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\author{Oren}
\title{SM}
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\begin{center}\large{\textbf{
Galvanometer\\
Submitted by: Oren Rosenblatt}}\end{center}
\textbf{Exercise 8484}\\ \\
A galvanometer at temperature $T$ has a mass $m$, a deflection spring with oscillation frequency $\omega_0$
and a damping resistance $R$. The position $x$ of the spring measures the current $I$ via the equation
\[\ddot{x}+\omega_0^2 x = -\gamma\dot{x}+A(t)+\alpha I\]
where $A(t)$ is a random force, $\gamma$ is the friction and $\alpha$ is an instrument parameter, converting the current into force.\\
\renewcommand{\theenumi}{\alph{enumi}}
\begin{enumerate}
\item Find the current condition for recording?
\item Evaluate dissipation rate for current I with frequency $\omega$.
\item Identify R by equating b. with $I^2R$.
\item Rewrite a. in terms of R in the limit $\omega\sim\omega_0$. 
\end{enumerate}
\newpage
\textbf{Solution}\\
\begin{enumerate}
\item
\begin{equation} \ddot{x}+\omega_0^2 x = -\gamma\dot{x}+A(t)+\alpha I\end{equation}
In equilibrium $\ddot{x}=\dot{x}=0$, and for the expectation value $<x>$ the random force $A(t)$ disappears and we get
\begin{equation}
<x>=\frac{\alpha I}{\omega_0^2}
\end{equation}
Now we will define $\tilde{x}=x-<x>$ and use the equipartition law
\begin{equation}
\frac{1}{2}m\omega_0^2 < \tilde{x}^2>=\frac{1}{2}k_bT
\end{equation}
for accurate measurement will demand 
\begin{equation}
<x^2> \ > \ <\tilde{x}^2>
\end{equation}
by using condition (4) in equation (3) we will get the recording condition for I.
\begin{equation}
\left( \frac{\alpha I}{\omega_0^2}\right)^2 \ > \ \left(\frac{k_bT}{m\omega_0^2} \right) \Rightarrow 
I^2 > \frac{k_bT\omega_0^2}{m\alpha^2}
\end{equation}
\item
The driving force is
\begin{equation}
f_0=m\alpha I
\end{equation}
and the dissipation is 
\begin{equation}
\frac{d\overline{E}}{dt}=\frac{1}{2}\omega |f_0|^2Im(\alpha_x(\omega))
\end{equation}
we shall use langevin equation in fourier space
\begin{equation}
(-m\omega^2-\dot\imath m\gamma\omega+m\omega_0^2)x_\omega=mA(\omega)+f(\omega)
\end{equation}
and again for the expection value A will fall.
\begin{equation}
<x_\omega >=\alpha_x(\omega)f(\omega)
\end{equation}
by using equation (8) and (9), and rearrange it we will get 
\begin{equation}
\alpha_x(\omega)=(-m\omega^2-\dot\imath m\gamma\omega+m\omega_0^2)^{-1}=\frac{1}{m(\omega_0^2-\omega^2-\dot\imath\gamma\omega)}
\end{equation}
from (10) will get $Im(\alpha_x(\omega))$
\begin{equation}
Im(\alpha_x(\omega))=\frac{\gamma\omega/m}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}
\end{equation}
Placing (11) in (7) will get
\begin{equation}
\frac{d\overline{E}}{dt}=\frac{1}{2}\frac{m\alpha^2\gamma\omega^2I_0^2}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}
\end{equation}
we used the fact that the only frequency that lasts is the current frequency and that the average mean square of the current is 
\begin{equation}
I=I_0\cos(\omega t)\Rightarrow \overline{I^2}=\frac{I_0^2}{2}
\end{equation}
\item
\begin{equation}
\frac{d\overline{E}}{dt}=R\overline{I^2}\Rightarrow\frac{RI_0^2}{2}=\frac{I_0^2}{2}\frac{m\alpha^2\gamma\omega^2}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}
\end{equation}
\begin{equation}
R=\frac{m\alpha^2\gamma\omega^2}{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}
\end{equation}
\item

Equation (15) in the limit $\omega\sim\omega_0$ becomes 
\begin{equation}
R\underset{\omega\rightarrow\omega_0}{=}\frac{\alpha^2m}{\gamma}
\end{equation}
we get the current condition in R terms by placing (16) in (5)
\begin{equation}
I^2 > \frac{k_BT\omega_0^2}{\gamma R}
\end{equation}
\end{enumerate}
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