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Differential equation: Spring/Mass system of driven motion with damping

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TeenieBopper
#1
Aug7-11, 02:37 PM
P: 29
1. The problem statement, all variables and given/known data
A 32 pound weight stretches a spring 2 feet. The mass is then released from an initial position of 1 foot below the equilibrium position. The surrounding medium offers a damping force of 8 times the instantaneous velocity. Find the equation of motion if the mass is driven by an external force of 2cos(5t).


2. Relevant equations
F=kx
m=W/g

m[itex]\frac{d^{2}x}{dt^{2}}[/itex]+[itex]\beta[/itex][itex]\frac{dx}{dt}[/itex]+kx=f(t)


3. The attempt at a solution

I found that k=16[itex]\frac{lb}{ft}[/itex] and m=1 slug. This gets me the following equation:

[itex]\frac{d^{2}x}{dt^{2}}[/itex]+[itex]\beta[/itex][itex]\frac{dx}{dt}[/itex]+16x=2cos(5t)

I'm at a loss for how to determine [itex]\beta[/itex], which is the damping force of 8 times the instantaneous velocity. I don't know how to determine instantaneous velocity. I know that once I have [itex]\beta[/itex], I can just use a LaPlace transform to find x(t). But [itex]\beta[/itex] is my stumbling block right now.

As I was writing this, it occured to me that [itex]\frac{dx}{dt}[/itex]=instantaneous velocity and that would make [itex]\beta[/itex]=8. That in turn makes the problem very easy to solve. Am I correct in this thinking?

We kind of rushed through this application in class the other day. Thanks in advance for any help you're able to provide.
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rock.freak667
#2
Aug7-11, 04:08 PM
HW Helper
P: 6,202
Quote Quote by TeenieBopper View Post

As I was writing this, it occured to me that [itex]\frac{dx}{dt}[/itex]=instantaneous velocity and that would make [itex]\beta[/itex]=8. That in turn makes the problem very easy to solve. Am I correct in this thinking?.
Yeah I would agree that β=8 here. Just make sure your initial conditions are correct with the proper signs.


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