By Rosu H.C.

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EXAMPLE We give here an example of employing the previous equation. Determine the final amplitude of oscillations of a system acted by an extenal force F0 = const. during a limited time T . For this time interval we have ξ = exp(iwt) ξ= ξ= F0 exp(iwt) m T exp(−iwt)dt , 0 F0 [1 − exp(−iwt)] exp(iwt) . iwm Using |ξ|2 = a2 w2 we obtain a= 2F0 1 sin( wT ) . 3 DAMPED HARMONIC OSCILLATOR Until now we have studied oscillatory motions in free space (the vacuum), or when the effects of the medium through which the oscillator moves are negligeable.

Since the coefficients kij and mij are real and symmetric, the quadratic forms of the numerator and denominator are real, and being essentially positive one concludes that w2 are equally positive. EXAMPLE As an example we model the equations of motion of a double pendulum. The potential energy of this system with two degrees of freedom is U = m1 gl1(1 − cos θ1 ) + m2 gl1(1 − cos θ1 ) + m2 gl2(1 − cos θ2 ) . Applying (9), one gets 1 1 U = (m1 + m2 )gl1θ12 + m2gl2θ22 . 2 2 57 Comparing with (10), we identify k11 = (m1 + m2)l12 k12 = k21 = 0 k22 = m2 gl2 .

4 NORMAL MODES Before defining the normal modes, we rewrite (15) as follows .. M X + K |X = 0 , where |X is the n-dimensional vector whose matrix representation is (19); M and K are two operators having the matrix representation given by (16) and (17), respectively. We have thus an operatorial equation. Since M is a nonsingular and symmetric operator, the inverse operator M−1 and the operators M1/2 and M−1/2 are well defined. In this case, we can express the operatorial equation in the form d2 1/2 M |X = −M −1/2 KM −1/2M 1/2 |X , dt2 or more compactly d2 X = −λ X dt2 , (32) where 1/2 X = M |X and λ = M −1/2KM −1/2 .