# hw1sol - HWl solution ﬁnal closing time — ﬁrst...

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

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

View Full Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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.‘ III-Illﬂl 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 .I-II-II-‘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“) = 0-1 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°)? = 22-5 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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern