RLC Circuit (lab session doubt)

[Phyba]

Homework Statement

I have to answer a few questions about a lab session I did (university) and I'm really stuck with this one.
The objective of the lab session was to find the self inductance (L) of a solenoid.

In order to do that, I had a function generator exciting a solenoid with a high self-inductance coefficient. (see image)

This one, at the same time, by magnetic coupling with the solenoid whose L we want to obtain, excites the RLC circuit, causing a transient response in the form of oscillatory damped tensions. We measured the period of those oscillations and with this formula
T=sqrt(L*C)

we can easily get the self-inductance L. No problem with that... Now, the question I'm having trouble with is the following
"Why do we use the high self-inductance solenoid instead of just connecting the function generator directly with the problem solenoid? Hint: Keep in mind that the internal resistance of the generator is 50 ohm "

I don't know why! :(

Homework Equations

T=sqrt(L*C)

The Attempt at a Solution

Perhaps it has something to do with limiting the current?

Thanks in advance!

 

[milesyoung]

Why do we use the high self-inductance solenoid instead of just connecting the function generator directly with the problem solenoid? Hint: Keep in mind that the internal resistance of the generator is 50 ohm

Do you know how to refer impedances from one side of a transformer to the other?

The internal impedance of the function generator and the impedance it sees from its terminals forms a voltage divider. If the two coils are tightly coupled and you have $L \ll L_\mathrm{exc}$, then your system has a nice property in terms of the voltage that appears across the oscilloscope terminals.

 

[milesyoung]

We measured the period of those oscillations and with this formula
T=sqrt(L*C)

I think you forgot a factor of 2π.