In summary: So the amplitude is unchanged, but the x(t) changes because the mass is moving.In summary, at equilibrium, the mass has an amplitude of d. When a spring is removed, the mass moves to the right and the amplitude changes to Asin(wt) + Bsin(wt).
One more time: how do you keep getting sqrt(3) inside sqrt() for amplitude? What is B for the 1-spring system?
#38
MeMoses
129
0
B is d/2
#39
voko
6,054
391
Then how do you get this: sqrt(6*d**2 / 4 + d/2*sqrt(3))?
#40
MeMoses
129
0
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.
#41
voko
6,054
391
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).