Hooke's Law Second Order Differential Equation

[Kernul]

Homework Statement

A mass $m$ on a frictionless table is connected to a spring with spring constant $k$ so that the force on it is $F_x = -kx$ where $x$ is the distance of the mass from its equilibrium position. It is then pulled so that the spring is stretched by a distance $x$ from its equilibrium position and at $t = 0$ is released.
Write Newton’s Second Law and solve for the acceleration. Solve for the acceleration and write the result as a second order, homogeneous differential equation of motion for this system.

Homework Equations

Newtons's Second Law.
Hooke's Law
Differential Equations

The Attempt at a Solution

I write the the Newton's Second Law and solve for the acceleration:
$$F = m a_x = - k x$$
$$a_x = -\frac{k}{m} x$$
Now it tells me to write the result as a second order, homogeneous differential equation of motion. I don't quite get how I should do this but I think this way:
I write $a_x = -\frac{k}{m} x$ as $\frac{d v_x}{d t} = -\frac{k}{m} x$ and multiplying both sides for $d t$ and integrating I have:
$$v_x(t) = -\frac{k}{m} x t + C$$
where $C$ is a constant and would actually be $v_{0x} = 0$
Same thing again with $\frac{d x}{d t} = -\frac{k}{m} x t$ and having:
$$x(t) = -\frac{k}{2m} x t^2 + C$$
Now, should I put all this like
$$x''(t) + x'(t) + x(t) = 0$$
and so
$$(-\frac{k}{m} x) + (-\frac{k}{m} x t) + (-\frac{k}{2m} x t^2) =0$$
Is this the correct way?

 

[TSny]

I write the the Newton's Second Law and solve for the acceleration:
$$F = m a_x = - k x$$
$$a_x = -\frac{k}{m} x$$
Now it tells me to write the result as a second order, homogeneous differential equation of motion. I don't quite get how I should do this

Can you express $a_x$ as a second derivative?

I write $a_x = -\frac{k}{m} x$ as $\frac{d v_x}{d t} = -\frac{k}{m} x$ and multiplying both sides for $d t$ and integrating I have:
$$v_x(t) = -\frac{k}{m} x t + C$$

Here you treated $x$ as a constant. But $x$ is a function of time. You are asked to find a second-order, homogeneous differential equation such that if you solved it, it would give you the answer for $x(t)$. But the statement of the question does not ask you to solve the differential equation.

 

[Kernul]

Can you express $a_x$ as a second derivative?
Here you treated $x$ as a constant. But $x$ is a function of time. You are asked to find a second-order, homogeneous differential equation such that if you solved it, it would give you the answer for $x(t)$. But the statement of the question does not ask you to solve the differential equation.

I can express $a_x$ as $\frac{d^2 x}{d t^2}$.
And yes, you're right, I wrongly treated it as a constant. So I should have something like $v_x = -\frac{k}{m} \int x(t) dt$.

 

[TSny]

I can express $a_x$ as $\frac{d^2 x}{d t^2}$.

Yes.

And yes, you're right, I wrongly treated it as a constant. So I should have something like $v_x = -\frac{k}{m} \int x(t) dt$.

A differential equation of motion for the mass would be a differential equation involving time derivatives of the unknown function $x(t)$. No need to bring in $v_x$.

 

[Chestermiller]

I can express $a_x$ as $\frac{d^2 x}{d t^2}$.

Good, now substitute that into your equation $$a_x = -\frac{k}{m} x$$What do you get?

 

[Kernul]

A differential equation of motion for the mass would be a differential equation involving time derivatives of the unknown function $x(t)$. No need to bring in $v_x$.

So I should write $\frac{d x}{d t}$ instead of $v_x$?

Good, now substitute that into your equation $$a_x = -\frac{k}{m} x$$What do you get?

I would get $$\frac{d^2 x}{d t^2} = -\frac{k}{m} x$$

 

[TSny]

So I should write $\frac{d x}{d t}$ instead of $v_x$?

I would get $$\frac{d^2 x}{d t^2} = -\frac{k}{m} x$$

Yes, that's essentially it! You can put into "standard form" by arranging it in the form $ax''(t) + bx'(t) + cx(t) = 0$. The zero on the right makes it homogenous.

http://mathworld.wolfram.com/HomogeneousOrdinaryDifferentialEquation.html

 

[Kernul]

Yes, that's essentially it! You can put into "standard form" by arranging it in the form $ax''(t) + bx'(t) + cx(t) = 0$. The zero on the right makes it homogenous.

http://mathworld.wolfram.com/HomogeneousOrdinaryDifferentialEquation.html

But substituting I would have something like this, right:
$$x''(t) + x'(t) + x(t) = 0$$
$$-\frac{k}{m} x(t) -\frac{k}{m} \int x(t) dt + x(t) = 0$$
$$x(t) = \frac{k}{m} x(t) + \frac{k}{m} \int x(t) dt$$

 

[TSny]

But substituting I would have something like this, right:
$$x''(t) + x'(t) + x(t) = 0$$

No. Why are you writing this equation?

You have already essentially answered the question. Can you put your result into the standard form $ax'' + bx' + cx = 0$, where $a, b$ and $c$ are constants?

 

[Kernul]

No. Why are you writing this equation?

You have already essentially answered the question. Can you put your result into the standard form $ax'' + bx' + cx = 0$, where $a, b$ and $c$ are constants?

Sorry, I just want to be sure to understand so I'm going to write all.
We know that the standard form is $ax''(t) + bx'(t) + cx(t) = 0$.
Then we know that:
$$x''(t) = \frac{d^2 x}{d t^2} = -\frac{k}{m} x(t)$$
$$x'(t) = \frac{d x}{d t} = -\frac{k}{m} \int x(t) d t$$
Substituting these into the standard form we get:
$$-\frac{k}{m} a x(t) -\frac{k}{m} b \int x(t) dt + c x(t) = 0$$
Am I right?

 

[TSny]

Sorry, I just want to be sure to understand so I'm going to write all.
We know that the standard form is $ax''(t) + bx'(t) + cx(t) = 0$.
Then we know that:
$$x''(t) = \frac{d^2 x}{d t^2} = -\frac{k}{m} x(t)$$
$$x'(t) = \frac{d x}{d t} = -\frac{k}{m} \int x(t) d t$$

No need to write the second equation. Rearrange the first equation to get 0 on the right and you're done.

 

[Kernul]

No need to write the second equation. Rearrange the first equation to get 0 on the right and you're done.

So simply this?
$$x''(t) + \frac{k}{m}x(t) = 0$$

 

[TSny]

Yes. Writing the answer as $x''(t) = -\frac{k}{m}x(t)$ is probably OK, too. Writing it with 0 on the right just puts the differential equation in standard form. The question statement is not clear on whether or not you need to put it into standard form. Since it says, "solve for the acceleration" it could be that they actually want $x''(t) = -\frac{k}{m}x(t)$. Who knows.

 

[Kernul]

Yes. Writing the answer as $x''(t) = -\frac{k}{m}x(t)$ is probably OK, too. Writing it with 0 on the right just puts the differential equation in standard form. The question statement is not clear on whether or not you need to put it into standard form.

Oh, I get it now. I think I got confused before.
Yeah, well, I'll use the standard form. Thank you very much!

 

[TSny]

OK. Good work.