Two mass-three spring system. Math 216 Differential Equations. Forces on mass m 1. Forces on mass m 2. Kenneth Harris

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1 Two mass-three spring system Math 6 Differential Equations Kenneth Harris kaharri@umich.edu m, m > 0, two masses k, k, k 3 > 0, spring elasticity t), t), displacement of m, m from equilibrium. Positive is right, negative left No friction or eternal forces Department of Mathematics University of Michigan November 3, 008 Kenneth Harris Math 6) Math 6 Differential Equations November 3, 008 / Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Forces on mass m Apply Newton s law to mass m. m = k + k ) = k + k ) + k. Forces on mass m Apply Newton s law to mass m. m = k ) k 3 = k k + k 3 ). k k k k k 3 k k 3 m m m m k stretched k compressed k stretched k compressed Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

2 Two mass-three spring system: Equations First method for solving second order system Two mass, three spring system as a system of equations: m = k + k ) + k m = k k + k 3 ). As a matri equation dividing by m, m ): k +k ) = The equation has the form m k m k m k +k 3 ) m = A. First of three ways to solve the second-order system: m = k + k ) + k m = k k + k 3 ). Method. Rewrite using differential operators: m D + k + k )I ) k = 0 k + m D + k + k 3 )I ) = 0. Solve using Method of Elimination from Section 4.. Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Second method for solving second order system Second of three ways to solve the second-order system: m = k + k ) + k m = k k + k 3 ). Method. Rewrite as a system of four equations and four unknowns Section 4.): y = y y = k + k ) y + k y 3 m m y 3 = y 4 y 4 = k y k + k 3 ) y 3 m m Solve using Eigenvalue Method of Section 5.. Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Third method for solving second order system Third of three ways to solve the second-order system: m = k + k ) + k m = k k + k 3 ). Method 3. Rewrite as a matri equation Section 5.): = k +k ) k m m k m k +k 3 ) m Solve using the etended) Eigenvalue Method of Section 5.3. Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

3 Etended eigenvalue method to second-order systems Etended eigenvalue method Etended eigenvalue method to second-order systems Mechanical systems Solve. = A, where A has real constant components. Guess a solution: t) = ve αt. Substitute into = A: since ve αt) = α ve αt, Cancel e αt α ve αt = Ave αt α v = Av Answer. t) = ve αt is a solution when α is an eigenvalue and v an associated eigenvector of A. Solution. If λ = α is an eigenvalue and v is an eigenvector, ve αt is a solution to = A. Application. Mass-spring problems are of the form = A where where the eigenvalues are negative. Let α = ω and v an associated real-valued) eigenvector. Then is a solution to = A. t) = ve iωt = vcos ωt + i sin ωt) Real Solutions. The real and imaginary parts t) = v cos ωt t) = v sin ωt are linearly independent real-valued solutions to = A. Kenneth Harris Math 6) Math 6 Differential Equations November 3, 008 / Kenneth Harris Math 6) Math 6 Differential Equations November 3, 008 / Etended eigenvalue method to second-order systems Second-order homogeneous linear systems The following is useful for mechanical systems. Theorem Suppose an n n matri A has distinct negative eigenvalues ω, ω,..., ω n with associated real eigenvectors v, v,..., v n. Then a general solution to = A is t) = n a j cos ω j t + b j sin ω j t) j= where a i and b i are parameters. Problem. Find a general solution to the mass-spring system: m = m =, two masses k = k 3 = 4, k = 6, spring elasticity t), t), displacement of m, m from equilibrium. No friction or outside forces The second-order equations are = = 6 0. Equivalently, as a matri equation: [ [ [ 0 6 = 6 0 Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

4 continued continued Compute eigenvalues. 0 λ λ = λ + 0λ + 64 = λ + 6)λ + 4) So, λ = 6, 4. There are four linearly independent real solutions v cos 4t, v sin 4t, v cos t, v sin t where v is an eigenvector for λ = 6 and v is an eigenvector for λ = 4. Compute eigenvector for λ = 6. These are solutions to [ [ [ = This reduces to a single equation Eigenvector. v = = 0 or =. [. Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, / continued concluded Compute eigenvector for λ = 4. These are solutions to [ [ [ = This reduces to a single equation Eigenvector. v = = 0 or =. [. General Solution. The mass-spring system = = 6 0. has a general solution as a vector equation) t) = [ [ a cos 4t +a sin 4t ) + [ [ b cos t +b sin t ) and as a scalar system: t) = a cos 4t + a sin 4t) + b cos t + b sin t) t) = a cos 4t + a sin 4t) + b cos t + b sin t) Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

