hw18 - PROBLEM 19.2 Determine the amplitude and maximum...

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Unformatted text preview: PROBLEM 19.2 Determine the amplitude and maximum velocity of a particle which moves in simple harmonic motion 1with a maximum eeeeleration of 21.6 W5" andafrcqueney offl Hz. SOLUTION Simple Harmonic Motion .1: = x,” sinlIwflr + pi} an” 2 21f” = 23.:[3 H21} = [63: Iade’s 5: = 1: = xmmu cos[w“r + 95} ‘r' =A' .l'fi‘ I'JT Eli M if = a = —xmm: sink)”: + gel] ”'1 ”m = <1?me 21.6 W33 = xmflo'rr mots}2 x", = 0.00355 ft (1.1925? in. xm = UJUEfi in. 4 v,” = xmm” = {0.00855 M152: IafilJISII = 0.429??? Pan’s 5.15372 ian PROBLEM 19.12 A Til-lb black attached to a spring of constant .17 = 9 kipse'fl can move without friction in a slot as shown. The black is given an initial 15-in. displacement downward fl‘c-In its equilibrium pesitien and released. ; {Ix Detennine 1.5 s afier the block has been released {a} the total distance i.”- traveled by the block, [b] the acceleratien cf the black. '9'- SU LUTIDN ___.- i.‘ = 'il Mpg-'1': Euuilihrium x = 3cm sin e1”? + 435} 911-01} lbi'ft (7011:] 11m: 15in. [32.1 11-1153} = 64.343 radfs 2.x 1 = — = asses s w H w” an}: a With the initial conditions: Jam} = 15 in. = 1.25 ft, 5cm} = Cl 1.25 ft = 3cm sin[fl + 95} as} = a = xmmneesw + a} :> a =% xm =1.2sa rs} = {1.25 fi}sin[fi4.343r + a t=1.5s 51.775st T fl-mflfirfl I" 1-5 5, the black CDmpletes 4}— $ = 15.361 cycles [LBW-55 sfcycle xiii} = {1.25 fi}sin[64.343[1.59}+§] = 41.3013? fl 5:111 .53. = n .25 fiHfi4.343}ees|:64.343[1.5 s] + g] = 451325 ms Efluiiib'rium PROBLEM 19.12 CONTINUED Sc, in one cycle, the block travels 4(125 fl} = 5 ft Fifteen cycles take 15[fl.fl9?65 sfcyclc} = 1464?? 5 Thus, the tntsl distance traveled is [51:5 fl] + 1.25 ft + {1.25 — 0.3fl13?}fi: = 'II‘T.1 1’: Total = Til fi ‘1 Ill?) 21:15}: —[1.25 fly-[454.3435]:sin[{s4.s43lsl{l.s s] + i] = 331163 fies? s = 3320 11st 4 PROBLEM 19.13 ‘lg? A Ell-kg block is supported by the spring arrangement shown. The block is moved vertically downward from its equilibrium position and released. KnowingIr that the amplitude of the resulting motion is 613 mm, determine {a} the period and frequency ot‘the motion. {:5} the maximum velocity and maximum acceleration of the block. e: lchihn SOLUTION [a] First, calculate the spring constant P = [24 mtmjs + [12 some + [12 swims = {4s mime; Ir = 43 kNa’rn Then .i- i at; = \lE = Lx 1“ mm = 30.934de3 m so kg 24 5 r” = E = 0.20219 5 E ‘1’" j], = i = 4,9312 Hz rl'l Elli 125 1:" =fl.2fl3s*l P fr, =4.93H2.~Il is} Now .r = x,” sin [or]?! + gift} And, since xv = {Lflfil} m x = {Uflfil} m}sin[{3fl.984 radish + a] x = [new m}[3fl.934 raws}cos[{3n.ss4 mars}: + a] r = {ones m}[3o.ss4 rsdfsiisinllmflfirt mats}: + a] Hence um“ = L859 int's '1 o = 516 mars: "i 1113K ...
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