# Hooke's Law and differential equations

1. Feb 17, 2007

### Gyroscope

1. The problem statement, all variables and given/known data
$$-kx=m\frac{d^2x}{dt^2}$$

I don't know how to solve differential equations, can someone show me how to do it, with this example.

2. Feb 17, 2007

### cristo

Staff Emeritus
This is an example of a second order differential equation; let's rewrite it as $$\frac{d^2x}{dt^2}+\frac{k}{m}x=0$$ This will have a general solution of the form x=Aeat. Plug this into the equation and we obtain the equation a2+k/m=0, which gives us the values for a; namely $a=\pm \sqrt{-\frac{k}{m}}=\pm i\sqrt{\frac{k}{m}}$. Letting $\sqrt{k/m}=\omega$ we obtain $x(t)=Ae^{i\omega t}+Be^{-i\omega t}$. Now, recognising that this is the form of a sum of cosine and sine functions, we obtain the general solution $x(t)=C\cos(\omega t)+D\sin(\omega t)$

I hope you follow that; however, if you have not studied differential equations, then I would suggest first learning about first order differential equations, and then moving onto second order equations (although, I suspect you want this solution for a specific purpose).

[edit: I didn't notice that this was in the homework forum, and so a full solution shouldn't be given; however, I suspect from previous posts that you may be self-learning the subject, and so this is not a homework question]

Last edited: Feb 18, 2007
3. Feb 17, 2007

### Gyroscope

Why do you need both solutions?

4. Feb 18, 2007

### cristo

Staff Emeritus
Since both $e^{i\omega t}$ and $e^{-i\omega t}$ are solutions to the equation, then the general solution will be a linear combination of the two, with the constants determined by the boundary conditions.

5. Feb 18, 2007

### Gyroscope

Thanks cristo. How can you pass from e^(something) to cosine and sine functions?

6. Feb 18, 2007

### cristo

Staff Emeritus
Use the definition of complex exponential, namely $e^{\pm i\theta}=\cos\theta \pm i\sin\theta$. With a bit of rearrangement, we find that $$\cos\theta =\frac{1}{2}(e^{i\theta}+e^{-i\theta})$$ and $$\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$$

7. Feb 18, 2007

### robphy

While the method described above is very useful and practical, don't forget what the equation is asking:

Can you determine which functions x(t) are proportional to minus their second derivatives?

You don't need a [crash] course in differential equations to answer that question.