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When a mass of 2 kilograms is attached to a spring whose constant is 32 N/m, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to f(t) = 102e-2t cos(4t) is applied to the sy Find the equation of motion in the absence of damping. x(t) = cos (41) + 15 sin (41) + -21 (3 cos (4t) - 3 sin (47)) m

When a mass of 2 kilograms is attached to a spring whose constant is 32 N/m, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to f(t) = 102e-2t cos(4t) is applied to the sy Find the equation of motion in the absence of damping. x(t) = cos (41) + 15 sin (41) + -21 (3 cos (4t) - 3 sin (47)) m

Advanced Engineering Mathematics
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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When a mass of 2 kilograms is attached to a spring whose constant is 32 N/m, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to f(t) = 102e-2t cos(4t) is applied to the system. Find the equation of motion in the absence of damping. 15 x(t) = - cos (41) + ¹5 sin (41) + -2t e - (3 cos ( 4t) — 3 sin ( 4t)) 2 X m
Transcribed Image Text:When a mass of 2 kilograms is attached to a spring whose constant is 32 N/m, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to f(t) = 102e-2t cos(4t) is applied to the system. Find the equation of motion in the absence of damping. 15 x(t) = - cos (41) + ¹5 sin (41) + -2t e - (3 cos ( 4t) — 3 sin ( 4t)) 2 X m
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