Why does dU/dt = 0 require v = dx/dt in the spring energy derivation?

[docholliday]

U = energy
In the book:
$\frac{dU}{dt} = \frac{d}{dt} (\frac{1}{2} mv^2 + \frac{1}{2} kx^2)$

then we have $m \frac{d^{2}x}{dt^2} + kx = 0$ because $v = \frac{dx}{dt}$

however they get rid of $\frac{dx}{dt}$ .

They are ignoring the case where v = 0, because then $m \frac{d^{2}x}{dt^2} + kx$ doesn't have to be zero, and it can still satisfy the equation.

 

[rock.freak667]

If you have y = (dx/dt)^2 and you put u = dx/dt

then y=u^2 such that dy/du = 2u and du/dt = d^2x/dt^2

So dy/dt = 2u*du/dt = 2(dx/dt)(d^2x/dt^2)

In your original equation, differentiating the KE term and the spring term will give you a dx/dt which can be canceled out since dU/dt= 0.

 

[Khashishi]

You should edit the post and replace [; ... ;] with [i tex] ... [/i tex]
(get rid of the space in [i tex]. I put that in so the parser wouldn't detect it.)

 

[docholliday]

yes, i get it but if dx/dt = 0, which it can, then the equation is satisfied and the other term doesn't have to be zero. However, we are saying the other term must always be zero.

 

[mfb]

dx/dt = 0 is true for a two point in time per period only, or for a non-moving spring in equilibrium. That is not relevant for the general case.