p0905

# P0905 -

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This note was uploaded on 12/16/2010 for the course AME 301 at USC.

Ask a homework question - tutors are online