Calculation of the Cyclotron frequency (non-relativistic)

Calculation:

The Lorentz force $F_{\rm{Lorentz}}$ is the centripetal force $F_{\rm{Zentripetal}}$ and causes the particles path to bend in a circle:$${F_{\rm{Lorentz}}=F_{\rm{Zentripetal}}}$$ $${\Rightarrow q\cdot v\cdot B=\frac{m\cdot v^2}{r}}{\Rightarrow v=\frac{r\cdot q\cdot B}{m}}$$ $$q=\text{charge of the particle}, v=\text{velocity of the particle}, B=\text{magnetic flux}, m=\text{mass of the particle}, r=\text{Radius of the circle}$$ With $v=\omega\cdot r=2\pi \cdot f\cdot r$ the Cyclotron frequency $f$ is$$ 2\pi \cdot f\cdot r=\frac{r\cdot q\cdot B}{m}\Rightarrow f=\frac{q\cdot B}{2\pi\cdot m}$$ $$\omega=\text{angular velocity}, f=\text{frequency}$$ and for the orbital period $T$$$T=\frac{1}{f}=\frac{2\pi\cdot m}{q\cdot B}$$ The cyclotron frequency $f$, the orbital period $T$ and so the time of a particle in a "dee" are independent of the radius $r$. So is possible to keep the frequency of the alternating voltage constant.

The kinetic energy $E_{\rm{kin}}$ grows with every passage of the electric field in the gap between the dees by $q\cdot V$:$$E_{kin}(n)=E_{kin_0}+n\cdot q\cdot V=\frac{1}{2}\cdot m\cdot {v_0}^2+n\cdot q\cdot V$$ $$E_{kin_0}=\text{Kinetic energy when leaving the paricle source}, n=\text{Number of passages through the electric field}, V=\text{Voltage between dees}$$ The velocity $v$ of the particle can be calculated with the initial speed $v_0$ when it leaves the particle source and the number $n$ of passages through the electric field between the "dees":$${\frac{1}{2}\cdot m\cdot v^2=\frac{1}{2}\cdot m\cdot {v_0}^2+n\cdot q\cdot V}$$ $${\Rightarrow v=\sqrt{\frac{2}{m}\left( \frac{1}{2}\cdot m\cdot {v_0}^2+q\cdot V\cdot n\right)}}$$ If the initial speed is $v_0=0$ the follwing formula can be used:$$v=\sqrt{\frac{2}{m}\cdot q\cdot V\cdot n}$$ next