{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

820_Dynamics 11ed Manual

820_Dynamics 11ed Manual - ⎛ ⎝ ⎞ ⎠ 1 m 2 k − sin...

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

View Full Document Right Arrow Icon
Engineering Mechanics - Dynamics Chapter 22 Problem 22-46 Use a block-and-spring model like that shown in Fig. 22-14 a but suspended from a vertical position and subjected to a periodic support displacement of δ = 0 sin ω t , determine the equation of motion for the system, and obtain its general solution. Define the displacement y measured from the static equilibrium position of the block when t = 0. Solution: For Static Equilibrium mg k st = Equation of Motion is then k δδ st + y () mg m y'' = m y'' k y + k 0 sin t () = y'' k m y + k m 0 sin t () = The solution consists of a homogeneous part and a particular part yt () A sin k m t B cos k m t
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ⎛ ⎝ ⎞ ⎠ + 1 m 2 k − sin t ( ) + = The constants A and B are determined from the initial conditions. Problem 22-47 A block of mass M is suspended from a spring having a stiffness k . If the block is acted upon by a vertical force F = F 0 sin t , determine the equation which describes the motion of the block when it is pulled down a distance d from the equilibrium position and released from rest at t = 0. Assume that positive displacement is downward. Given: M 5 kg = k 300 N m = F 7 N = 818...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online