Q2BSol - AM 250 Name Quiz#2 ID[Mitsslile 29th 2009 ®[11 1...

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Unformatted text preview: AM 250 Name: Quiz #2 ID# [Mitsslile 29th, 2009 ® [11] 1. Solve the initial value problem: 3;” + 411’ + By = 48—3: + 1, y(0) = 2, y'(0) = 0. ‘1): ‘5“ i ‘36 Fm ‘31,.“ .mfittmsw (MV3)(P‘H)= 0 , ,se “Or" ~3Ae‘3i—g eurfim’ce 4%“ =9 ”6A 9:“ +‘fiA-t61’t + HA 'z‘t — 6:5“ “3(0‘11'1? g1: (‘Jrcflfilg :_ C‘i-Lzzgé \j‘ivFO #7 OT— -3C"L1"3' a M —;1£. ‘ u/s Cr- “V4 C1 7/; I41 2. In many applications a response can be used to determine properties of the system. To demonstrate the principle, consider the equation y" + :13," +133; = U, with the initial conditions y(0) = 1, y’([}) = 0, and determine the values of o: and {3 if the solution is found to be y = e"3* cos %t. -—-—7 «3+ cam +l3=0 m: -q 1 W you; it in, Salem, osdllmdc the I» we be a _ 2:7 ' ‘Eflc [C (.05 i wilt l (—3 9"" hrs-«113] 3} e ‘ "“1— We Fuel innue- ‘D-al—(T-‘ZL‘; 5° “:6. (9 M1 tip—«131) so “Wren“ 3 cl {2) l-‘r a 37 he 3 9‘ :2. a I} © . In class we discussed the mechanical oscillator with sinusoidal forcing: my” +cy’ +ky = F0 cos wt. We first. reduced it to the standard oscillator DE: y” + 2wgCy’ + wgy 2 f0 cos wt, and then to fully non-dimensional form: Y" + 20” + Y = cos 97-. I’d like to see if you’ve understood what we’ve done. There are 9 constants here: m, c, 1:, F0, to, we, g, f0, and Q. Write the correct one beside its description below (please write clearly; if we can‘t tel] if you’re writing f9 or F0 we‘ll have to mark your answer wrong): L damping coefficient (dimensionless) * \ In! ELM-l lnLOr‘Qck [9,"): . damping constant 5; {fit A \ ‘mnw kg“ 9.. ram amplitude of reduced external force natural frequency ratio of forcing frequency to natural frequency spring constant ...
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