If > 2√/km, the system mx" + px' + kx = 0 is over- damped. The system is allowed to come to equilibrium. Then the mass is given a sharp tap, imparting an instanta- neous downward velocity vo (a) Show that the position of the mass is given by where Vo x(t) = e-ur(2m) sinh yt. u²-4mk Y = 2m (b) Show that the mass reaches its lowest point at 2my н t= tanh1 Y a time independent of the initial conditions. (c) Show that, in the critically damped case, the time it takes the mass to reach its lowest point is given by t = 2m/u.
If > 2√/km, the system mx" + px' + kx = 0 is over- damped. The system is allowed to come to equilibrium. Then the mass is given a sharp tap, imparting an instanta- neous downward velocity vo (a) Show that the position of the mass is given by where Vo x(t) = e-ur(2m) sinh yt. u²-4mk Y = 2m (b) Show that the mass reaches its lowest point at 2my н t= tanh1 Y a time independent of the initial conditions. (c) Show that, in the critically damped case, the time it takes the mass to reach its lowest point is given by t = 2m/u.
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Transcribed Image Text:4.4 21
21. If > 2√/km, the system mx" + ux' + kx = 0 is over-
damped. The system is allowed to come to equilibrium.
Then the mass is given a sharp tap, imparting an instanta-
neous downward velocity vo
(a) Show that the position of the mass is given by
where
Vo
x(t) = e-ur/(2m) sinhạt,
Y
u² - 4mk
Y=
2m
(b) Show that the mass reaches its lowest point at
t =
tanh-1
2my
н
Y
a time independent of the initial conditions.
(c) Show that, in the critically damped case, the time it
takes the mass to reach its lowest point is given by
= 2m/u.
=
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