I'm having difficulty comprehending the response in this solution, regarding the question, 'How large must u be so that the particle will pass through the origin?' Could you please clarify how the solution justifies this aspect?

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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I'm having difficulty comprehending the response in this solution, regarding the question, 'How large must u be so that the particle will pass through the origin?' Could you please clarify how the solution justifies this aspect?

that is, if
e6t
u -
u -
-8°
Such a value of t will exist if u is such that F(u) > 1, where
F
FIGURE 5.1 The function F(u).
и
F
=
-
2
u -
.
-8°
น
The graph of F is shown in Figure 5.1. The condition F > 1 is satisfied if
u > 8, but not otherwise. Hence the particle will pass through the origin if u > 8.
Transcribed Image Text:that is, if e6t u - u - -8° Such a value of t will exist if u is such that F(u) > 1, where F FIGURE 5.1 The function F(u). и F = - 2 u - . -8° น The graph of F is shown in Figure 5.1. The condition F > 1 is satisfied if u > 8, but not otherwise. Hence the particle will pass through the origin if u > 8.
Problem 5.4
An overdamped harmonic oscillator satisfies the equation
x+10x + 16x = : 0.
At time t = O the particle is projected from the point x = 1 towards the origin with
speed u. Find x in the subsequent motion.
Show that the particle will reach the origin at some later time t if
и
2
= e
e6t
-
и 8
How large must u be so that the particle will pass through the origin?
Solution
The equation of motion is solved in the standard manner by seeking solutions of the
form x = eat. Then λ must satisfy the equation
the roots of which are λ
==
22+102+16 = 0,
-2, -8. We have thus found the pair of solutions
x =
-2t
e-8t
The general solution of the equation of motion is therefore
x = Ae
-2t
+ Be-8t
'
where A and B are arbitrary constants. The initial conditions x = 1 and x = -u
when t = 0 give the equations
A + B = 1,
2A + 8B = u,
from which it follows that A =
particle is therefore given by
-
— (u – 8), B
=
(u 2). The motion of the
x = ½ (u − 2)e¯8 – ½ (u — 8)e¯² ¸
The particle is at the origin at time t if
-
(u− 2)e8t — (u - 8)e¯²t = 0,
Transcribed Image Text:Problem 5.4 An overdamped harmonic oscillator satisfies the equation x+10x + 16x = : 0. At time t = O the particle is projected from the point x = 1 towards the origin with speed u. Find x in the subsequent motion. Show that the particle will reach the origin at some later time t if и 2 = e e6t - и 8 How large must u be so that the particle will pass through the origin? Solution The equation of motion is solved in the standard manner by seeking solutions of the form x = eat. Then λ must satisfy the equation the roots of which are λ == 22+102+16 = 0, -2, -8. We have thus found the pair of solutions x = -2t e-8t The general solution of the equation of motion is therefore x = Ae -2t + Be-8t ' where A and B are arbitrary constants. The initial conditions x = 1 and x = -u when t = 0 give the equations A + B = 1, 2A + 8B = u, from which it follows that A = particle is therefore given by - — (u – 8), B = (u 2). The motion of the x = ½ (u − 2)e¯8 – ½ (u — 8)e¯² ¸ The particle is at the origin at time t if - (u− 2)e8t — (u - 8)e¯²t = 0,
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