The position xft) of an object of mass m that is attached to a vertical spring with restitution constant k, can be determined by solving the following initial value problem ma" + kr =0 z(0) =a>0 2(0) = b >0, where the initial conditions indicate that the object is meters below the equilibrium position and is released with a downward velocity of b meters per second. Suppose an object of mass unitary is attached to a vertical spring and is initially 1 meter below the position balance when propelled downward with a speed of 2 meters per second. If the spring constant of restitution is 1. Determine the position of the object at time t, with t in seconds. To find the position of the object at time t, a student poses the value problem initial: "(t) + z(t) = 0 2(0) – 2. (21) z(0) - 1 where you can see a differential equation with an absent independent variable, for this it takes out the change of variable ") - 2'(4) → v(1) = 2(0), which transforms the differential equation of second order in the differential equation: - (1) + z(1) = 0 At the end of the process, she obtains the solution of the initial value problem (21), which is: z(t) = V5sen (t +0,4636) (22) In relation to the scheme of the solution proposed by the student to solve the problem of initial value (21), does the following (a) Determine, using only the techniques for solving linear differential equations of higher order, the solution of the initial value problem (21). Compare your obtained answer with the solution given in (22). (b) Make a comparison between the solution process used by the student and the one carried out at point (a): with what method is the initial value problem given in (21) simpler than solve? what advantages and disadvantages does each of them have? Explain your answer fully.
The position xft) of an object of mass m that is attached to a vertical spring with restitution constant k, can be determined by solving the following initial value problem ma" + kr =0 z(0) =a>0 2(0) = b >0, where the initial conditions indicate that the object is meters below the equilibrium position and is released with a downward velocity of b meters per second. Suppose an object of mass unitary is attached to a vertical spring and is initially 1 meter below the position balance when propelled downward with a speed of 2 meters per second. If the spring constant of restitution is 1. Determine the position of the object at time t, with t in seconds. To find the position of the object at time t, a student poses the value problem initial: "(t) + z(t) = 0 2(0) – 2. (21) z(0) - 1 where you can see a differential equation with an absent independent variable, for this it takes out the change of variable ") - 2'(4) → v(1) = 2(0), which transforms the differential equation of second order in the differential equation: - (1) + z(1) = 0 At the end of the process, she obtains the solution of the initial value problem (21), which is: z(t) = V5sen (t +0,4636) (22) In relation to the scheme of the solution proposed by the student to solve the problem of initial value (21), does the following (a) Determine, using only the techniques for solving linear differential equations of higher order, the solution of the initial value problem (21). Compare your obtained answer with the solution given in (22). (b) Make a comparison between the solution process used by the student and the one carried out at point (a): with what method is the initial value problem given in (21) simpler than solve? what advantages and disadvantages does each of them have? Explain your answer fully.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:The position x(t) of an object of mass m that is attached to a vertical spring with restitution
constant k, can be determined by solving the following initial value problem
mz" + kr = 0
r(0) = a >0 r(0) -b>0,
where the initial conditions indicate that the object is meters below the equilibrium position and is
released with a downward velocity of b meters per second. Suppose an object of mass unitary is
attached to a vertical spring and is initially 1 meter below the position balance when propelled
downward with a speed of 2 meters per second. If the spring constant of restitution is 1.
Determine the position of the object at time t, with t in seconds.
To find the position of the object at time t, a student poses the value problem initial:
"(t) +z(t) - 0
(0) - 2.
(21)
r(0) -1
Where you can see a differential equation with an absent independent variable, for this it takes
out the change of variable (t) = 1() v(t) = 1"(1), which transforms the differential
equation of second order in the differential equation:
7 v(t) + r(t) =0
At the end of the process, she obtains the solution of the initial value problem (21), which is:
z(t) = V5sen (21 +0,4636)
In relation to the scheme of the solution proposed by the student to solve the problem of
(22)
initial value (21), does the following:
(a) Determine, using only the techniques for solving linear differential equations of
higher order, the solution of the initial value problem (21). Compare your obtained answer with
the solution given in (22).
(b) Make a comparison between the solution process used by the student and the one carried out
at point (a): with what method is the initial value problem given in (21) simpler than
solve? What advantages and disadvantages does each of them have? Explain your answer fully.
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