Differentiability of the acceleration

[Sriman Dutta]

Hi, I'm back after a lot of days......

Just came back with a thought...

Suppose there is a block of mass m that has compressed a light ideal spring having spring constant k. Now, if we write the equation of motion for this system, then

ma=-kx or a=-kx/m

It's evident that a is a function of x. So we can differentiate it wrt x to gt the jerk j.

So, j=-kv/m.

But v is variable and still differentiable.

And continuing it, we have an endless chain.

So is it infinitely differentiable??

[mathematic]

Your development has x as a function of something, so that you get x'=v.  In order to answer your question you need an explicit statement of the dependence of x or v on the something (time?).

[Sriman Dutta]

x isn't a function of any variable. It's the position vector of the block from the mean position.

[Country Boy]

 If f is not a function of any variable, then it makes no sense to talk about its derivative.  If this is a "deterministic problem', which it would be if you are using Newtonian physics as in you first post, x will have a specific value for any specific time.  In that case, x is a function of the time.  Perhaps you are not clear on what the word "function" means.

[Dubbelosix]

If you are though, don't be ashamed to admit it, I am sure Ivy doesn't mean to cause offence. 

[Sriman Dutta]

If I consider x=Asin(wt) ??

[John Cuthber]

If you have something whose position varies sinusoidally with time then the acceleration is a cosine curve and the derivative of that is a sine function so you are back to where you started.

You can keep differentiating as often as you like- the functions are all well behaved.

There's a similar situation with an object moving in a circle. The acceleration is always towards the centre of the circle and the rate of change of the acceleration is circular too.

You can differentiate as much as you want.

[Country Boy]

On ‎10‎/‎15‎/‎2017 at 8:18 AM, Sriman Dutta said:

If I consider x=Asin(wt) ??

 So x is a function of t.  v= dx/dt= Aw cos(wt) and a= dv/dt= -Aw^2 sin(wt)= -w^2 x. 

In your original post you had a=-kx/m  so w^2= k/m.

[Sriman Dutta]

Thank you for all the help.