What Went Wrong in My SHM Proof Using Calculus?

[Silimay]

In a block-spring system with a block of mass m and a spring of spring constant k, prove that the angular velocity ω of the block = (k/m)^(1/2).

I can prove this easily in the following manner:

F = ma Newton's law
F = -kx Hooke's law
a = -ω^2x
ma = -kx = -mω^2x
k = mω^2
ω^2 = (k/m)
ω = (k/m)^(1/2)

But when I try to prove it using calculus (as my teacher instructed me to do) something goes wrong:

F = ma
F = -kx
X = Acos(ωt + φ) SHM
F = ma = -kx = -kAcos(ωt + φ)

I integrated to get:

mv = -kωAsin(ωt + φ)

Did I do something wrong here? I kept integrating (so that there was a ω^2 term on the right side) and substitued for x = Acos(ωt + φ) and canceled out x; but then I got ω = (m/k)^1/2. I don't understand why---is there a flaw in the math somewhere? I think I can probably do the proof by simply integrating the SHM equation and substituting it for acceleration, but I'd like to know what I did wrong above.

Thanks for any help :-)

 

[Justin Lazear]

You messed up your integral. What you've found is the derivative without the negative sign.

What's easier is to realize that a = $\Ddot{x}$, and then just differentiate x twice with respect to time and plug it in for a, then compare terms.

--J

 

[Silimay]

Thanks! I didn't realize I was integrating with the chain rule :-) I have to do integration by substitution.