Simple harmonic motion problem

  • Thread starter Silimay
  • Start date
  • #1
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 cancelled 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 :-)
 

Answers and Replies

  • #2
You messed up your integral. What you've found is the derivative without the negative sign.

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

--J
 
  • #3
Silimay
Thanks! I didn't realize I was integrating with the chain rule :-) I have to do integration by substitution.
 

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