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Unformatted text preview: HWl solution January 16, 2008 ﬁnal closing time — ﬁrst closing time T: 4 _ 15.6 — (—9.3)
" 4 =62210100sec =— It is indeed close to the period value we found from
maximum opening times. d) frequency f is a) maximum opening times, topen are
tape" = {—12.5, —6.0,0, —6.5, —12.5} ﬁlo—0m 2 b)  a:(t) = 1.8 + 0.5 sin(t) — 0.2 sin(2t) + 1.5 cos(3t) (1) closing times, tame are tale“ = {—9.3, —3.0,3.3,9.5, 15.6} 1—01553ec * These time values does not have to be accurate. As a)
long as the values are similar to what I wrote here, it is ﬁne The period of sin(t) is 27r The period of sin(2t) is 17
c) The period of cos(3t) is 2% x(t) is a function of sin(t), sin(2t), and sin(3t). Thus,
Period from tclose the period of x(t) is the least common multiple of the
Since the ﬁrst closing, there are four closing times. periods of these sin functions, which is T = 211’. You
Thus, can also show this graphically. Let T = 21r. 63*! 3 Since the object undergoes simple harmonic motion,
it satisﬁes following equation. I I.‘
IIIIllﬂl III—III—Illn d2rc 2
IIIIIIIIIIIIIIIIIIIIIIIIIIIIII 3325 = ”w m
. III IIIIIIIIIIIIIIIIIIIIIIIIIIII 2
IIIIIIIIIIIIIIIIIIIIIIIIIIIIII = (27rf) 06
IIIIII’IIIIIIIII IIIIIII III'III III _ _4 2 2 2
.IIIII‘II  7' f 9’ < I
The solution of this equation is
x(t) = Asin(wt + (1)) (3)
a)
From the given conditions,
A = 0.1m am” = 2m/sea:2 Since a(t) = Z—:§, we can get a(t) by plugging (3) into  (2).
 IIImIIIIIIIIIIIIIIIIIIIIIIIII'I d2
x .
l'lIIlI'lII Il'llll III'IIII IIIII 3;; = —4vr2f2w = ~4vr2f2As1ant + 45) (4)
.I'’ n‘m'—’
7r, f and A are constants and maximum value of sin()
is 1. Thus,
am“ = IszI (5)
amam
w = (6)
b) A
__ 1 “man:
_ 27r A (7) We can prove that x(t) is periodic function by show
ing that x(t) = x(t + T) x(t + T) = x(t + 21r) _ _ since v(t) = ,
= 1.8 + 0.5 sm(t + 27r) — 0.2 sm(2(t + 21r)) ‘
+ 1.500863“ + 271'» ”It) = (:3 = Aw sin(wt + ¢) (8)
= 1.8 + 0.5 sin(t + 21r) — 0.2 sin(2t + 471') t
+ 1.5 cos(3t + 61r) Then,
: 1.8 + 0.5 sin(t) — O.2sin(2t) + 1.5 cos(3t) ”m“ = A” = 2““ (9)
= x(t) 21rfA = 21r x A? x 0.1m/sec = 3?: m/sec C)
13(0) = 5m
plug (10) into (3) Asin(¢) = 0.05m sin(¢) : 0.1m 2% sincew=2\/§,¢=%orég‘andA=0.1m, W) = 0.1 sin(2\/§t + g) = 0.1 cos(2\/5t — g) m
0
33“) = 01 sin(2\/§t + 57") = 0.1 cos(2\/5t + g) m 4 Follow the same steps as 31 3. The solution is :z:(t) = Asin(21rft + (15) (11)
Given conditions are
A=0.2m f=8Hz :I:(0)=A (12) plug the initial conditions into (11) to get the 43.
Asin(¢>) = A
71
¢ ' 5 Finally, the solution is
1r $(t) = 0.2 sin(167rt + 2 ) = 0.2 cos(167rt) m (13) a) w(t) = 0.2 sin(167rt + g) m = 0.2 cos(161rt) m d
v(t) = if: = 3.21rcos(161rt + g) m/sec = —3.27r sin(167rt) m/sec = 0
d
a(t) = (1—: == —51.27r2 sin(161rt + '72:) m/sec2 = ——51.271'2 c0s(161rt) m/sec2 (14) (15) (16) 58H * Here, I plot %% instead of a(t) in order to plot all three graphs on one graph sheet 5 Initial conditions are A = 1 cm and w = 2T" = 1.
