Mass between two springs: what happens when one spring is removed?

[MeMoses]

sqrt(3*d**2 + d/2*sqrt(3))

 

[voko]

I don't understand how you got sqrt(3) inside sqrt().

By the way, I think you made a mistake early on when you got x = A sin 2wt + B cos 2wt. How did you get the 2 there?

 

[MeMoses]

damn it should be sqrt(2) not 2 in there, and before that mistake I had A=d*sqrt(3) and B=d/2, A^2 = 3*d^2 and B^2 d/2*sqrt(3)

 

[MeMoses]

And fixing that mistake I get A in the 1 spring to equal d/2*sqrt(6). I then get amplitude = sqrt(6*d**2 / 4 + d/2*sqrt(3))

 

[voko]

So the final result is?

 

[MeMoses]

sqrt(6*d**2 / 4 + d/2*sqrt(3))

 

[voko]

One more time: how do you keep getting sqrt(3) inside sqrt() for amplitude? What is B for the 1-spring system?

 

[MeMoses]

B is d/2

 

[voko]

Then how do you get this: sqrt(6*d**2 / 4 + d/2*sqrt(3))?

 

[MeMoses]

I am an idiot. So for amplitude I get d/2*sqrt(7). I don't know why I added a sqrt(3) as it came from nowhere so ignore that and you should get this.

 

[voko]

I think that concludes this exercise. What you should really try to remember is how you can always convert $A \sin \omega t + B \cos \omega t$ into $C \sin (\omega t + \alpha)$. In many cases, you can directly solve $x'' + \omega^2x = 0$ as $C \sin (\omega t + \alpha)$ and determine $C$ and $\alpha$ from initial conditions. Also, the relationship between $A$, $B$ and $C$ is also a useful one (as you have surely noticed).