5 analysis analysis Analysis. The general solution t) = a cos 4t + a sin 4t) + b cos t + b sin t) t) = a cos 4t + a sin 4t) + b cos t + b sin t) can be simplified to t) = c cos4t α ) + c cost α ) t) = c cos4t α ) + c cost α ) where c = a + a, tan α = a c = b + b, tan α = b b a, The displacements of mass m and of mass m : t) = c cos4t α ) + c cost α ) t) = c cos4t α ) + c cost α ) A linear combination of two natural modes of oscillation: the natural frequencies ω = 4 and ω =. The natural mode ω = t) = c cost α ) t) = c cost α ) a free oscillation no damping) in which the masses move in synchrony in the same direction and with the same frequency ω = ) and equal amplitudes of oscillation c ). Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, 008 / analysis Forced oscillations Mass-spring systems with eternal forces The natural mode ω = 4 t) = c cos4t α ) t) = c cos4 ) a free oscillation no damping) in which the masses move in synchrony in the opposite directions and with the same frequency ω = 4) and equal amplitudes of oscillation c ). Suppose we apply eternal forces F to mass m and F to mass m in the mass-spring system. We now have a nonhomogeneous system m = k + k ) t) + k t) + F t) m = k t) k + k 3 ) t) + F t) t Kenneth Harris Math 6) Math 6 Differential Equations November 3, 008 / Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

6 Forced oscillations Mechanical systems with forced oscillations Forced oscillations. We are interested in periodic eternal forces applied to the masses F 0 is a constant vector) [ F0 F 0 cos ωt = cos ωt F As a system of equations: As a matri equation = 6 4 m = k + k ) + k + F 0 cos ωt m = k k + k 3 ) + F cos ωt k +k ) k m m k k +k 3 ) m m Note the form = A + F 0 cos ωt F 0 F 3 5 cos ωt, Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Forced oscillations First method for solving second order system First of two ways to solve the second-order system: m = k + k ) + k + F 0 cos ωt m = k k + k 3 ) + F cos ωt Method. Rewrite using differential operators: m D + k + k )I ) k = F 0 cos ωt k + m D + k + k 3 )I ) = F cos ωt Solve using Method of Elimination from Section 4.. Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Forced oscillations Second method for solving second order system Second of two ways to solve the second-order system: m = k + k ) + k + F 0 cos ωt m = k k + k 3 ) + F cos ωt Method. Rewrite as a matri equation Section 5.): k +k ) k m m F 0 = + cos ωt k m k +k 3 ) m which is of the form = A + F 0 cos ωt. Guess a particular solution of the form p = c cos ωt, and determine the values of the parameter c. F Forced oscillations Second method for solving second order system Solve the second-order system = A + F 0 cos ωt Guess a particular solution of the form p = c cos ωt. Substitute p where p = ω c cos ωt and cancel the common cos ωt ω c cos ωt = Ac cos ωt + F 0 cos ωt ω c = Ac + F 0. New problem. We want a solution c to A + ω I ) c = F 0. Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

7 Forced oscillations Second method for solving second order system Eample Eample Solve for unknown c A + ω I ) c = F 0. Observation. As long as ω is not an eigenvalue for A, then A + ω I is invertible. Solution. If ω is not an eigenvalue for A, then A + ω I ) c = F 0. has a unique solution. Observation. If ω is an eigenvalue for A, then we have resonance. In this case, we can use techniques of Section 5.6 which generalize the Method of Undetermined Coefficients and Method of Variation of Parameters to vector systems. Problem. Find a general solution to the mass-spring system of when: Eternal forces: F = 30 cos t, F = 60 cos t The second-order equations are = cos t = cos t Equivalently, as a matri equation: [ [ 0 6 = 6 0 [ + [ 30 cos t 60 cos t Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Eample continued Eample Compute a particular solution for [ [ 0 6 = 6 0 [ + [ 30 cos t 60 cos t Guess. p = c cos t. Let ω =. Since ω = is not an eigenvalue for the system λ = 6, 4), there is no resonance. Substitute the guess p = c cos t into the equation. This reduces to [ c c cos t = [ [ c c cos t + [ [ [ [ c 30 = c 60 cos t Eample continued Solve the system of equations So, c = 4 and c = 6. Particular solution Eample 0 = 3c + c 0 = c 3c p = [ 4 cos t 6 General solution. c + p, where c is from : t) = a cos 4t + a sin 4t) + b cos t + b sin t) + 4 cos t t) = a cos 4t + a sin 4t) + b cos t + b sin t) + 6 cos t Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

8 Eample continued Eample Eample continued Eample Problem. Find the motion when the two masses begin at the equillibrium position at rest. The initial conditions are 0) = 0, 0) = 0, 0) = 0, 0) = 0 The general solution and derivatives are t) = a cos 4t + a sin 4t) + b cos t + b sin t) + 4 cos t t) = 4a sin 4t + 4a cos t) + b sin t + b cos t) 6 sin t t) = a cos 4t + a sin 4t) + b cos t + b sin t) + 6 cos t t) = 4a sin 4t + 4a cos 4t) + b sin t + b cos t) 4 sin t Substitute. 0 = a + b = 4a + b 0 = a + b = 4a + b Solve. The solutions are 0 = a + b = 4a + b 0 = a + b = 4a + b a = b = 5 a = b = 0 Solution. The general solution and derivatives are t) = cos 4t 5 cos t + 4 cos t t) = cos 4t 5 cos t + 6 cos t Kenneth Harris Math 6) Math 6 Differential Equations November 3, / Kenneth Harris Math 6) Math 6 Differential Equations November 3, /

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