Why is kx=25N instead of -kx=25N?

- anonymous

Why is kx=25N instead of -kx=25N?

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- anonymous

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- anonymous

@Outkast3r09

- anonymous

shouldn't F=-kx?

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## More answers

- anonymous

@CliffSedge ?

- anonymous

Ahh this my physics teacher went over

- anonymous

cool! Now you can explain it to me :)

- anonymous

the negative explains how that force is acting upon the spring

- anonymous

it depends on the what you're trying to find

- anonymous

if you're trying to find the force the spring is exerting on the mass or the mass exerting on the spring

- anonymous

Hmmm let's see|dw:1353379482698:dw|

- anonymous

|dw:1353379572781:dw|

- anonymous

well I guess we're stretching to a certain length, so the first figure?

- anonymous

I mean the first figure you drew

- anonymous

the figures are the same

- anonymous

it depends on whether you're trying to find the length of the spring or the position of the mass

- anonymous

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

- anonymous

if it asked abouthe spring itself it'd be negative

- anonymous

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

- anonymous

so since it's asking for position of the mass it's this|dw:1353379915405:dw|

- anonymous

why is it positive? because it is going in the direction that we designated as positive?

- anonymous

yes

- anonymous

technically this question can have 2 answers
+ answer and a - answer depending on how you set up your axis

- anonymous

OH I see!
|dw:1353380068540:dw|
Like so?

- anonymous

Now I can say that F=kx

- anonymous

right?

- anonymous

no the other way around, I should probably make the restoring force positive...

- anonymous

|dw:1353380229463:dw|

- anonymous

yeah the picture they used was probably a free falling mass

- anonymous

and saying that the sum of the force of gravity + any other force = F

- anonymous

Yeah that makes more sense. so they probably had up as positive and down as negative.

- anonymous

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

- anonymous

yep

- anonymous

so why is k(0.2) why 0.2? How did they come up with x=0.2?

- anonymous

never mind

- anonymous

LOL

- anonymous

that's the displacement haha

- anonymous

yes the displacement will still be posive

- anonymous

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....

- anonymous

It's convention to designate potential energy as negative. The reaction force of the spring is opposite the positive applied force.

- anonymous

More clearly: The negative sign of F=-kx means that the spring force is opposite to the applied force on the spring.

- anonymous

potential energy as positive...the restorative force?

- anonymous

Yes that makes sense

- anonymous

I had to reread the last sentence you wrote several times, but it makes sense now

- anonymous

but where did they get the 8 in sin and cos 8t

- anonymous

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.

- anonymous

The 8 is based on the period of the spring - that is determined by the spring constant, k, and the attached mass.

- anonymous

\(\large T=2\pi \sqrt{\frac{m}{k}}\)
T^-1 is the frequency of oscillation.

- anonymous

yep, sounds good. I googled it LOL \[T=2\pi \sqrt{\frac mk}\]

- anonymous

The "8" in the cosine function usually goes by the name, omega, \(\large \omega\). Might want to look that up too.

- anonymous

so the formula is \[x(t)=c_1cosTt+c_2sinTt\]

- anonymous

so T is \[\omega\]

- anonymous

\(\large . . . cos(\omega t) . . \)
No, ω is related to T.

- anonymous

Oh \[\omega=\sqrt{\frac km}\]

- anonymous

Right!

- anonymous

\[x(t)=c_1cos\omega t+c_2sin\omega t\]

- anonymous

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 ;-) )

- anonymous

leaning more towards engineering....physics formula sheet :P

- anonymous

omega doesn't seem to have units

- anonymous

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..)

- anonymous

omega should have units of s^-1, it's a frequency.

- anonymous

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.

- anonymous

radians per second

- anonymous

according to my friend wiki

- anonymous

so s^-1 should be sufficient, i can leave the radians part out in my notation?

- anonymous

Yes, because radians are dimensionless (it's derived from length ÷ length), that's what the 2π conversion factor is in there for.

- anonymous

Oh I see

- anonymous

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

- anonymous

Thanks everyone!!!!

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