The capacitor discharges in around 5RC, when we have an ac signal of 40Hz,  and we want to match the frequency, we can calculate the following:

$$
\begin{flalign}
&\tau = RC &\\\
&f = 40Hz = \frac{1}{40}ds \\
&\textit{Lets put the resistor to 1k}\ohm\\
&\frac{1}{80} = 1000*C & | \div 1000 \\
&\frac{1}{80000} = C\\
&C = 12.5 \micro F\\
\end{flalign}
$$
```circuitjs
$ 1 0.000005 9.78399845368213 50 5 43 5e-11
R 80 224 0 224 0 2 40 5 0 0 0.5
c 80 224 192 224 0 0.00001 -1.936025478465976 0.001
r 192 224 192 304 0 1000
g 192 304 192 336 0 0
O 192 224 304 224 0 0
o 4 64 0 151554 10 0.1 0 1
o 0 64 0 151554 10 0.1 0 2 0 3
```
![[spike-circuit-rc-relation.png]]