To determine The harmonic motion which starts from y 0 where the initial velocity is v 0 . Explanation Result used: Case (i): If the auxiliary equation has distinct real roots λ 1 , and λ 2 of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e λ 1 x , e λ 2 x and the general solution of y ″ + a y ′ + b y = 0 is y = c 1 e λ 1 x + c 2 e λ 2 x . Case (ii): If the auxiliary equation has real double root λ = − a of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e − a x , x e − a x and the general solution of y ″ + a y ′ + b y = 0 is y = ( c 1 + c 2 ) e − a x . Case (iii): If the auxiliary equation has complex conjugate root λ 1 = a + i ω , λ 2 = a − i ω of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e a x cos ω x , e a x sin ω x and the general solution of y ″ + a y ′ + b y = 0 is y = e − a x ( A cos ω x + B sin ω x ) . Calculation: It is known that, the Newton’s law for F = − F 1 is m y ″ = − F 1 = − k y . That is, m y ″ + k y = 0 which represents a homogenous ODE with a constant coefficient. Compute the auxiliary equation of the above ODE as follows. Let y = e λ x . That implies, the first and second derivative of y is y ′ = λ e λ x and y ″ = λ 2 e λ x , respectively. Substitute y ″ = λ 2 e λ x and y = e λ x in (1) and simplify further. m λ 2 e λ x + k e λ x = 0 e λ x ( m λ 2 + k ) = 0 m λ 2 + k = 0 λ 2 + k m = 0 That is, the characteristic equation is λ 2 + k m = 0 . Compute the roots of the above characteristic equation as follows. λ 2 + k m = 0 λ 2 = − k m λ = − k m λ = ± i k m λ = ± i ω 0 [ ∵ ω 0 = k m ] From the above calculation, it is observed that λ 2 + k m = 0 has complex conjugate roots. Then by case (iii), the basis ϕ 1 ( t ) and ϕ 2 ( t ) of λ 2 + k m = 0 is e 0 ⋅ x cos ω 0 x , e 0 ⋅ x sin ω 0 x . The general solution y ( t ) = A ϕ 1 ( t ) + B ϕ 2 ( t ) of λ 2 + k m = 0 is, y ( t ) = A cos ( ω 0 t ) + B sin ( ω 0 t ) To determine To graph: The harmonic motion y = y 0 cos ( ω 0 t ) + v 0 ω 0 sin ( ω 0 t ) where w 0 = π , and y 0 = 1 ; The value of t where all the curves intersect and explain why the curves intersect at that point.

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.4, Problem 1P

To determine

The harmonic motion which starts from y0 where the initial velocity is v0.

To determine

To graph: The harmonic motion y=y0cos(ω0t)+v0ω0sin(ω0t) where w0=π, and y0=1; The value of t where all the curves intersect and explain why the curves intersect at that point.

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Chapter 2.3, Problem 15P

Chapter 2.4, Problem 2P

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