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A spring with a 2-kg mass and a damping constant 2 can be held stretched 2.5 meters beyond its natural length by a force of 7.5 newtons. Suppose the spring is stretched 5 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c² - 4mk? -20 m²kg²/sec² Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form c₁eat cos(pt) + c₂e* sin(8t) α = -1/2 B = Y = 8 = sqrt5 C1 = 5 C2 = sqrt5/2 sqrt5 -1/2

A spring with a 2-kg mass and a damping constant 2 can be held stretched 2.5 meters beyond its natural length by a force of 7.5 newtons. Suppose the spring is stretched 5 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c² - 4mk? -20 m²kg²/sec² Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form c₁eat cos(pt) + c₂e* sin(8t) α = -1/2 B = Y = 8 = sqrt5 C1 = 5 C2 = sqrt5/2 sqrt5 -1/2

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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A spring with a 2-kg mass and a damping constant 2 can be held stretched 2.5 meters beyond its natural length by a force of 7.5 newtons. Suppose the spring is stretched 5 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c² - 4mk? -20 m²kg²/sec² Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form C₁eat cos(pt) + c₂ert sin(8t) α= -1/2 B = sqrt5 Y = -1/2 8 = sqrt5 C1 = 5 C₂ = sqrt5/2
Transcribed Image Text:A spring with a 2-kg mass and a damping constant 2 can be held stretched 2.5 meters beyond its natural length by a force of 7.5 newtons. Suppose the spring is stretched 5 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c² - 4mk? -20 m²kg²/sec² Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form C₁eat cos(pt) + c₂ert sin(8t) α= -1/2 B = sqrt5 Y = -1/2 8 = sqrt5 C1 = 5 C₂ = sqrt5/2
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