Integrating a second derivative-involving solution for Simple Harmonic Motion

AI Thread Summary
The discussion focuses on integrating the equation m(d²x/dt²) = -kx to find the general solution for simple harmonic motion. The user is struggling with the second derivative of acceleration and has attempted to separate variables, which has complicated the process. A suggestion is made to rearrange the second derivative as v(dv/dx) and substitute this into the equation for easier integration. The user is advised to integrate the resulting expression, which leads to a relationship involving v² and x², and to determine the constant C using initial conditions. Ultimately, the user is encouraged to convert v back to dx/dt and integrate again to reach the general solution x = Acos(wt + c).
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Hello, i am now in the process of integrating m(d^2x/dt^2)=-kx which i know i will have to do twice in order to obtain the general solution to simple harmonic motion, x= Acos(wt+c) c=phi

but I'm just having problems with the second derivative of acceleration (d^2*x/dt^2) when it comes to integrating, I tried separating them into dv/dt and then dv/dx and dx/dt, so i obtain v, etc... but this complicates things even more for me! It would be so great if someone could give me a hint as to where i can start in integrating with this second derivative...
Thank you very much for your time.
 
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insertnamehere said:
Hello, i am now in the process of integrating m(d^2x/dt^2)=-kx which i know i will have to do twice in order to obtain the general solution to simple harmonic motion, x= Acos(wt+c) c=phi

but I'm just having problems with the second derivative of acceleration (d^2*x/dt^2) when it comes to integrating, I tried separating them into dv/dt and then dv/dx and dx/dt, so i obtain v, etc... but this complicates things even more for me! It would be so great if someone could give me a hint as to where i can start in integrating with this second derivative...
Thank you very much for your time.
The 2nd differential is often rearranged like this,

\frac{d^2x}{dt^2}= v\frac{dv}{dx}

and you get that this way,

\frac{d^2x}{dt^2}=\frac{d}{dt}(dx/dt)=\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}= \frac{dv}{dx}v = v\frac{dv}{dx}

substitute for \frac{d^2x}{dt^2} to v\frac{dv}{dx} then do the integration.
 
Yes, i tried doing that, but then eventually i get stuck with
(v^2)/2= (-k/m)(x^2)/2 + C
Would i have to integrate again to obtain the general solution of
x=Acos(wt+c)? What can i do with the v^2 and the x^2, and the negative k/m?
 
Your halfway there.
From the initial conditions of the problem, work out a value for C.
rewrite the eqn as v = whatever ...
convert v to dx/dt and integrate agan.
 
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