Assume that there is no damping outside and take as the value for the acceleration of the gravity g = 9.8 m/s². Also, if x(t) represents the displacement of the mass, in meters, from equilibrium position with respect to time t, in seconds, then the function x(t) satisfies the following differential equation: Az" (t) + 6x(t) = Ct + D + (−3zt + E) u (t − F), Where is the unit step function (Heaviside function). Question: Determine the value of E and F.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Assume that there is no damping outside and take as the value for the acceleration of the
gravity g = 9.8 m/s². Also, if x(t) represents the displacement of the mass, in meters, from
equilibrium position with respect to time t, in seconds, then the function x(t) satisfies the
following differential equation:
Ax" (t) + 6x(t) = Ct + D +
- (- / + + + E) a
- E) U (t – F),
Where is the unit step function (Heaviside function).
Question: Determine the value of E and F.
Transcribed Image Text:Assume that there is no damping outside and take as the value for the acceleration of the gravity g = 9.8 m/s². Also, if x(t) represents the displacement of the mass, in meters, from equilibrium position with respect to time t, in seconds, then the function x(t) satisfies the following differential equation: Ax" (t) + 6x(t) = Ct + D + - (- / + + + E) a - E) U (t – F), Where is the unit step function (Heaviside function). Question: Determine the value of E and F.
8. Consider a vertical spring, with a fixed end, of constant 6 N/m, to which is attached a mass
of p kilograms, which stretches the spring 39.2 meters, to the equilibrium position. Being the mass
in equilibrium and at rest, from t = 0 an external force f(t) is activated, whose graph is shown
next:
f(t)
6
4
3
Transcribed Image Text:8. Consider a vertical spring, with a fixed end, of constant 6 N/m, to which is attached a mass of p kilograms, which stretches the spring 39.2 meters, to the equilibrium position. Being the mass in equilibrium and at rest, from t = 0 an external force f(t) is activated, whose graph is shown next: f(t) 6 4 3
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