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# HW#6sol - Mechanical Vibrations Homework#6 Due by end of...

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Unformatted text preview: Mechanical Vibrations Homework #6 Due by end of week: Friday, lﬂDecﬂﬂﬂf-l A. shaker is used harmonicaflg.r excite the string shown in Figure 1. Use ﬁxed boundary cnnditir. at both ends and the following parameters for the string: 3.; = I12 [Kgfin], string length L = 2 [r T = 20 [N]. fillso1 use the following partial differential equation. 5% (,mt] _T52y{s1t] ’ as 53:3 = dim _ LiiFaE-‘é'm M——# Determine an eJquession for the response amplitude of the string: as a. function of the forei frequency \$11 that would be measured by a non—contact displacement sensor shown below. St: with equation 1 and show your work. Next! plot the response amplitude for the ﬁrst three natu frequencies. Force Non-oontsct Transducer Displacement / Strlng Sensor Metal Plate 0.3 m 0.6 m Figure 1: Schematic of a harmonicaiiy excited string experiment with ﬁxed end conditions. ET" q: ETR 22441 :SIJEE-[EETS 224:1? MHSHE 5 5*?1 1 “ I’H‘HE iﬁﬁﬁan‘ ..:1.. : Hut 6 5mm va-ﬁh: Force-c1 “if-*5 up 0"- cuﬂmuohs ##1qu TC) ISL-{hut '_ E!) \$94 el'V-L-Nt fWrt'hﬁ-‘puh‘ ‘FDT Ht (IF-Err gum till 5 0. £Uu-HC hum c: ‘F .fL 1"} HDLﬁir- ﬁrs? 3 Hit-{MAI #Pﬁlufmcgfq : 11-: git-{4' 4“: Ehnk'vr' Lz-{rayﬂ pi: Ehnj. a aillﬁhﬂh ‘ F 130 i T1 '2: F-‘xf: wt; 1 - 9 “£34 E0” ”:9— T-aia: Ham £Exii§ FIB-rial E-C-E 1-) yrojfj'dﬁ 1‘] \fCLii‘} =0 H-ESSmwae. Salim-‘- \J'C‘J-nj-JL‘E': Wﬂx3¢f4c32 '11 Eyﬁxﬁétg‘l _ “Fl—W1 LLPQJdC-Uli Raﬁ El“ cm x} . 1I.C¥. - ‘- 0“) Mean“) ”14“”-1-Ma 93%, #40:) n: +Lc wade: Whig} E Pal-a 21'“ng Ca? fin)“ Ewthuuq’) F1715: UIL-E'E +1: 51+ Knits. SLaVEE‘L mac. 1 U. *2 { FVoHQ'} «Kt-Fig) iijlj'T-diﬁ‘iq 1%1k) -—_: O 32¢{ﬁ 4: git-ENE .ML‘lré“) 7t?- - Tcﬁﬁ’l 35:1 - l .1 .l_ ‘1 — 1 M -— : =- ‘1': ct: w L" ‘5 '-.'| H-IEL-"E 1"“? ENE E'TE': 224-2: Pan SHEET?“ 7! 724-1 F .u' 1: I} _ . dreams-“:3 r; ii 4*: :1_T .1 (9 ﬂ '03 *- §Pl¢ :0 --==- ¢5+ ”'25 to a: 1-4- 8 ' '1 \$1. Lit—O a q" :0 _—=.'J q ’1' (:15) a w + E gab-Ii th'g .- [,qu 1} = A¢JL ~11: L} : Akeitgg tie - 42,45.le ULS-L B. if: +936- i=ﬁ£OA+B c3 A=-B E392) 44 em“ + 36,66!” = -E‘> e‘lﬁL + ESP" 2: c. i - Sims pL-I-SL‘nﬁL-J +5 [mpt— Sr'mptj - “1’8 ﬁr'nﬂL = O B=o L5. +n’ua‘al md 5m 13L =0 :3, £3:th what n := 1,11%, .. @ wn=13n :ar'npn‘i. *(Qr‘ﬂw purpose- m“ ﬂu Fmbkem ﬁnal»:— Hw'e- 96am LPH- 1&5. 5m m =2: ---/ 9’1: 151 hI-n ‘55:} :5 W3;- 133 amp; x =‘:- W J G L- Thcﬁa- mode shapes on. Hm Ernst] antler; or 1,58% hound-fans.