Unformatted text preview: e, it is possible to reduce the solution to a linear approximation by using sin θ ≈ θ as follows. (J cg + mL2 )θ + mgLθ = 0 Lecture Notes -32- 06/16/06 12:25 PM Mechanical Vibrations I Energy Example A
In order to solve the Energy Equation, m x(t) d (T + U ) = 0 , dt
for the given single degree-of-freedom system, it is first necessary to formulate the energy equations: kinetic and potential. Formulating the kinetic energy yields, T = 1 mx 2 , 2 and formulating the potential energy yields, U = 1 kx 2 + mgx . 2 Evaluating the Energy equation yields, k d d (T + U ) = ( 1 mx 2 + 1 kx 2 + mgx ) = mxx + kxx + mgx = 0 2 2 dt dt
Observe the odd form of the equation. There is an extra velocity term, x , that must be canceled out to yield the actual equation of motion. mx + kx + mg = 0 mx + kx = −mg
Since the mg term ends up on the right hand side with the external force, the coordinate x can be redefined to be from the static equilibrium position. As a result, the mg term is eliminated as shown.
mx + kx = 0 NOTE: Comparing the solution generated by Lagrange Method to that of the Energy Method* shows that the Energy Method is not simply a single degree-of-freedom Lagrange Solution, but rather a related, by not equivalent technique. Recognize also that the application of the Energy Method is extremely limited, i.e. single degree-of-freedom, conservative and un-forced. The Lagrange Method has no such limitations and is in fact a completely general solution method, comparable in usefulness to Newton’s Method. Note that the Energy Method is presented for completeness. You may encounter it in your studies of other reference material, but because it is not a general technique, it will not be used in this course. Lecture Notes -3306/16/06 12:25 PM * Mechanical Vibrations I #7 - Steady-State Solution of Equation of Motion
The general solution to the equation of motion involves both the homogeneous and the particular solution. While the form of the homogeneous solution is dictated by the nature of the governing second order constant matrix coefficient differential equation, the form of the particular solution is controlled by the form of the forcing function (excitation). Mathematically, there are infinite possibilities, but for practical applications, the form can be restricted to harmonic functions. By limiting the form of the excitation to harmonic functions, the concept of steady-state solution can be defined. For vibration purposes, the steady-state solution is the resulting response after all the initial condition transients have decayed. When solving the response for the steady-state, the solution is most easily formulated in either the Laplace Domain (s) or the Frequency Domain ( ω ). (NOTE: it will be shown later that the Frequency Domain solution can be developed from the Laplace Domain by letting s = jω .) Working with a representative single degree-of –freedom system (SDOF), the equation of motion is: m mx(t ) + cx(t ) + k x(t ) = f (t )
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This note was uploaded on 09/29/2013 for the course MECHANICAL ME taught by Professor Regalla during the Fall '11 term at Birla Institute of Technology & Science, Pilani - Hyderabad.
- Fall '11