Indeed, I too have realised this and I started wondering and searching on the web didn't bring up a satisfactory answer for me. The sites I got were vague and not explicit enough.

Wait, now I'm thinking about something...

Okay, let's say the bob is raised (keeping the thread of the pendulum taut) through 1 m and the length of the string is 100 m (for ease of working).

PE = mgh = 1*10*1 = 10 J

(taking m = 1 kg XD and g = 10 m/s^2 again, for calculations purposes)

Max KE = 10 J

so, max v = \sqrt(10*2/1) = 2\sqrt(5)

We can also find omega.

ω = \sqrt{g/l} = \sqrt{10/100} = \sqrt{10}/10

we know that v = ω \sqrt{x_o^2 - x^2}

2\sqrt{5} = \sqrt{10}/10 x_o

squaring,

20/ \sqrt{2} = x_o

Now, from trigonometry (assuming that amplitude is displacement between the two points), the amplitude would be given by:

Amplitude = \sqrt{x^2 + y^2} = \sqrt{ (\sqrt{100^2-99^2})^2 + 1^2}

(A sketch shows it all)

So, displacement = \sqrt{199 + 1} = \sqrt{200} = 10\sqrt{2}

*looks up*

Hm... 20/ \sqrt{2} = 20\sqrt{2} / 2 = 10\sqrt{2}

Okay, now I'm satisfied.

I'm not sure why I didn't think of this earlier >.<

Maybe it's your answer which triggered it...