# Hooke's Law Second Order Differential Equation

1. Dec 16, 2016

### Kernul

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Newtons's Second Law.
Hooke's Law
Differential Equations

3. 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?

2. Dec 16, 2016

### TSny

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.

3. Dec 16, 2016

### Kernul

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 $v_x = -\frac{k}{m} \int x(t) dt$.

4. Dec 16, 2016

### TSny

Yes.
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 $v_x$.

5. Dec 16, 2016

### Staff: Mentor

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

6. Dec 16, 2016

### Kernul

So I should write $\frac{d x}{d t}$ instead of $v_x$?
I would get $$\frac{d^2 x}{d t^2} = -\frac{k}{m} x$$

7. Dec 16, 2016

### TSny

8. Dec 16, 2016

### Kernul

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

9. Dec 16, 2016

### TSny

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?

10. Dec 16, 2016

### Kernul

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?

11. Dec 16, 2016

### TSny

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

12. Dec 16, 2016

### Kernul

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

13. Dec 16, 2016

### 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.

14. Dec 16, 2016

### Kernul

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

15. Dec 16, 2016

### TSny

OK. Good work.