oscill_sum

# oscill_sum - Some Basic Information about M-S-D Systems 1...

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Unformatted text preview: Some Basic Information about M-S-D Systems 1 Introduction We want to give some summary of the facts concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillators governed by second-order, constant coeffi- cient systems. Recall that the homogeneous differential equation that we are dealing with may be written in the form m d 2 dt 2 ) x + 2 γ d dt x + ω 2 o x = 0 , which has characteristic equation m λ 2 + c λ + k = 0 , or λ 2 + 2 γλ + ω 2 o = 0 , where 2 γ = c/m and ω 2 o = k/m . Using these values of the constants, the roots are λ ± = − 2 γ ± radicalbig 4 γ 2 − 4 ω 2 o 2 = − γ ± radicalbig γ 2 − ω 2 o = ± radicalbig − ω 2 o = ± i ω o , γ = 0 γ 1 ± radicalBigg 1 − parenleftbigg ω o γ parenrightbigg 2 , γ negationslash = 0 . 2 The Harmonic Oscillator The first alternative leads to the Harmonic Oscillator which has general solution x ( t ) = C 1 cos ( ω o t ) + C 2 sin ( ω o t ) . We make the folowing remarks: (a) ω o is called the natural frequency . This is the number of waves passing a fixed observer in unit time. 1 (b) The quantity 2 π ω o is called the wave length . It is usually called λ and is the distance at which the profile repeats itself. (c) The time taken for one complete wave to pass any particular point is called the period , T , of the wave. Since the cosine and sine functions are periodic of period 2 π , the period is given by T = λ/ 2 π = 1 /ω o . (d) x ( t ) = A cos ( ω o t − δ ) with A = radicalbig C 2 1 + C 2 2 and cos ( δ ) = C 1 /A, sin ( δ ) = C 2 /A or tan ( δ ) = C 2 /C 1 is called the phase-amplitude form of the solution; A is the amplitude and δ is the phase shift. Example 2.1 : Let m = 1 and k = 36. Then ω o = 6 and the solution of the initial value problem ¨ x + 36 x = 0 with initial conditions x ( t ) = 1 2 , ˙ x (0) = 1 can be determined from the general form of the solution x ( t ) = C 1 cos (6 t ) + C 2 sin (6 t ) ˙ x ( t ) = − 6 C 1 sin (6 t ) + 6 C 2 cos (6 t ) Then x (0) = 1 / 2 = C 1 and ˙ x (0) = 1 · C 2 or C 2 = 1 / 6. Hence the required solution is x ( t ) = 1 2 cos (6 t ) + 1 6 sin (6 t ) . The AMPLITUDE of the vibration is A = radicalBig ( 1 2 ) 2 + ( 1 6 ) 2 = radicalBig 5 18 ≈ . 5270. The PHASE SHIFT is computed from the definition of δ taking care that the correct branch of the inverse tangent is chosen. Here C 1 = 1 / 2 and C 2 = 1 / 6 so sin ( δ ) > 0 and cos ( δ ) > 0 which implies that δ lies between 0 and π/ 2. tan ( δ ) = ( 1 2 ) ( 1 6 ) = 6 2 = 3 , and we may conclude that, approximately, δ = 1 . 249 radians or, again approximately, 71 . 56 degrees....
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## This note was uploaded on 12/02/2009 for the course MATH 352 taught by Professor Staff during the Spring '08 term at University of Delaware.

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oscill_sum - Some Basic Information about M-S-D Systems 1...

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