# Is spring's restoring force action reaction pair?

Let's imagine a spring is attached to a rigid wall.One end is free to move. Now I exert force on that end and make some elongation there. According to Newton's 3rd law, there should be a action reaction pair. So a restoring force will be occured in the spring. But when I loose that end there shouldn't be any restoring force(because I no longer exert any force) in the spring and the spring should remain in such an elongated state. But reality differs . Why?

Second question is that, if restoring force and external force are action reaction pair then they are equal in magnitude. which means that we cannot exert a constant force I mean external force can't be constant because restoring force being -kx depends on x which is a variable. But common sense says it's easily possible to exert constant force.

However I am literally stuck in these two conceptual questions and will be glad to see detailed answer.

• You exert a force on the spring. The spring exerts a force on you. When you let go, the force exerted by spring on you is zero as you would expect. The restoring force still exists and hence the springs goes back.
– sku
Mar 14 at 6:19
• @sku why does restoring force exist then?When and why is actually it produced in the spring ?
– user325381
Mar 14 at 6:26
• A spring wants to be in the relaxed state. If you compress or pull, you are creating a restoring force which is working against you.
– sku
Mar 14 at 6:40
• If it was an ideal spring, and you exclude gravity, the spring will continue to oscillate with frequency $\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ But real springs are not ideal. That is why reality differs from theory. And although the force varies with $x$, for each value of $x$, $F_{\text{exerted}}=kx$ In other words, you have to keep increasing the exerted force for increasing values of $x$ and Newton's third law always hold. Mar 14 at 6:48
• @josephh your frequency is for a massless spring with a mass m on the end. That is not what we have here. The only mass is the spring. If only it were that simple. Mar 14 at 7:22

But when I loose that end there shouldn't be any restoring force(because I no longer exert any force) in the spring and the spring should remain in such an elongated state. But reality differs . Why?

You can imagine a spring as a chain (collection) of particles. First particle in the chain pushes the second, the second pushes the third and so on.. You are pushing the last particle in the chain, and the same particle is pushing you back (action-reaction). Once you stop pushing, the last particle also stops pushing you back, but is still being pushed by other particles in the chain until spring restores to its original position.

if restoring force and external force are action reaction pair then they are equal in magnitude.

restoring force being -kx depends on x which is a variable.

Correct and correct.

But common sense says it's easily possible to exert constant force.

Of course it is possible to exert a constant force. For a spring to compress, there must be two external forces acting on the spring - one from the left and one from the right. If one of these two forces is larger than the other, the spring will compress but it will also move (accelerate) in the direction of the larger force.

If one end of the spring is attached to a fixed wall and you apply some force while spring is relaxed, the wall will (almost) immediately apply the same force to the other end. The pushback that you feel from the spring at $$\Delta x = 0$$ is not restoring force by the spring but the normal force by the wall propagated through the spring.

• Can we say this? As I exert force the 'x' gets larger . Does it mean that I am applying variable force? Or it says if I apply a particular force there will be a particular amount of elongation or say value of 'x' . Again we apply another particular force which gives another particular elongation. And doing this for different constant forces everytime (not at a time) gives some values. Now if I graph those numbers it's linear. Hence we get F=Kx. But this relationship doesn't mean that as I continue to pull, x gets larger and F becomes variable.
– user325381
Mar 14 at 8:42
• @VedflewquestaPercycloxirieta What it means is that, in equilibrium, for a certain constant force you exert on the spring it will elongate $\Delta x_0$ from its relaxed state. If you want to elongate it more than $\Delta x_0$ you need to provide (exert) more force. Mar 14 at 8:54