# Hooke's Law integration

1. Aug 6, 2012

### g.lemaitre

Do you see where it says

m(dv/dt)v = (d/dt)(.5mv^2)

If it's an integration which I don't think it is then I would think it should be

(d/dt)((mv^3)/3) because you're taking the two v's and adding an additional power. I don't think it is an integration because it says right there in the book that you can't integrate, so if it's now then where does the 1/2 come from?

2. Aug 6, 2012

### oli4

Hi g.lemaitre
It is indeed an integration and it is correct
just derive (1/2mv²) and you will see that you get back to mvv' or mv dv/dt
Cheers...

3. Aug 6, 2012

### Philip Wood

It's a case of the chain rule:

$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)\:=\:\frac{d}{dv}\left(\frac{1}{2}mv^2\right)\:\times\frac{dv}{dt}\:=\:mv\frac{dv}{dt}$.

Your textbook isn't very clear. You could integrate the left hand side of the equation wrt t, before multiplying by v. It's the right hand side that you can't integrate wrt to t until you've multiplied by v. That's why you need to multiply through by v. Then the left hand side integrates (wrt t) to give $\frac{1}{2}mv^2$. The right hand integrates to give $-\frac{1}{2}kx^2$ + constant. Try differentiating this wrt t using the chain rule!

Last edited: Aug 6, 2012
4. Aug 6, 2012

### Emilyjoint

The spring constant 'k' is not a measure of the STRENGTH of the spring.
It is a measure of the STIFFNESS.... units are N/m
The strength is measured by ultimate tensile stress
edit....is this question about simple harmonic motion?
or hookes law (elasticity)

Last edited: Aug 6, 2012