Solving m(d²x/dt²) + kx = 0 for displacement vs time

[hallic]

I've created a second order homogeneous equation from my orginal data
m(d^2x/dt^2) + kx = 0
how can I turn it into a expression of displacement relevant to time?

 

[Identity]

$$\frac{d^2x}{dt^2} = \frac{-kx}{m}$$

Now,

$$\frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} = \frac{d(\frac{1}{2}v^2)}{dv}\frac{dv}{dx} = \frac{d(\frac{1}{2}v^2)}{dx}=\frac{d}{dx}\left(\frac{1}{2}\left(\frac{dx}{dt}\right)^2\right)$$

By applying the chain rule twice.

So,

$$\frac{d}{dx}\left(\frac{1}{2}\left(\frac{dx}{dt}\right)^2\right) = \frac{-kx}{m}$$

Hopefully you can now solve this as a differential equation

 

[gato_]

Isn't it simpler to solve the first one, it being standard? $$x=ACos(\sqrt{\frac{k}{m}}t+\phi_{0})$$

 

[Identity]

Haha yeah, I guess so. But since I haven't learned the theory for second order differential equations, this is a way using high school calculus.

If you go through it, it comes out to be the same answer you have