Hooke's Law and Restoring/Applied Force

[Frawx]

According to Hooke's Law, F=-kx where F is the restoring force, k is the spring constant and x is the length of extension/compression.

When an applied force compresses a spring, a restoring force will act in the opposite direction.
When a spring is compressed and is in equilibrium (not extending/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending/compressing)

However, the act of compressing a spring will require the applied force to be greater than the restoring force even at an infinitesimally small time interval, am I wrong?

Is it more correct to say that F=kx where F is the applied force or F=-kx where F is the restoring force? I think that F=-kx is the better option, but I'm pretty confused here. Does it matter at all?

 

[Dewgale]

Well, the $F$ in $F = -kx$ is the force that causes the spring to move, the restoring force. So no, $F_{app} \neq kx$ except for at the exact situation where the spring is extended or compressed, and being held in that position by the applied force -- but there's nothing fundamental about that equation, it's just due to the fact that the forces must be equal and opposite.

So, if you assume that the applied force is constant, then the restoring force will increase as the string is stretched/compressed, until it is equal and opposite to the constant applied force. That is the point where your system will be in mechanical equilibrium.

 

[Ibix]

That is the point where your system will be in mechanical equilibrium.

To be clear, this is where the forces are in equilibrium. In the circumstances you describe the end of the spring will have a non-zero velocity, so the system will overshoot and simple harmonic motion will happen. You need damping or a changing force in order to reach rest at that point.

 

[Dewgale]

To be clear, this is where the forces are in equilibrium. In the circumstances you describe the end of the spring will have a non-zero velocity, so the system will overshoot and simple harmonic motion will happen. You need damping or a changing force in order to reach rest at that point.

That's true. If you consider it to be constant right up until that point, and then briefly change the force to stop the spring in place, then you'll be in equilibrium.

 

[CWatters]

the act of compressing a spring will require the applied force to be greater than the restoring force even at an infinitesimally small time interval, am I wrong?

Is it more correct to say that F=kx where F is the applied force or F=-kx where F is the restoring force? I think that F=-kx is the better option, but I'm pretty confused here. Does it matter at all?

The F in F=-kx is the force created by the spring. The clue is in the direction. The spring force acts in the opposite direction to the displacement. The force required to compress the spring has the same direction as x so that would be F=+kx.

The equation ignores any force needed to accelerate the mass of the spring. Eg you can only apply it to massless springs or when the inertia of the spring can be ignored, perhaps because the acceleration is low.

So I would say you are correct. If you are going to compress a spring that is initially at rest the applied force must be (at least briefly) greater than kx in order to accelerate the spring itself. While compressing the spring with constant velocity the force will be equal to kx. While decelerating to rest the applied force must be (at least briefly) less than kx.

Come to think of it I've never actually tried to calculate the total force when the mass or inertia of the spring can't be ignored. You can't just write F=-(kx + ma) because the mass m of the spring is distributed and its not all accelerating at the same rate.

 

[Frawx]

Thanks for all the responses! :)
I guess I will sum things up to be sure.
For F=kx where F is applied force,
When the spring accelerates initially, F>kx (restoring force < applied force)
When the spring decelerates finally, F<kx (restoring force > applied force)
When the spring is at equilibrium/moving at constant velocity, F=kx (restoring force = applied force)
And the thing applies to F=-kx where F is the restoring force. (except you reverse the inequalities)

 

[Chestermiller]

Thanks for all the responses! :)
I guess I will sum things up to be sure.
For F=kx where F is applied force,
When the spring accelerates initially, F>kx (restoring force < applied force)
When the spring decelerates finally, F<kx (restoring force > applied force)
When the spring is at equilibrium/moving at constant velocity, F=kx (restoring force = applied force)
And the thing applies to F=-kx where F is the restoring force. (except you reverse the inequalities)

If the spring is assumed to have zero mass (i.e., negligible mass), it is referred to as an ideal spring. In such a situation, the force in the spring responds instantaneously to the displacement, and F is always equal to -kx. For a spring of finite mass, you can do a force balance on a differential section of the spring. This leads to a wave equation (partial differential equation) for the displacement as a function of position along the spring and time.