“ ﬂnwsubshﬁuh Q) (m Q} (Q /U E LPN“ n [H— TE ﬂax] ¢nﬁu=pﬁult) NH JTHEEh - 3.‘ 'IDB SHEET: SHEETL 22-1-11 5 .1 2214'! 512-141 :15? :M’ME zu- _ _.__.___— ___. __i_ Huh d- warn. and Ham riRQT‘J bcﬁer I U51 -ﬂ.. ﬂu}: WW) 2 ”(ship-j @ % j:fg}(g)dx Thu 69%}: ‘9 FEE” a M‘Hu‘m 463 {ix + TIEZW' )HUHwH“ LwMJdu mln'l =j< 5",..th {tn—1,) Fa 5““ ear I? I": ‘(0 I f I m “05°”J mm @ ”3mm j‘ﬂcxﬁdwcr whim (TBEWRQ _p i ifmzojwj u; 0U: + 775: MI?” [9004— MJF; E 3n inch-m T (Ruth) +1“ 'ﬂn ¢“&}=ﬂ\$1’:t;}du Rem 4'1”, +03%: ONE ””3 qﬁm =Lm amt (in: 2.1!: Cm amt we “do a... w =05. ﬂ C... = “in”? F0 3} "J1- mt 6“) EU SHEETS-1 22-142 mm SHEET}: EU'II SHEET-‘- 22-141 22-144 43:: . 5 ﬂing—f: n ' Mm in arder 'hlﬁ 0134!?!th “IL IF Bum di'mH-t HUI-E) €13 Mat t'npun‘ 'pordnj Rand-Lb!) . H: Shﬁ) 2 Magwpm _ H7 F EiJ‘L‘b a ﬂ —————————————————————————————————————————————————— % HW E. SOLUTIONS KEY % TA ANSWERS a __________________________________________________ close all: clear all; Clo; % Constants li = .3; % shaker location lo = ; % sensor location mu = .2: a [kgfm] L = 2; % total string length T - 20; % [N] tension num_modes = 3: a loop over first 3 modes hetaveo = [1:3].*pifL; % heta values E = l; s amplitude freq = 0:0.0ﬂ01:10; % beam forcing frequency units of [Hertz] Omega - 2*pi*freq; % beam forcing frequency units of [radfs] GmegaE = [2*pi*freq}.*2: numfreq = length{freqh; % number of frequency points FD = l; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3%%%%%%%%%%%%%%%%%%%%%%%% % SUH UP TO THE nth MGDE FOR THE TRANSFER FUNCTIONS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%E%%%%%%%%%%%%%%%%%%%%%%%%% for n=l:num_modes heta_n betaveoin}; % nth eigenvalue wn = sqrttTfmu*heta_n“2): % nth natural freq % ————————————————————————————————————————————————————————————————— a Unoomment this section to find the expression for the integral of psi % syms x heta_n L % psi = B*sin[heta_n*x}*L; e Psi2 = int{psi*psi,x,U,L} % _________________________________________________________________ % Parameters for Dm Psi_lo = B*sintheta_n*lo]; Psi_li = B*sintheta_n*li}; Psi2 = 132*L*2*{-costheta_n*LJ*sin[heta_n*LJ+beta_n*Llfbeta n: Dm = Psi_liftmu*Psi£}; _ Hntn,l:numfreqj w Dm.*Psi_lD.f[wn‘E-Dmegazj; end Hn = sumEHn}; figure suhplot[2,l,l] plottfreq,Hn,'linewidth',2] ylahel['Real TF'] xlahelE'Freq [Hz]'J axisiEﬂ 10 —5 5]] subplotﬁ2,1,2} plot{freq,ahs[Hn},'linewidth',2] ylabeliTTF Magnitude'] xlabelE'Freq [Hz]'] axis{[ﬂ lﬂ D 5]] 1E} Freq [Hz] 5 d. 3 Egan: n: 113 2 1 n... Freq [Hz] ...
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