What does AC sq wave do to a DC solenoid?

I want to do an electronic crossbow release. Don't ask me why, long story, but it's a light crossbow, under 60 lbs. It will be hard to predict just how much force is needed to release the mechanism. It can't be too lose or it will go off by itself when you knock it and the solenoid core's own momentum could let the release go.

The biggest battery I plan to use is 2x 3v CR123A lithium batteries. These are the big ones, not coin cells.

I'll have a microcontroller, I can drive in a lot of different ways. Now the more pull the better, it also needs to be quick which is why I'd not use any kind of motor. I would like to drive a solenoid. Now most solenoids I see in the surplus catalog are 12V and I have 6V. I wouldn't mind overdriving it to get more pull either, since it's only a pulse.

Anyways, the question: if I invert 6V to the solenoid, so it's got +6V and GND one moment then the terminals reverse, so it then has -6V. It's a 12V p-p square wave as far as the solenoid sees. Does this pull stronger or weaker than DC?? Just how much stronger?

If it doesn't reload, you could pull a pin out with a solenoid? Or you could charge a capacitor to get a bigger "current hit". That doesn't really answer your question, though :?

There's more to it than this but at any instant a 6 volt pulse is still a 6 volt pulse. RMS voltage maximum is still 6 volts. Inductance will change the shape of the pulse so it won't be a square wave anyway. It would seem that the steady application of 6 vdc will do more for you - unless you use the principle you describe to charge a capacitor to 12 volts. Overdriving a solenoid designed for a given voltage may not net you the increase you expect. You'd have to know something about the amount of pull at various levels of power.