Yes, I got it now.
The series one adds as in series therefore 1/stotal=1/s1+1/s2 and the other one like parallel therefore stotal=s1+s2.
Wouldn't have thought this because experience actually told me that it shouldn't be so much of a difference... I mean ii) has 4 times more total energy than i).
But now I'm stuck at a new problem.
A damped oscillator of mass m=1,6 kg and spring constant s=20N/m
has a damped frequency omega' htat is 99% of the undamped frequency omega.
a) What is the damping constant r?
Attempt:
As far as I know \omega ' = \sqrt{ \omega ^2 - \frac{b}{2m}}
But were to go from here?
edith: Ok, got that now.
therefore b = (0.99^2-1)*(s/m)*(2m)*(-1) which in this case is 0.796.
b)What is the Q of the system?
Attempt:
Q=\sqrt{mass * Spring constant} / r
But again I would have to find r from a) which I can't really figure out.
Ok, with a) answered I got Q to beeing 7.1066
c) Confirm that the system is lightly damped.
I think a system is lightly damped if omega' is about euqal to omega, but this can't really be the answer here. Because that's what's stated anyways since 99% is "about equal" isn't it?
d) What new damping constant r_{new} is required to make the system critically damped?
Again, I couldn't find any definition on the web or in my notes what critically damped means.
f)Using r_{new} calculate the displacement at t=1s given that the displacement is zero and the velocity is 5 ms^-1 at t=0.
No clue here.
Note

ow can I insert Tex in this forum?