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

by docholliday
Tags: springs mechanics
 P: 5 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.
 HW Helper P: 6,214 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.
 P: 833 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.)
P: 5

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

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.
 Mentor P: 10,840 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.

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