A mass 3 lb hangs from a spring constant of 12 lb/sec². Suppose the mass is set in motion from its equilibrium position, and then it is released with an initial velocity of 2 ft/sec downwards. There is no damping. Then u(t) satisfies the initial value problem: 32 u" +12u=0, 3 32 3u" + 12u = 0, 3u" + 12u = 0, 3u" + 12u = 0, u" + 12u = = 0, u(0)=0, u(0) = 0, u(0) = 0, u(0) = 0, u(0) = 2, u'(0) = 2 u'(0) =-2 u' (0) = 2 u' (0) =-2 u' (0) = 0

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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A mass 3 16 hangs from a spring constant of 12 lb/sec². Suppose the mass is set in
motion from its equilibrium position, and then it is released with an initial velocity
of 2 ft/sec downwards. There is no damping. Then u(t) satisfies the initial value
problem:
¡u” +12u=0, u(0) = 0,
32
u(0) = 0,
3
32
u" + 12u = 0,
3u" + 12u: =
0,
3u" + 12u: 0,
3u" + 12u: =
0,
u(0) = 0,
u(0) = 0,
u(0) = 2,
u' (0) = 2
u'(0) =-2
u'(0) = 2
u' (0)
u'(0) : = 0
=
Transcribed Image Text:A mass 3 16 hangs from a spring constant of 12 lb/sec². Suppose the mass is set in motion from its equilibrium position, and then it is released with an initial velocity of 2 ft/sec downwards. There is no damping. Then u(t) satisfies the initial value problem: ¡u” +12u=0, u(0) = 0, 32 u(0) = 0, 3 32 u" + 12u = 0, 3u" + 12u: = 0, 3u" + 12u: 0, 3u" + 12u: = 0, u(0) = 0, u(0) = 0, u(0) = 2, u' (0) = 2 u'(0) =-2 u'(0) = 2 u' (0) u'(0) : = 0 =
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