Materials Material Science

Posted On April 11, 2017 at 3:59 pm by / Comments Off on Download Energy methods in dynamics by Khanh Chau Le PDF

By Khanh Chau Le

Half I. Linear thought: 1. unmarried oscillator -- 2. Coupled oscillators -- three. non-stop oscillators -- four. Linear waves -- half II. Nonlinear thought: five. self sufficient unmarried oscillator -- 6. Non-autonomous unmarried oscillator -- 7. Coupled oscillators -- Nonlinear waves

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Extra info for Energy methods in dynamics

Sample text

22) Harmonic excitations. Equations of motion derived above are the inhomogeneous linear differential equations of second order. The solution of these linear equations is the sum of any particular solution of the inhomogeneous equations and the general solution of the homogeneous equations which has been found in previous Section. Thus, it is enough to find any particular solution of the inhomogeneous equations. For the harmonic excitations this can be done directly. We consider two cases. Conservative oscillators.

The elimination of forced vibration can be explained physically as follows. At the frequency ω = ν2 the eliminator and the excitation vibrate in counter-phases such that the spring force acting on m1 from the eliminator is equal and opposite to the exciting force k1 xˆe cos ω t. 26) at ω = ν2 yields ν 2 xˆe k1 xˆ2 = − 10 2 = − xˆe . k2 μν2 Thus, the resultant force acting on m1 is zero and therefore that mass does not vibrate. To eliminate the unwanted forced vibration of m1 caused by some excitation source with the fixed frequency ω we must therefore choose the mass and the spring of the eliminator in such a relation that k2 /m2 = ω .

26). 26) by q˙ j and summing up over j from 1 to n we obtain n ∑ j=1 d ∂L ∂L q˙ j − q˙ j dt ∂ q˙ j ∂qj = 0. Using the product and chain rules of differentiation we get n d ∑ dt j=1 n ∂L ∂L ∂L q˙ j − ∑ q¨ j + q˙ j ∂ q˙ j ∂qj j=1 ∂ q˙ j = d n ∂L (∑ q˙ j − L) = 0. 27) Taking into account the property of K, we see that the expression in parentheses is equal to 2K − L = K + U. Thus, the total energy E = K + U = E0 is conserved. 25) by replacing the variations δ q j with the real velocities q˙ j . 25) to t1 t0 d (K + U) dt = 0 dt ⇒ K + U = E0 .