## Quick Conceptual Spring Question

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### Quick Conceptual Spring Question

F = -k deltax

At time = 0 "I apply an instantaneous force".

I can't grasp how you can get an F at t=0, because I'd assume deltax at t=0 is 0, correct? How does this work?
GTM

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### Re: Quick Conceptual Spring Question

Either the force suddenly begins and continues for some time, like igniting a rocket engine, or the force suddenly begins and ends in a very short time interval, like striking with a hammer.
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### Re: Quick Conceptual Spring Question

But how can there be a force if there has been no change in the length of the spring yet at t=0?
GTM

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### Re: Quick Conceptual Spring Question

The equation you've got there is only the force imparted by the spring itself. It says nothing about any other forces you might choose to apply, such as via rocket engine or hammer strike.
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gmalivuk
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### Re: Quick Conceptual Spring Question

Suppose at t=0 you give your spring an initial force, so it starts accelerating in a direction along the x axis. The restoring force of the spring is then
F=-kx.
Hooke's law doesn't give you the instantaneous force that sets your spring in motion, "you" do.
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### Re: Quick Conceptual Spring Question

That makes sense!

Thanks
GTM

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### Re: Quick Conceptual Spring Question

You need to note that the spring has a mass, otherwise your equations of motions becomes acceleration = infinite. Actually, this gets a bit complicated, since the mass is distributed. In physics classes, they simplify this by using an "ideal spring" (e.g. a spring with no mass) and attaching a mass to the end of it.

So generally, if you apply a force on the spring F, your equation of motion should look like:

a = (F-kx)/m

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