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Unformatted text preview: Some Basic Information about MSD Systems 1 Introduction We want to give some summary of the facts concerning unforced (homogeneous) and forced (nonhomogeneous) models for linear oscillators governed by secondorder, 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 phaseamplitude 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.
 Spring '08
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