Homework Statement
Solve
(1-4x^2)y''+34x\cdot y'-70y=0
Homework Equations
Basically, I found the recurrence relationship to be:
a_{n+2}=\frac{2 (-7 + n) (-5 + 2 n)}{(n+1)(n+2)}a_n}
Now, I solve for y1 where y1 had a_0=0 and a_1 = 1. It is a simple polynomial of degree 7...
Statement
Let S be a linear operator S: U-> U on a finite dimensional vector space U.
Prove that Ker(S) = Ker(S^2) if and only if Im(S) = Im(S^2)
So, I'm really not sure about how to prove this properly. I have a few ideas, but this one seemed to make sens intuitively to me. So, I'm...
Homework Statement
A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end,on both pulling in opposite directions. The springs have constant k and at equilibrium, their pull is perpendicular to the rod. Find the frequency of small...
Fun times.
I remember doing this in CEGEP :)
What you can do is just remember that there are no horizontal acceleration, only vertical. So you can find the initial horizontal V with distance / time right?
And since you can measure your angle, you can find your actual initial velocity.
The rest...
What are the equations that you are using?
Anyways,
A safe sanity check is that for any object on Earth, if it is just falling, then the linear acceleration cannot be greater than 9.81 m/s^2
It's very simple:
The change in potential energy is equal to the kinetic energy.
eg: \delta U = K_E in this case since you're initial KE = 0.
But more generally, (if there are no friction forces)
\DELTA U = \DELTA K_E
So, in a sens, you are right. You would have to write the PE (Potential...
Cool, will check it out.
I just hate how they called it imaginary and how people truly believe that you cannot "imaginary" numbers in forced damped oscillators
:S
True, but from what I recall, for the quadratic, they always rewrote the equations such that there were only positive terms in...
Hey Guys,
I am looking for a book that talks about the history of imaginary numbers and how people came to need it.
I'm just having trouble imagining how people could have accepted imaginary numbers before negative numbers (at least in the western world).
Thank you,
Eric
P.S.
If...
But the disk is upright, not flat.
So based on that diagram,
it's z would be my y.
x is x
and why would be z.
so, 1/2+1/4 = 3/4.
Hm,
but it does seem to give what I want if I do Iy=Iz-Ix.
But I highly doubt that it would be something I could do. :S
But at least if I just took Iy to begin...
Actually, it is not making sens to me.
There must be something wrong I am doing.
Frrom the perpendicular axis theorem.
Iz_o =Ix+IyWe have
Ix= 1/2 mb^2.
Iy = 1/4 mb^2.
so, Iz _o = 3/4 mb^2.
Now, using the parallel axis theorem,
Iz=I_zo + mr^2
Iz = 3/4 mb^2 + mr^2.
Now I was interested at...