P0905 -

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Unformatted text preview: 9.5. CHAPTER 9, PROBLEM 5 353 9.5 Chapter 9, Problem 5 Problem: A block of mass m is attached to a spring with constant k from below. It is also connected to a spring with constant 5k from above by a cable and pulley as shown. At time t = 0 the block is released from z = 0 where both springs are unstretched. You can neglect the mass of the pulley. (a) Determine the equilibrium displacement, zeq , of the block. (b) Letting T denote cable tension, verify that mz = mg − k(z + zeq ) − T ¨ and T= 5 k (z + zeq ) 4 (c) Derive a single differential equation for the system and determine the natural frequency. ............................................................. .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... .................................................................................... ............................... . . . . . ................. .......................................................................................................... ........................................................................................... . . ... .... .. . .. . . .... .... .... .... . . . .. . . . .. . .. . .... . .... . . .. . . . .. ... . . .. .... ... .... ... .. . . .. .. .... . .... . . ... ... . . . . . .. . . .. ....... . ......... . .. . .. .. ................... .......... .. ... ....... .. .. .. .......................... ... .. . .. . . ........................ .......................... . . ..................... . .... .. . ....................... ......................... . ... .. . ........ . . ..... . . . . . . . . . . . . . ......................... . . . ......................... . . ......................... . . ......................... . ......................... . . ......................... . . ......................... . . ......................... . . ......................... .. . . . . .. . .. . . . .. ..... . .... . . ... . ...... . .. .. . . .. ..... ..... . .. . .. . . .... . ... . .. . . . ... ..... .. . . . . .. .. .. .. .. .. . ........................ . . . . . . . .............................................................................................. ....................... .. .. .. .. .. . . . . . . .. ............. ... .. .... ................................................................................................ . ............................................................................................................... .......................................................................................................................... .......................................................................................................................... .......................................................................................................................... • 5k g • m • • z k •• Solution: (a) The forces acting on the block and the pulley are as shown in the following figure. Clearly, when the block moves downward a distance z , conservation of cable length tells us that the upper spring will be stretched a distance z/2. So, in equilibrium, we have mg − kzeq − T = 0 . . . .. 5 . . . . . . .2 . . . . . . . . . . . . . . . . . .. . ... ... .. .............. .................... ................... . ....................... . ......................... ............................ ............... ... ...................... . .. .............. ...................... . ........ . . ...... . . . .......... .. .. . . . . . . . . . . . . .. .. . .. .. . . . and 2T − 5k . .. . .. . . . . . . . . . . . ......... . ........................ ........................ ......................... ............... ......................... ......................... .. ......................... .. ......................... ....................... . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. .. . zeq =0 2 kzeq T m • kzeq T T mg We can solve for T from the force balance for the pulley, viz., T= 5 kzeq 4 Substituting into the force balance equation for the block yields 5 mg − kzeq − kzeq = 0 4 =⇒ 9 mg − kzeq = 0 4 354 Therefore, the equilibrium displacement is zeq = (b) The basic equations of motion for this system are mz = mg − k (z + zeq ) − T ¨ Thus, the equations of motion are mz = mg − k (z + zeq ) − T ¨ and and 4 mg 9k CHAPTER 9. MECHANICAL VIBRATIONS 2T = 5 k (z + zeq ) 2 T= 5 k (z + zeq ) 4 (c) Combining the equations of motion derived in Part (b), we have 5 9 mz = mg − k (z + zeq ) − k (z + zeq ) = mg − k (z + zeq ) ¨ 4 4 But, we know from Part (a) that 9 kzeq 4 Therefore, canceling like terms, the equation of motion simplifies to mg = 9 mz + kz = 0 ¨ 4 By inspection, the natural frequency of this system is given by 2 ωn = 9 4k m =⇒ ωn = 3 2 k m ...
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This note was uploaded on 12/16/2010 for the course AME 301 at USC.

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