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The block is being pulled by a constant force $F$. Now,to find the time period of the spring, the following method is used. First the equilibrium/mean position is determined and then the block is displaced slightly and then net force on the block is calculated. Here is where i am all fuzzy.

First of all, at mean position,only the net force has to be $0$ meaning $F$ must be equal to $-kx$. But according to the teachers,at this mean position now,the block is at rest and is doing no motion. I don't understand this at all. Net force being $0$ doesn't mean the block has to be at rest,it can move with constant velocity as well,it is not at rest!!! Then the teachers say as the block is now at rest,by giving some impulse or energy,the block is displaced slightly and it will start to perform oscillation about the newly obtained mean position. I need to know if the * marked lines are used are correct or not.Please enlighten me to build my concepts.

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  • $\begingroup$ How is this different than a block hanging vertically on a spring? $\endgroup$
    – David White
    Commented Dec 17, 2021 at 20:22
  • $\begingroup$ If you notice my previous post properly,there were two cases:the first case was gently lowering down where the block was actually at rest and the second was letting go suddenly. But in this case i don't know how the phenomena is occuring according to the teachers. $\endgroup$
    – madness
    Commented Dec 17, 2021 at 20:34
  • $\begingroup$ If the horizontal block is moving on a frictionless surface, it will act just like a block hanging vertically on a spring. $\endgroup$
    – David White
    Commented Dec 17, 2021 at 22:17
  • $\begingroup$ @DavidWhite yes, David and not wanting to split hairs but the EoM is slightly different: offset by the equilibrium position, in the case of a 'hanging' mass, actually. $\endgroup$
    – Gert
    Commented Dec 17, 2021 at 23:03
  • $\begingroup$ @Gert, the exact same experiment can be done if you pull a hanging mass down slightly from its equilibrium position. You would get the same sinusoidal periodic motion, with an amplitude that is equal to the displacement from the equilibrium position. $\endgroup$
    – David White
    Commented Dec 18, 2021 at 1:27

2 Answers 2

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Your teacher has stated an initial condition for the system which is that the block is at rest at what might be called the static equilibrium position.

Although in theory with no net force acting on the block and it could be moving with a constant velocity your teacher has decided to consider the block having a constant velocity of zero, ie the block is at rest.

The block at its initial condition is given is then given a kick and the subsequent motion of the block is then considered.

With the kick the system is given some extra energy and this results in the block moving away from where it started and undergoing oscillatory motion.
When the moving block passes through the static equilibrium position the net force on the block is zero and it has an instantaneous velocity and zero acceleration.
The inertia of the block moves it passed the static equilibrium position which results in there being a net force on the block which as a consequence undergoes an acceleration towards the static equilibrium point.

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  • $\begingroup$ Thanks a lot for the answer. A question:By static equilibrium,do you mean the initial state which is shown in the figure above(which is actually the natural length of spring) or the calculated equilibrium position? Also i think it would be better for me to clarify on the part that my teacher told that through the constant force,the block will keep getting displaced and at equilibrium position,it can no longer get further due to net force being $0$,thats all he said without mentioning initial static condition.Please kindly shed some light on it. $\endgroup$
    – madness
    Commented Dec 18, 2021 at 13:16
  • $\begingroup$ The position when the force has been applied and the block is still. If the block is not moving in the static equilibrium position it will stay there for ever more. If the block is kicked it will then undergo oscillatory motion and assuming that there is no friction that oscillatory motion will go on for ever. $\endgroup$
    – Farcher
    Commented Dec 18, 2021 at 13:25
  • $\begingroup$ Actually my exact question is where and how the block could be still,the figure i gave is certainly not the equilibrium position since $F$ is there and no spring force is acting,but as you said at the equilibrium poisition it becomes still,thats exactly i want to know how.Sorry for being annoying. $\endgroup$
    – madness
    Commented Dec 18, 2021 at 13:31
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I think you are getting confused a bit by $t=0$ and what happens before it.

Say you place the mass plus spring on that surface as depicted in your figure. Assume this system will not be at equilibrium but now we let it (the mass) move for a bit. 'A bit' here means that after a while, due to even very minus friction (internal to the spring, block/surface, block/air etc) it will come to a halt.

At that point no net force acts on it in the horizontal, i.e. $x$-axis. The mass will now be in a position we conventionally denote as $x=0$, aka the equilibrium position.

Assume now that at $t=0$ (the start of our experiment) we move the mass a little to the right by an amount $x_0(>0)$.

Now the spring will exert a restoring force $F$:

$$F=-kx_0$$ And on release of the mass, again at $t=0$, the Newtonian Equation of Motion will be:

$$ma=-kx$$

Solving this with $x=x_0$ at $t=0$ will give you the right equation of motion $x(t)$.

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  • $\begingroup$ Thanks for the answer,actually my main concern was if the block will come to rest at the equilibrium position as the teachers said and why that would happen as i already said net force 0 at equilibrium position could also mean the block is moving with constant velocity. $\endgroup$
    – madness
    Commented Dec 18, 2021 at 2:54

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