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Unformatted text preview: Solutions: Homework Set 2 1. Problem 2.4. Assuming small rotation (clockwise because of mg ) and measuring m s displacement x positive downward, spring forces are f S 1 = k 1 ( x- a ), f S 2 = k 2 b . From FBD of (massless) rigid bar, M O = k 1 ( x- a ) a- k 2 b 2 = 0. From FBD of m , F x = m x = mg- k 1 ( x- a ). Eliminate to get EOM: m x + k eq x = mg, k eq = k 1 k 2 b 2 k 1 a 2 + k 2 b 2 Equilibrium position (particular solution): x eq m = mg k eq . If eq m is of interest, obtain from M O equation: eq m = k 1 a k 1 a 2 + k 2 b 2 x eq m . Natural frequency: n = radicalbig k eq /m . 2. Problem 2.8. From Problem 1.8, EOM: 3( R- r ) +2 g sin = 0. For small , sin . Natural frequency: n = radicalbig 8 g/ 9 R . 3. Problem 2.15. EOM: m x + k eq x = 0 where k eq = parenleftbigg 1 k 1 + k 2 + 1 k 3 parenrightbigg- 1 . Therefore n = radicalbig k eq /m . Response to ICs: x ( t ) = x cos n t + v n...
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This note was uploaded on 09/18/2011 for the course ASE 365 taught by Professor Staff during the Spring '10 term at University of Texas at Austin.
- Spring '10