2nd order ODE: modeling a spring

[ReidMerrill]

Suppose a spring with spring constant 6N/m is horizontal and has one end attached to the wall and the other end attached to a 3 kg mass. Suppose the friction/damping constant is 1 N s/m

Set up a differential equation that describes this system with x denoting displacement of the mass from equilibrium position and give your answer in terms of x,x', and x''. Assume that positive displacement means the mass is farther from the wall than equalibrium

I'm not sure what to do here. I'm guessing I'll need the equation

mx'' + bx' + kx = F Where mx'' = Fdamping + F spring + F inertia b is damping coefficient: 1 N m/s k is spring coefficient: 6 N/m

Where do I get m? Fdamping is 1 and Fspring is 6 right? Would I get Finertia from F=ma? Because a is not given.

 

[BvU]

Hi,

You already have the acceleration a when you write $\ddot x$.

If there is no spring, your equation of motion reads $F = ma = m\ddot x$. In that case, what is F ?

Are there other (external) forces at work, except for the spring and the damping force ?

 

[Dr.D]

Why do you think you need a "Finertia" term? Newton's Second Law only involves the sum of real forces; no fictional forces are to be included. Draw the FBD and include in your force sum on the actual forces present.

 

[ReidMerrill]

Hi,

You already have the acceleration a when you write $\ddot x$.

If there is no spring, your equation of motion reads $F = ma = m\ddot x$. In that case, what is F ?

Are there other (external) forces at work, except for the spring and the damping force ?

I don't think so.

 

[Dr.D]

You asked "where do I get m?" but the problem statement said that one end of the spring was attached to a 3 kg mass. That is where you get m.

 

[BvU]

You missed the 'in that case, what is F?' . I was fishing for the answer: 'the externally applied force'

I don't think so.

So that means in your differential equation the right-hand side can be set to 0 and $m\ddot x =$ the two forces in the story ($\beta\dot x$ and $kx$) , each with the proper sign

 

[ReidMerrill]

You missed the 'in that case, what is F?' . I was fishing for the answer: 'the externally applied force'
So that means in your differential equation the right-hand side can be set to 0 and $m\ddot x =$ the two forces in the story ($\beta\dot x$ and $kx$) , each with the proper sign

So the equation would be 3x'' + 1x' + 6x =0 ?

 

[BvU]

Only if you willingly omit the dimensions -- generally not a wise thing to do . Any doubt remaining ?