
#1
Dec612, 06:50 PM

P: 5

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. 



#2
Dec612, 07:00 PM

HW Helper
P: 6,213

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. 



#3
Dec612, 07:01 PM

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.) 



#4
Dec612, 07:08 PM

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.




#5
Dec712, 07:50 AM

Mentor
P: 10,813

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



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