6. Finding the Differential Equation An 8-lb weight is attached to a frictionless spring, which in turn is suspended from the ceiling. The weight stretches the spring 6 inches and comes to rest in its equilibrium position. The weight is then pulled down an additional 3 inches and released with a downward velocity of 1 ft/sec. (a)Find the initial-value problem that describes the motion of the weight. (b)Find the motion of the weight. (c)Find the amplitude, phase angle, frequency, and period of the motion. 6. (а) й + 64u 0 и(0) %3D V5 (2V5 V5 sin 81) = 0.28 cos (8t - 0.46) 5 (b) и cos 81 + %3D Wo 4 (c) R = 0.28 ft, 8 = 0.46, f 2т oscillations/sec, T %3D %3D sec т 4

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The problem is attached in the image below and is regarding differential equations of the second order. Also attached is the answers the textbook came to. I'm just having a difficult time actually getting started and carrying through the problem. 

6. Finding the Differential Equation An 8-lb weight is attached to a frictionless spring, which in
turn is suspended from the ceiling. The weight stretches the spring 6 inches and comes to rest in
its equilibrium position. The weight is then pulled down an additional 3 inches and released with
a downward velocity of 1 ft/sec.
(a)Find the initial-value problem that describes the motion of the weight.
(b)Find the motion of the weight.
(c)Find the amplitude, phase angle, frequency, and period of the motion.
Transcribed Image Text:6. Finding the Differential Equation An 8-lb weight is attached to a frictionless spring, which in turn is suspended from the ceiling. The weight stretches the spring 6 inches and comes to rest in its equilibrium position. The weight is then pulled down an additional 3 inches and released with a downward velocity of 1 ft/sec. (a)Find the initial-value problem that describes the motion of the weight. (b)Find the motion of the weight. (c)Find the amplitude, phase angle, frequency, and period of the motion.
6. (а) й + 64u
0 и(0)
%3D
V5 (2V5
V5
sin 81) = 0.28 cos (8t - 0.46)
5
(b) и
cos 81 +
%3D
Wo
4
(c) R = 0.28 ft, 8 = 0.46, f
2т
oscillations/sec, T
%3D
%3D
sec
т
4
Transcribed Image Text:6. (а) й + 64u 0 и(0) %3D V5 (2V5 V5 sin 81) = 0.28 cos (8t - 0.46) 5 (b) и cos 81 + %3D Wo 4 (c) R = 0.28 ft, 8 = 0.46, f 2т oscillations/sec, T %3D %3D sec т 4
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