should be 1/4000 m‘1 . note: The parabola has its focus from the definition. The parabola is locus of points which are equidistant from the focus and the directrix. Here, we will prove that the focus point is the focal point of light by using geometry. Fiom the definition of parabola, W = 737 = l (95) W = m = h (96) (97) Fiom triangle AABC and AABD, Line A—B is a common line (98) 4,403 = AADB = 90° (99) W = 7373 — C_P (100) = l — hcosfi (101) W = W — ET (102) = l — h (103) _AABC and AABD are not congruent since F5 79 BC. Now, we move the point A to B as close as possi- ble. Then, 37‘ becomes tangential line of parabola at B and the angle theta becomes zero as A approaches to B. W = ginsu —— hcos 9) (104) = z — h (105) = m (106) Born (98), (99) and (106), AABC' and AABD are congruent. As a result, ' (ABC = AABD (107) = ZEBF (108) 103G = AEBG (109) Eye [CBG = AEBG, if E isythe incident light, PB must be the reflected light. Therefore, the focus is the focal point of the light. ...
View Full Document

  • Winter '08
  • Bruinsma

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern