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**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. ...

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