Acceleration as a function of position

[fog37]

Hello Forum,

If a force is a function of position x only, like Hooks's spring force F=-kx, the acceleration is also only dependent on the position x, i.e. a(x).
At every instant of time t, the object has a specific velocity v(t), position x(t) and acceleration a(t), which all depend on the time variable t.

How do we reconcile the fact that the acceleration should be only dependent on x, i.e. a(x), when the acceleration becomes dependent of time, i.e. a(t), for the objects motion?

Clearly, there is a relation between x and t: x(t) or even t(x). So is the position dependence of the acceleration a, from Hooks's spring force F=-kx, a "primary" dependence while the time dependence is "secondary"?

Thanks!
Fog37

 

[Orodruin]

Since acceleration is the second derivative of the position with respect to time, what you get is a second order differential equation.

 

[fog37]

Thanks Orodruin.

But my dilemma is that the force provides an acceleration that seems to depend only on position: a(x).

However, when we consider the object's motion, we discover that the object's acceleration can be expressed as a function of time t, i.e. a(t), since a( t(x) ) through parametrization. So does the acceleration a truly depend on x or on t?

 

[Ibix]

Which way is more convenient for you personally to think about it in the problem you have in front of you? That's the one it truly is.

 

[fog37]

Well,
a force that depends on position gives an acceleration that depends on position and a force that depends on time gives an acceleration that depends on time.

I think that the two motions would be very different since the motion differential equations would be different. Or not?

 

[Orodruin]

It is a function of time since it is a function of position and position is a function of time. However, it only depends on time through the dependence on x, not explicitly on t as an independent variable.

 

[fog37]

Thanks Orodruin.

However, could we not see it the other way: the acceleration depends on x through the dependence on t, not explicitly on x as an independent variable? What does it really mean explicitly or implicitly in this context?

 

[Orodruin]

That only works properly as long as time is in a one-to-one correspondence with position. For the harmonic oscillator you described, it is not (except for at most half a period).

 

[A.T.]

So does the acceleration a truly depend on x or on t?

What does "truly" mean here, and how is the distinction to "not truly" relevant?

 

[Dale]

But my dilemma is that the force provides an acceleration that seems to depend only on position: a(x).

a(x) = a(x(t)) = a(t)
It is all the same thing

Edit: perhaps more importantly $a=\frac{d^2}{dt^2}x$

 

[fog37]

Thanks everyone. I may be making a big deal about nothing.
I know that a general force may be of the form F(x,v,t) and there are forces F(t) described as time-dependent and not space dependent while other forces are exclusively space dependent, like F= - kx or the gravitational force. They seem to specify if the force depends on one variable or the other so it is not all the same. What determines that choice? why do we say that Hookes law is a function of position and not a function of time?

 

[A.T.]

why do we say that Hookes law is a function of position and not a function of time?

Because that force always depends on position, but doesn't always depend on time.

 

[vanhees71]

Thanks Orodruin.

But my dilemma is that the force provides an acceleration that seems to depend only on position: a(x).

However, when we consider the object's motion, we discover that the object's acceleration can be expressed as a function of time t, i.e. a(t), since a( t(x) ) through parametrization. So does the acceleration a truly depend on x or on t?

Yes, after you've solved the equations of motion you have
$$\vec{a}(t)=\vec{a}[x(t)]=\frac{1}{m} \vec{F}[\vec{x}(t)].$$
The very purpose of the equation of motion is to find the trajectory of the particle subject to the force, $\vec{F}$, and the initial conditions $\vec{x}(t_0)=\vec{x}_0$, $\dot{\vec{x}}(t_0)=\vec{v}_0$.