Suddenly changing k in a 2 spring system

[MeMoses]

Homework Statement

Two springs each have a spring constant k and equilibrium length l. They are both stretched a distance l and attached to a mass m and two walls. At a given instant the right spring suddenly changes its spring constant to 3k. What is the resulting x(t)? Take initial position to be x=0
Basically 2 springs stretched between 2 walls with mass m in the middle and a spring constant suddenly changes to 3k.

Homework Equations

I imagine I just have to get the equation for force and solve the resulting differential equation.

The Attempt at a Solution

I started out drawing a diagram and each spring is exerting a force of k*l in opposite directions. When the spring constant changes the net force is equal to 2k*l towards the right. I can't seem to get F in terms of x though. Any help is appreciated.

 

[MeMoses]

Okay I simplified the equation to F(x) = 2k(l*x - 2x) = mx'' Can some one show me how guessing that x = e**(alpha*t) works here. I tried seperating the variables which works but is overly complicated and i know there is a simpler approach

 

[nasu]

The elastic force depends on the deformation of the spring and not on its equilibrium length. Your expression for force is not correct.
Try to draw a diagram and identify by how much is each spring stretched or compressed when the ball is displaced by "x" from the original position.

 

[MeMoses]

I have the force at x=0 is 2kl If e1 and e2 are the extensions of the springs then -k*e1 + 3k*e2 = 2kl.
And my force = -ke1 - kx + 3ke2 -3kx which simplifies to 2k(l-2x), i accidently added an x after the l which might have added a problem. I i don't know if its clear but at the start the mass is 2*l from each wall since each spring is stretched an additional l.

 

[voko]

From the left spring's perspective, what is its extension from the unstretched state at certain x? When x = 0, it is l. What is it when x < 0, x > 0? How about x = -l, or +l? What is the general expression for it? Once you know the general expression for the extension, plug it into Hook's law, but mind the signs. Then do the same for the right spring. The net force is the sum of the two forces.

 

[MeMoses]

So the force from the left spring is -k(l+x) and the right is 3k(l-x) and then add these, correct? I assume it is because this is the same as I got before.

 

[voko]

That seems good so far.

 

[nasu]

The elastic force depends on the deformation of the spring and not on its equilibrium length. Your expression for force is not correct.
Try to draw a diagram and identify by how much is each spring stretched or compressed when the ball is displaced by "x" from the original position.

Newer mind. I did not see that l is both the natural length and the the amount of stretch.

 

[MeMoses]

Thanks i figured out using the guess that x=e^(alpha*t) and I am fairly certain I solved it correctly. I got x(t) = l/2 + sin(2wt) + cos(2wt) with w = sqrt(k/m). Thanks for the help

 

[voko]

This does not look right. What is the general solution to the equation?

 

[MeMoses]

I'm sorry I am not the best with terminology but isn't x(t) = l/2 + sin(2wt) + cos(2wt) the general solution? Also I double checked the solution to my differential equation on wolframalpha which has the same answer (I know its probably not the best way to verify but its quick and easy).

 

[voko]

The general solution to a second order differential equation must have two indefinite constants.

Your solution does not satisfy the condition that at t = 0, x = 0 and v = x' = 0.

 

[MeMoses]

Oh that's my bad I cannot copy my own notes. Its Asin + Bsin

 

[voko]

You need to determine A and B from the initial conditions.

 

[MeMoses]

How can I do that with the given information?

 

[voko]

See #12. Can you explain why these values?

 

[MeMoses]

oh ok I completely missed that. Yea they make sense, so A = 0(from x'(0)) and B = -l/2(from x(0))

 

[voko]

Very well.