Swing High -- Swing Low?

Question

A certain pendulum swings through an arc of one degree in one second. Next, the same pendulum is made to swing through an arc of two degrees. The time required to swing through the two-degree arc is:

a) one-half second
b) one second
c) two seconds

The answer is ?. Why not try this one out like Galileo did? Use a long string and a heavy weight. Do not time single swings. Take the time for ten swings. The swings will die a little during the timing, but don't worry about that.

After you do the experiment think about this. Pull the pendulum back to two degrees (we can call it position B)and let it swing back to a perpendicular position (rest, we can call it position C). Then pull it further back to say three degrees (we can call it position A) and let it swing back to perpendicular (position C). The trip from A is longer, but faster. It is length of a pendulum, not its displacement, that determines its period of vibration.

Answer 1

linlyons's speculation tweaked me again! The restoring torque of a pendulum is proportional to the sine of the displacement, not the displacement itself. With a torque proportional to displacement, we'd have the equivalent of a spring-mass system, always a constant frequency. But we are dealing with the old "small angle approximation" error instead, up to 90 deg swing. (Beyond that it really gets flaky.) And since elliptic integrals are involved, the math gets quite hairy. The correction factor for the period is an infinite series:
[ 1 + (1/2)^2 sin^2(θ/2) + (1 * 3 /(2 * 4) )^2 sin^4(θ/2) + ... ]
So the rules change for even the smallest amplitudes, but this can be overlooked by mere mortals (all but the National Bureau of Standards) up to 10 or 20 deg.
For a better explanation, see the ref.

Answer 2

One second

Always the same for the same pendulum.
That is why an old clock works.

Answer 3

one secnd............