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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 Tillb 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 15in.
displacement downward ﬂ‘cIn 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}
91101} lbi'ft (7011:]
11m: 15in. [32.1 111153} = 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
ﬂmﬂﬁrﬂ I" 15 5, the black CDmpletes 4}— $ = 15.361 cycles [LBW55 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 Ellkg 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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 Summer '06
 

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