# I have been given the potential of a simple harmonic oscillator: V=1/2kx^2 I want to calculate the value x(t) of a particle moving in this potential, with initial conditions x(t=0)=0 and v(t=0)=v_0. How would I go about doing this in simple classical mechanics?

I have been given the potential of a simple harmonic oscillator:
$V=\frac{1}{2}k{x}^{2}$
I want to calculate the value $x\left(t\right)$ of a particle moving in this potential, with initial conditions $x\left(t=0\right)=0$ and $v\left(t=0\right)={v}_{0}$. How would I go about doing this in simple classical mechanics?
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LilsGroolonip86
Solving the equation:
You can guess the solution $x=A\mathrm{cos}\sqrt{k/m}t+B\mathrm{sin}\sqrt{k/m}t$ or you can use substitution $x={e}^{\lambda t}$ and find $\lambda$
Anyway:
$x\left(t=0\right)=A\mathrm{cos}0+B\mathrm{sin}0=A=0$
$v\left(t=0\right)=-A\sqrt{k/m}\mathrm{sin}0+B\sqrt{k/m}\mathrm{cos}0=B\sqrt{k/m}={v}_{0}$
therefore:
$x\left(t\right)=\sqrt{\frac{m}{k}}{v}_{0}\mathrm{sin}\left(\sqrt{\frac{k}{m}}t\right)$
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batejavizb
With given potential you can define force acting on your oscillator as
$F=-\frac{dV}{dx}=-kx$
This gives you Hooks law (no surprise). Than just use Newton's second law:
$ma=-kx$
Than create second order differential equation replacing a with $\stackrel{¨}{x}$ and solve it with your initial conditions.