dU/dt = 0 for oscillating spring, help with derivation

docholliday

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

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

however they get rid of [itex] \frac{dx}{dt} [/itex] .

They are ignoring the case where v = 0, because then [itex] m \frac{d^{2}x}{dt^2} + kx [/itex] 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.