p0902 - 9.2. CHAPTER 9, PROBLEM 2 343 9.2 Chapter 9,...

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Unformatted text preview: 9.2. CHAPTER 9, PROBLEM 2 343 9.2 Chapter 9, Problem 2 Problem: A weight of mass M is suspended from a horizontal surface with a three-spring arrangement as shown. The springs above the bar have constants k1 = k and k2 = 3 k. The lower spring has 2 constant k3 = 12k and is attached to the center of a bar of negligible mass midway between the points of attachment of the upper springs. Determine the natural frequency of the system. . . . ............................. . .. ............................... ............................... .............................. .. .. .. . . . . ................................................................................................................................................................................................................................................. ....................................................................................................................................................................................................................................... ........................................................ . . ....... . . . . . . ......... . .................................................. ...................... ....... 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.......................................................................................... ...... ........................................................................... .. .... ...... . . ................. ....... . . ... .. ....... . .. . ............................ . . .... . ... . . . . . .. . .. . .... .... .... .. . 3 . .. .... ..... . . ..... . .. .. ...... ... .. . . .. . .. . .... . .. . . . . ....... . . . ... . . . .. .. . . ... . .... . . . ... . . .. . . .3 . ...... . . . . .. . ..... . .. . .. .. . . . .. . ...... ..... . . .......... ... . ...................... .. ......................... . .. . .. ........................................... . ...... .. ......... .. . . .................................... . . . ............................ . . . ............................. . . . ............................. ............................. . . . .......................... . . . .. . . ............................ . .. .. .............................. . . ...... .. .. .. ...... .. ...................... . .... . . . ........................... . .. .. . . . ... 2 • • • k k g ...... ... .. . . . .. . . . . . . . . . . . . x x • • ∆x . . . . . . . . . . . . . .. . . ...... ..... . k x M • HINT: Determine the effective spring constant for an equivalent single springmass system by balancing forces and moments in the equilibrium position. Solution: Following the hint, if we can find an effective spring constant, ke , for an equivalent single spring-mass system, the natural frequency is simply ωn = ke /M . The figure includes sufficient detail to indicate the various spring displacements from unstretched positions. From the symmetry of the bar, clearly the displacement of its center is 1 xbar = (x1 + x2 ) 2 Hence, the overall displacement in equilibrium, ∆x, is ∆x = x3 + xbar = x3 + 1 (x1 + x2 ) 2 We must now determine the equilibrium spring displacements, x1 , x2 and x3 . The figure below shows the forces acting on the bar and on the weight. Focusing first on the weight, obviously the spring force from Spring 3 balances the downward force of gravity, i.e., k3 x3 = M g . . . .. . .. ... ... . . . . . . .. 2 . ... ... . . . ... .. . . . . .............. ................ ......... . . .... . .. ... ...... ...... .... ...................... . . ... 1. . . ....................................................... ............................................................... . .. ... . . . . . . . .. .......... . ... . . . . . .......... ......................................................................................................................................... .. . . . ....................... ..... . ...................................... . ................... ................. . . . . ... .............................................................................................................. ....... .. . . . .... ......... ............................................................ ...... .. .. . ............................................................................................................................... ....................................................................................................................................... . . .......... . ... . .. ................. .. . . ................................... ........ .. . . . . . ... . . . ................................... . . . . . . . . . . .. .. .. 3 3 . .... .. . . ... . .. .. . ... . ... . .. . .. .. . .. . .. .. . 33 . . . . . . . . . ...... ...... ..................... . .... .. .. .. ... . ...................... . .... ............................ .. . . . . . ............................. ............................. . . ............................. . . ............................. . . ............................. . . ............................. . . ............................. . . ............................. . . ............................. . . ...................... . .. . .. . . .. ...................... ..................... . . . .. . . . .. . . . . . . . . . . . . . . . . . . . ... ... .. .. .. . . k x2 k1 x kx kx M Mg 344 CHAPTER 9. MECHANICAL VIBRATIONS Turning to the bar, taking moments about its center tells us that 3 i=1 Mi = −k1 x1 + k2 x2 + k3 x3 · 0 = 0 =⇒ k1 x1 = k2 x2 Finally, balancing forces on the bar and noting from above that k3 x3 = M g , we have 3 i=1 Fi = k1 x1 + k2 x2 − k3 x3 = 0 =⇒ k1 x1 + k2 x2 = M g Therefore, the three spring forces are k1 x1 = 1 M g, 2 k2 x2 = 1 M g, 2 k3 x3 = M g This means the equilibrium spring displacements are x1 = Mg , 2k1 x2 = Mg , 2k2 x3 = Mg k3 So, the overall displacement in equilibrium is ∆x = Mg 1 + k3 2 Mg Mg + 2k1 2k2 = Mg 1 1 1 4k1 k2 + k1 k3 + k2 k3 + + = Mg k3 4k1 4k2 4k1 k2 k3 An equivalent single spring-mass system would be such that, in equilibrium, ke ∆x = M g Combining these two equations tells us that the effective spring constant, ke , is ke = 4k1 k2 k3 4k1 k2 + k1 k3 + k2 k3 Finally, for the given values of k1 = k , k2 = 3 k and k3 = 12k , we have 2 ke = 3 2k 4k · 4k · 3 k · 12k 72k3 2 = 2k =2 3 6k + 12k 2 + 18k 2 + k · 12k + 2 k · 12k Therefore, the natural frequency of the system is ωn = 2k M ...
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