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JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
@Outkast3r09
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
shouldn't F=kx?
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
@CliffSedge ?
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
Ahh this my physics teacher went over
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
cool! Now you can explain it to me :)
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
the negative explains how that force is acting upon the spring
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
it depends on the what you're trying to find
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
if you're trying to find the force the spring is exerting on the mass or the mass exerting on the spring
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Hmmm let's seedw:1353379482698:dw
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
dw:1353379572781:dw
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
well I guess we're stretching to a certain length, so the first figure?
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
I mean the first figure you drew
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
the figures are the same
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
it depends on whether you're trying to find the length of the spring or the position of the mass
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Uhm I think we want my figure because In your figure we have to consider the gravitational force and the restoring force. I don't think they want to consider the gravitational force
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
if it asked abouthe spring itself it'd be negative
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
there is two ways this question can be answered 1) How long did the spring stretch from initial 2) What is the position of the mass after initial. They're the same thing
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
so since it's asking for position of the mass it's thisdw:1353379915405:dw
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
why is it positive? because it is going in the direction that we designated as positive?
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
technically this question can have 2 answers + answer and a  answer depending on how you set up your axis
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
OH I see! dw:1353380068540:dw Like so?
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Now I can say that F=kx
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
right?
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
no the other way around, I should probably make the restoring force positive...
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
dw:1353380229463:dw
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
yeah the picture they used was probably a free falling mass
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
and saying that the sum of the force of gravity + any other force = F
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Yeah that makes more sense. so they probably had up as positive and down as negative.
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
yes but it really just depends on what you define it... if your teacher marks you wrong and your diagram says otherwise just clear it up with him
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
so why is k(0.2) why 0.2? How did they come up with x=0.2?
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
never mind
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
that's the displacement haha
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
yes the displacement will still be posive
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
hmm interesting: (this the DE part of my calc II book) \[2\frac{d^2x}{dt^2}+128x=0 \text{ has the solution }x(t)=c_1cos8t+c_2sin8t \] yeah that's easier that doing it manually....
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
It's convention to designate potential energy as negative. The reaction force of the spring is opposite the positive applied force.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
More clearly: The negative sign of F=kx means that the spring force is opposite to the applied force on the spring.
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
potential energy as positive...the restorative force?
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Yes that makes sense
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
I had to reread the last sentence you wrote several times, but it makes sense now
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
but where did they get the 8 in sin and cos 8t
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
As others mentioned above, it's a sign convention. You define what direction you want to be positive and anything opposite to that is negative. The norm, though is to treat spring force as negative/opposite to any force acting *on* the spring.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
The 8 is based on the period of the spring  that is determined by the spring constant, k, and the attached mass.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
\(\large T=2\pi \sqrt{\frac{m}{k}}\) T^1 is the frequency of oscillation.
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
yep, sounds good. I googled it LOL \[T=2\pi \sqrt{\frac mk}\]
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
The "8" in the cosine function usually goes by the name, omega, \(\large \omega\). Might want to look that up too.
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
so the formula is \[x(t)=c_1cosTt+c_2sinTt\]
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
so T is \[\omega\]
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
\(\large . . . cos(\omega t) . . \) No, ω is related to T.
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Oh \[\omega=\sqrt{\frac km}\]
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
\[x(t)=c_1cos\omega t+c_2sin\omega t\]
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
Yes, you can either derive that using the Diff.Eq. or look it up on a physics formula sheet. (Depends on if you are more mathematician or engineer ;) )
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
leaning more towards engineering....physics formula sheet :P
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
omega doesn't seem to have units
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
That would be my choice too, but know that the way the physicists got those formulas in the first place was to solve the DEs (or rather to get their math dept. grad students and TAs to do the solving for them..)
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
omega should have units of s^1, it's a frequency.
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
Or .. is at least based on a frequency, something in the derivation might make the units cancel, but I'm pretty sure it's just the reciprocal of the time period, T.
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
radians per second
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
according to my friend wiki
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
so s^1 should be sufficient, i can leave the radians part out in my notation?
 2 years ago

CliffSedge Group TitleBest ResponseYou've already chosen the best response.0
Yes, because radians are dimensionless (it's derived from length ÷ length), that's what the 2π conversion factor is in there for.
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Oh I see
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
why is c_2 zero? shouldn't it be ...so when we say\[ x(0)=8c_2 cos8t\] does that mean \[0=8c_2 cos8t=8c_2cos(0)=8c_2(1)=c_2\] yep that seems right LOL
 2 years ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.1
Thanks everyone!!!!
 2 years ago
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