Average Period

We begin with Newton's second law as we find the acceleration of the system. From $\sum_{j} F_{ij} = ma_i,$ we get the system of equations $T-Mg=Ma$ , and $2Mg-T=2Ma$ . Solving for $a$ gives $a=g/3$ .

Now we find the time it takes for mass $2M$ to hit the ground. Since it's accelerating, we use $\frac{1}{2}at^{2} = d$ . Solving for time, and using $a=g/3$ , we get a time of $t=\sqrt{6d/g}$ .

We know that the period of a non-accelerating pendulum is $T=2{\pi} \sqrt{\frac{l}{g}}$ . In this problem, $l$ changes with time, and because of the upward acceleration, we replace $g$ with $g_{eff} = g + g/3 = 4g/3$ (think equivalence principle).

Now let's find how $l$ changes with time. $l$ decreases in length at the same rate mass $2M$ falls, so we can write $l_{t} = l_{o} - \frac{g}{6} t^{2}$ . This means the instantaneous period at time t is $T_{t} = 2{\pi} \sqrt{\frac{l_{o} - \frac{g}{6} t^{2}}{4g/3}}$ .

To find the average period from this equation, just integrate $T_{t}$ with respect to time from $t=0$ to $t=\sqrt{6d/g}$ , then divide by the total time.

This comes down to evaluating $T_{avg} = \frac{2{\pi} \displaystyle \int_{0}^{\sqrt{6d/g}} \sqrt{\frac{l_{o} - \frac{g}{6} t^{2}}{4g/3}} dt}{\sqrt{6d/g}}.$

To evaluate, manipulate it to get it in a form of a constant times $\displaystyle \int_{0}^{x} \sqrt{u^{2}-t^{2}} dt = \frac{u^{2}\sin^{-1}{(\frac{x}{u})} + x\sqrt{u^{2} - x^{2}}}{2}$ .

This comes out to be, with a great deal of simplifying, $\frac{\pi}{2} \left( \sqrt{\frac{3(l_{o}-d)}{g}} + \sqrt{\frac{3}{gd}}l_{o}\sin^{-1}{\sqrt{\frac{d}{l_{o}}}} \right)$ .

Hence, $a=1/2$ , and $b=c=3$ , so $a*(b+c)=\boxed{3}$ .

Edit: An arithmetical error has been fixed.