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Test1_sol_vib_2003 - Mechanical Vibrations Exam 1 Student...

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Unformatted text preview: Mechanical Vibrations Exam 1 Student Name: fl So i n 1. [5 pts] The response of a single degree Reéponse Amplitude of freedom mass, spring, damper system is recorded over a range of frequencies. This test is performed for three different dampers and the measured amplitudes . for each damper are shown in Figure , ----------------- 7 ---------------------- 1. If the damping ratios are (1 = 0.02, 1 C2 = 0.04, C3 = 0.08, label the ampli— tude curve that corresponds to the each damping ratio. Derivation is not re- quired. IX(Q)| (m/N) 5 10 15 20 25 so frequency (Hertz) Figure 1: Response amplitude from three different vibration test; each test used a different damper. 2. {5 pts] Finish the following sentence: F i c r (2r 1‘9 S Any periodic function can be expressed as a 0" r ' Write one form of this expression. j((9: f (w CHI/2 “71%) T - JH c ‘: L M " r/‘Wé 3. [15 pts] The position of a point mass, shown as point B in Figure 2, is written with respect to the origin (A) of a rotating coordinate system. The vector describing the position Within the rotating coordinate frame is FB/A = 31527; + 103', where t is the time in seconds. The (cc, 3/, 2) frame rotates at Q = 10% + 53'. Find the velocity of the mass, Fem as measured in the fixed frame at t = 1 second. U Figure 2: Fixed coordinate system (U, V, W) and a rotating coordinate system (as, y, z). (b: rpL rm g: 7‘; + d M 1&4?” +JH§ \J » dt' 91* K” A 4. [15 pts] Estimate the natural frequency (nun), damping ratio (g), and the mass (m) of the single degree of freedom transfer function, H (0), shown in Figure 3. The value of H ((2 = 0) x 2.2517 X 10—4. x104 Transfer Function Real Hm) (m/N) 5 1o 15 20 25 30 frequency (Hertz) 4'01sz 1o :5 20 25 so frequency (Hertz) rug—J r , m Figure 3: SDOF Transfer function 2 j, G wit u _ r 2 , 7’? Xp— Dem rm rm “309/” effi— "’f .. APU (D H(fL -— eorpof : ___1_r1\____________. fl . I ' Wf Wat—filflc'anjL 415w! ‘ (9/ K 3 HM”): 1 1.1517100” 5. [30 pts] Develop the equation of motion for the system shown in Figure 4, Where f(t) = Aces Qt. Use either 213%: = % (ml/E) or Lagranges method. Solve for the transient and steady—state displacement of the system. (771 = 1 Kg, 01 = 0.5 Ns/meter, c2 = 1.5 Ns/meter, k1 = 3 N/meter, k2 = 6 N/meter). The initial conditions are :50 = 0 and v0 = 1 meter/s. x(t) gt— FR b Ft! ”.7 {[4 ) & e—~—- F“ (—d ¥ Figure 4: Single degree of freedom system F“ (2 " ’f’fanrifi'fl'ofléi Moria/1 F,“ T F. K/ {‘72 ;- {/7 X A‘f: I‘ \_ .I (\‘t .I _ ,‘t‘ n : A e 2 "KM/4% / Xk- We f SUB W007 ©2+2ffilnd+ wn2)ACI‘=C) é—= /\ _ >{wn ream/{Ll if? " _ 1""13 7W”! 4 '1—4ii2‘r7: m) : A.)r+)\1B+ifl'D= l/f’l f) /\1(fl{Q'B'D)1\1B+iflD:I «"3: Limo + XD 6. [30 pts] Derive the equation of motion and transfer function for the system shown in Figure 5. Assume a motor is used to apply an external moment, Ma, about point 0 (Where Ma 2 Acos Qt (Nm)). Use the following symbols: k3 is a torsional spring (N/radian), mass of object 1 is m1, mass of object 2 is mg, the moment of inertial about each members center of gravity I91 and I32. Include gravity and assume small angles: sin9 m 9. in W Figure 5: Rotating system with torsional spring, kg, and external moment M,1 about point 0. A {hKefig'lme CM,+M1)j(-3)/+ng 9 I0 5 l - d] +rum FB 0/ e _: m, [0/2 4* M2,? (“1+ m2} ...
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