Second order differential equation

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?
 
[tex]\frac{d^2x}{dt^2} = \frac{-kx}{m}[/tex]

Now,

[tex]\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)[/tex]

By applying the chain rule twice.

So,

[tex]\frac{d}{dx}\left(\frac{1}{2}\left(\frac{dx}{dt}\right)^2\right) = \frac{-kx}{m}[/tex]

Hopefully you can now solve this as a differential equation
 
116
0
Isn't it simpler to solve the first one, it being standard? [tex]x=ACos(\sqrt{\frac{k}{m}}t+\phi_{0})[/tex]
 
Haha yeah, I guess so. But since I haven't learnt the theory for second order differential equations, this is a way using highschool calculus.

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

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top