Is ## \frac{V_T}{\beta} ## the Crossover Frequency in Josephson Junctions?

[LagrangeEuler]

Homework Statement

$\alpha \frac{d^2\theta}{dt^2}+\beta\frac{d\theta}{dt}+V'(\theta)=V(t)$
Inertial effects are negligible at frequencies of up to several hundred megahertz, so the first therm can be neglected.
I'm not sure if that means that
$\beta\frac{d\theta}{dt}+V'(\theta)=V(t)$ (1)
or
$\frac{\beta}{\alpha}\frac{d\theta}{dt}+\frac{V'(\theta)}{\alpha}=\frac{V(t)}{\alpha}$ (2)
With using $V'(\theta)=V_T\sin(\theta)$
authors get
$\frac{d \theta}{dt}=\omega_{co}(\frac{V}{V_t}-\sin(\theta))$
where $\omega_{co}$ is classical crossover frequency.

Homework Equations

The Attempt at a Solution

From (1) I get
$\frac{d\theta}{dt}=\frac{V_T}{\beta}[\frac{V(t)}{V_T}-\sin(\theta)]$
so is $\frac{V_T}{\beta}$ crossover frequency? Tnx for the answer.

 

[UltrafastPED]

The inertial term is the one with the second derivative wrt time, so (1) is correct.

 

[LagrangeEuler]

Tnx.