Position of mass on spring w/o energy equations

[yosimba2000]

This is not a HW/coursework problem. Just something I randomly thought about.

1. Homework Statement
There is a horizontal spring with a mass at the end of it. The mass has an initial velocity V that compresses the spring. Find the equation of motion for this mass without using energy equations.

Homework Equations

At any displacement X, force by spring is F = -kx
Force on mass by spring = -m* (applied acceleration on mass by spring) = -ma

Equation of motions:
Vf^2 - Vi^2 = 2ax

The Attempt at a Solution

As the displacement changes, then the force and hence acceleration provided by the spring against the mass is also changing. So acceleration against mass is function of distance.

F = -kx = -ma
a = kx/m

Vf^2 - Vi^2 = 2ax <-- plug in a

Vf^2 - Vi^2 = 2 (kx/m)x

Vf^2 = 2 (kx/m)x + Vi^2

Did I do it right?

 

[Simon Bridge]

F = -kx = -ma

This is not correct.
Newton's law is $\vec F=m\vec a$, and you have $\vec F=-k\vec x$ ... so substitute the second into the first.
You will also need $\vec a = \ddot{\vec x}$

Vf^2 - Vi^2 = 2ax

This equation is only valid for a specific situation: what is that situation and is that situation the one you have in your problem?

Lastly: what does "equation of motion" mean for this problem?

 

[yosimba2000]

My equation of motion is that I want to solve for the final velocity of the mass.

Vf^2 - Vi^2 = 2ax is used in analyzing free fall, but I don't know why it couldn't be used here. In free fall, we assume constant acceleration, but couldn't this formula also be used in a situation where acceleration is not constant, like it changes as a function of time?

I don't understand why F = -kx = -ma is not true. If the spring is compressed, the spring force is pushing back the opposite way of compression, so there's a negative sign to get -kx. And the spring causes an acceleration on the mass, where the acceleration is opposite the compression, so there should be a negative sign, so the magnitude of the force is ma, but the direction is -ma?

 

[Simon Bridge]

If the equation is valid only for the situation where the acceleration is a constant, then it is NOT valid for when the acceleration is not a constant.
In this situation, the acceleration is not a constant, therefore the equation will not work. You can easily see why this must be true if you consider the equation for when the velocity is a constant compared with the situation when the velocity is not a constant... it is the same for acceleration or anything else.

F=-kx=-ma is not true because the acceleration is physically in the opposite direction to the displacement. So when displacement is positive, acceleration is negative. What you wrote gave a positive acceleration for a positive displacement. This is simply not what happens in Nature. Physics equations are required to describe what happens in Nature. Therefore it is not true.

You should be able to see this from the algebra. You know F=ma, and you know that F=-kx ... if you substitute the second one into the first one, you do NOT get $-kx=-ma$. It is not ture in Nature and it is not true mathematically. There is no way you start with F=ma and F=-kx and end up with -kx=-ma.

What "final velocity"? There is no final velocity mentioned in the problem statement. Try again.

You clearly do not know what an "equation fo motion" is... you need to review your course notes or look it up online. ie.
https://en.wikipedia.org/wiki/Equations_of_motion

 

[yosimba2000]

I did not realize the free fall equations were derived by assuming constant acceleration, but now I know.

I know that the acceleration is supposed to be negative to the displacement and mathematically what I typed doesn't match what nature does, but my mistake was that I was trying to force the acceleration to be negative so I put a negative sign in front of it.

 

[Simon Bridge]

That's right - you need $\vec a \propto -\vec x$ ... that is a positive acceleration comes from a negative displacement.
It is the relationship that is imortant.

Do you see what "equation of motion" means now?