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 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 variablet with the general form C₁eat cos(ßt) + c₂en¹ sin(t)
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 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 variablet with the general form C₁eat cos(ßt) + c₂en¹ sin(t)
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 of 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 \( c^2 - 4mk \)?
\[
c^2 - 4mk = -20
\]
\[
m\frac{{d^2x}}{{dt^2}} + cx + kx = 0
\]
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:
\[
c_1 e^{\alpha t} \cos(\beta t) + c_2 e^{\gamma t} \sin(\delta t)
\]
- \(\alpha = -1/2\)
- \(\beta = \sqrt{5}\)
- \(\gamma = -1/2\)
- \(\delta = \sqrt{5}\)
\[
c_1 = 5
\]
\[
c_2 = \sqrt{5}/2
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1f1c68a1-c113-41cc-b0cf-42f2666e5687%2F80cca9f9-4a70-4b03-aece-f8154beaa3b3%2Fhur8gy_processed.png&w=3840&q=75)
Transcribed Image Text:A spring with a 2-kg mass and a damping constant of 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 \( c^2 - 4mk \)?
\[
c^2 - 4mk = -20
\]
\[
m\frac{{d^2x}}{{dt^2}} + cx + kx = 0
\]
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:
\[
c_1 e^{\alpha t} \cos(\beta t) + c_2 e^{\gamma t} \sin(\delta t)
\]
- \(\alpha = -1/2\)
- \(\beta = \sqrt{5}\)
- \(\gamma = -1/2\)
- \(\delta = \sqrt{5}\)
\[
c_1 = 5
\]
\[
c_2 = \sqrt{5}/2
\]
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