The solution is m(t) = 1 cm x sin(t + (b) for (1) = 57",§§75,% and 21r
When (15 = 27r, the graph is identical to the graph
with 4) = 0. The sin value does not change when we subtract or add 2mr from/to ¢>. Thus, we can
always ﬁnd qb = [0, 27f] for any ¢ 2 27r such that
sin(¢’>) = sin(¢). * Here, I do not draw graphs, but you should graph
all graphs with different ¢s given. (17) 6
a) When we consider SI units The unit of a is m/secz.
The unit of A is m. The unit of T is sec. Let’s assume acceleration depends only on the T and
A. Then, we can see that the unit of $5 is m/secz,
which is equal to the unit of a. Thus, A
acc— T, (18) ,which is consistent with the Eq. 1.15 in the lecture
note. b)
6.0.1 We need to change physical characteristics of the
harmonic oscillator system in oder to change its
maximum acceleration.(i.e. spring constant, maxi
mum amplitude ...) However, we can change the
phase value Without changing the physical system.
while the system undergoes harmonic motion, we can
change the starting time of observation. This will
change the phase value. Maximum acceleration value
should not depend on a variable that can vary even
when the harmonic oscillating system is unchanged. 6.0.2 you can also show that the maximum acceleration
does not depends on the phase mathematically. a(t) = d—‘5 = —(—’—')2Asin( 2 21rt
dt2 T T + <15) (19) 7r, T and A are constants and maximum value of sin
is 1. Thus, am,” = w2A (20) It does not depend on d). 7 since the wing tip undergoes simple harmonic motion.
The motion is described by following equation. $(t) = Asin(wt + (15) (21)
Then, from the picture, A=4 cmx sin(%) =2\/§ cm Other interesting values are. T = 0.02366
f = 31, = 50 Hz 0.: = 21r f = 1001r radian/sec
vmac = wA = 200x51 cm/sec
am” = sz = 20000x/571’2 cm/sec2 6K” 9
The equation of motion of a spring is
F = ma = —ka:
J29:
(1251: k
_ = __ 2
dt2 mm ( 3) recall that the equation of motion of a simple har—
monic motion is, dzx
——— = —w2a: dt2 By comparing these two equation, we can conclude
that a spring undergoes simple harmonic motion with k
2—
w —— (24) From the previous problems we know that
vmaw = wA (25) plug (24) into (25) and we get ”max = A E vmaa:2 A2— 3?r I0 vmaa: A2 (26) k=m From the given condition 11 = 3 m/ sec, m = 0.1 Kg
and A = 0.2 m. plug these values into (26). 0.1 Kg >< (ii—553°)? = 225 Wm 10 From the (23),
k
—— _ —;x (27) a) From the given condition, 3:1 2 0.2 m when F = 30
N. From the spring force equation. F = —k.’121 (28)
_ _E
‘. (l?
—30N
— — 0 m = 150 N/m (29) Now, when 5kg mass is attached to the spring, the
equilibrium position :30 is F=—kxo+mg=0
“FE k
plug in given values and the spring constant found _ 5 Kg >< 9.8m/sec2 (30) 150N/m
= 0.33 m
'0)
From (24),
k:
1 k
f — 57; :7; (32) plug in given values f = ﬁ‘llﬁ—ggm = 0.87Hz
11 68;) At equilibrium, F = ~k1$1 + [€232 = 0 (33) When we move the object to right side by distance
x, the equation of the motion is, F = k1(a:1 + (1:) + k2(a:2 — as) = ma (34)
ma 2 —(k1 + k2)$ — [912:1 + [$2.122 (35)
plug (33) into (35).
ma = —(k1 + k2):L' (36)
This is the equation of motion of the mass M
b)
From a),
dzil? k1 + 162
a;  —( m )9” (37) This is the simple harmonic equation. Since the mo—
tion satisﬁes simple harmonic equation. It will per form harmonic motion. Moreover, a.)2 = 515151
Then
1
f = 500
1 k1 + k2
= 2—7; m (38) The frequency is, f = i hﬂz 21r m ...
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 Winter '08
 Bruinsma

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