Explanation Given: The spring/mass system is m x ″ + 2 x ′ + k x = 0 with the initial condition as x ( 0 ) = 0 and x ′ ( 0 ) = v 0 . The value of k is 10 . Calculation: The second order linear differential equation which represents the model of spring/mass system is as follows: m x ″ + 2 x ′ + k x = 0 … … ( 1 ) It is given that the mass ( m ) passes through the equilibrium position at t = 1 second. The equation formed from the above statement is as follows: x ( 1 ) = 0 The initial conditions are as follows: x ( 0 ) = 0 x ′ ( 0 ) = v 0 It is given that the value of k is 10 . Substitute 10 for k in equation ( 1 ) . m x ″ + 2 x ′ + 10 x = 0 x ″ + 2 x ′ m + 10 x m = 0 The auxiliary equation for the above differential equation is as follows: γ 2 + 2 γ m + 10 m = 0 The solution of the above equation is as follows: γ = ( − 2 m ) ± ( − 2 m ) 2 − 40 m 2 = ( − 2 m ) ± 2 1 m 2 − 10 m 2 = ( − 1 m ) ± 1 m 2 − 10 m = ( − 1 m ) ± 1 − 10 m m Label the above equation as follows: γ = ( − 1 m ) ± 1 − 10 m m … … ( 2 ) The value of γ depends on the value of m . For the case m = 1 10 . Substitute 1 10 for m in equation ( 2 ) . γ = ( − 10 ) ± 10 1 − ( 10 ⋅ 1 10 ) = − 10 ± 0 = − 10 Therefore, the value of γ is as follows: γ 1 = − 10 γ 2 = − 10 So, for m = 1 10 the solution of the given differential equation is as follows: x ( t ) = c 1 e − 10 t + c 2 t e − 10 t … … ( 3 ) As per the given initial condition x ( 0 ) = 0 , the result obtained is as follows: x ( 0 ) = c 1 e 0 + ( c 2 ⋅ 0 ⋅ e ) c 1 = 0 Substitute the value of c 1 in equation ( 3 ) . x ( t ) = c 2 t e − 10 t Differentiate the above equation with respect to t . x ′ ( t ) = c 2 e − 10 t − 10 c 2 t e − 10 t Substitute 0 for t in the above equation. x ′ ( 0 ) = c 2 e 0 − ( 10 ⋅ c 2 ⋅ t ⋅ e 0 ) c 2 = v 0 Substitute the value of c 2 in x ( t ) = c 2 t e − 10 t . x ( t ) = v 0 t e − 10 t As per the given initial condition x ( 0 ) = 0 , the result obtained is as follows: x ( 1 ) = v 0 e − 10 v 0 e − 10 = 0 v 0 = 0 The above equation is not valid because v 0 ≠ 0 . This implies that for m = 1 10 non-trivial solution does not exist. For the case 0 < m < 1 10 . The roots of the auxiliary equation as obtained above are as follows: γ 1 = ( − 1 m ) − 1 − 10 m m γ 2 = ( − 1 m ) + 1 − 10 m m The solution of the given differential equation is as follows: x ( t ) = c 3 e γ 1 t + c 4 e γ 2 t … … ( 4 ) As per the given initial condition x ( 0 ) = 0 , the result obtained is as follows: x ( 0 ) = c 3 e 0 + c 4 e 0 c 3 + c 4 = 0 c 3 = − c 4 Differentiate equation ( 4 ) with respect to t . x ′ ( t ) = c 3 γ 1 e γ 1 t + c 4 γ 2 e γ 2 t As per the given initial condition x ′ ( 0 ) = v 0 , the result obtained is as follows: x ′ ( 0 ) = c 3 γ 1 e 0 + c 4 γ 2 e 0 c 3 γ 1 + c 4 γ 2 = 0 Substitute c 3 = − c 4 in the above equation. − c 4 γ 1 + c 4 γ 2 = v 0 c 4 ( γ 2 − γ 1 ) = v 0 c 4 = v 0 ( γ 2 − γ 1 ) Substitute the value of c 4 in c 3 = − c 4 . c 3 = − v 0 ( γ 2 − γ 1 ) Substitute the value of c 3 and c 4 in equation ( 4 ) . x ( t ) = − v 0 ( γ 2 − γ 1 ) e γ 1 t + v 0 ( γ 2 − γ 1 ) e γ 2 t = v 0 ( γ 2 − γ 1 ) ( e γ 2 t − e γ 1 t ) As per the given initial condition x ′ ( 0 ) = v 0 , the result obtained is as follows: x ( 0 ) = v 0 ( γ 2 − γ 1 ) ( e 0 − e 0 ) v 0 = v 0 ( γ 2 − γ 1 ) ( 1 − 1 ) v 0 = 0 The above equation is not valid because v 0 ≠ 0

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 5.2, Problem 34E

To determine

The value of the mass (m) which passes through the equilibrium position at t=1 second and it is provided that model for spring/mass system is mx″+2x′+kx=0 with the initial condition as x(0)=0 and x′(0)=v0.

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Chapter 5 Solutions

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

Ch. 5.1 - 5.1.1 Spring/Mass systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A force...Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - 5.1.1Spring/Mass Systems: Free Undamped Motion A...

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The value of the mass ( m ) which passes through the equilibrium position at t = 1 second and it is provided that model for spring/mass system is m x ″ + 2 x ′ + k x = 0 with the initial condition as x ( 0 ) = 0 and x ′ ( 0 ) = v 0 .

Solution Summary: The author describes the second order linear differential equation which represents the model of spring/mass system.

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The value of the mass (m) which passes through the equilibrium position at t=1 second and it is provided that model for spring/mass system is mx″+2x′+kx=0 with the initial condition as x(0)=0 and x′(0)=v0.

Chapter 5 Solutions