Electronic – Measuring RMS power output from audio amplifier channel

None of the options is good. The speaker produces a sound pressure level for a certain wattage inputted and this is the criteria you need to look at. Some speakers are more effective than others and this effectiveness is largely determined by their enclosure. Luckily, for any given size and shape of speaker the enclosure is roughly constant.

Decide what SPL you need and look at speaker specs to shortlist a handful then design your enclosure to suit the shortlisted items then do comparative studies. You are obviously looking for bandwidth, loudness and distortion as the key ingredients of the selected speaker. Choose also, the speaker that is capable of withstanding the power output that can be thrown at it. Choose an amplifier that won't melt down ie has protection built in.

With some assumptions (pure sine wave, resistive load, etc.) you can simply apply your formula.

$$ 50 = \frac{V^2}{8} $$

leads to

$$ V = \sqrt{400} $$

hence \$V = 20\$. This is the DC voltage that puts 50 W in an 8 Ω load.

For an AC sine wave the peak (with respect to the average) is \$\sqrt{2}\$ times this value, or approximately 28 V. These peaks appear both in positive and negative direction, so the peak-to-peak voltage is twice this value, approximately 56 V.

This is the minimal DC supply a capacitor-coupled push-pull amplifier stage would need. A direct-coupled push-pull amplifier would need one positive and one negative 28 V supplies, a bridge amplifier would need one 28 V supply. (In each case, more is needed due to losses, margins, etc.)