Infinite Atwood’s machine

The effective mass (or you can call it reduced mass) of 2 blocks of masses $M_1$ and $M_2$ hanging from a pulley is given by $M_{\text{effective}} = \dfrac{4M_1M_2}{M_1+M_2}$

So, attach a block mass $x$ on the string and note that it should be the effective mass of the whole system, so we have the equation $x=\dfrac{4Mx}{M+x}$ and from here we get that $x=3M$

For the acceleration in this type of system, it is given by the ratio of difference in masses and the sum of the masses times the acceleration due to gravity, so $a=\left(\dfrac{3M-M}{3M+M}\right)g=\dfrac{2M}{4M}g=\dfrac{g}{2}$ , hence answer is $\boxed{0.5}$

I am writing the derivation of the effective mass expression as given by @Mark Hennings in the comments section.

So, If a pulley system is suspended with masses $M_1,M_2$ on each side, then we have $M_1g - T \; = \; M_1a \hspace{2cm} T - M_2g \; = \; M_2a$ where $T$ is the tension in the string, and $a$ is the acceleration of the particles. Eliminating $a$ , we deduce that $T \; = \;\frac{2M_1M_2g}{M_1+M_2}$ Thus the tension in any string supporting the pulley must be $2T = \frac{4M_1M_2g}{M_1+M_2}$ and so the effect on the system above the pulley would be the same if the pulley system was replaced by a mass of $\frac{4M_1M_2}{M_1+M_2}$ .

For $a$ , eliminate $T$ from these equations to get $a=\dfrac{M_1-M_2}{M_1+M_2}g$

Suppose that all the strings have length $L$ , and let $x_j$ be the distance of the $j$ th mass (for $j \ge 0$ ) from the $j$ th pulley. Then the distance of the $j$ th mass below the top ( $0$ th) pulley is $d_j \; = \; x_j + \sum_{k=0}^{j-1}(L - x_k) \; = \; jL + x_j - \sum_{k=0}^{j-1}x_k$ for $j \ge 1$ , so that $\ddot{d}_j \; = \; \ddot{x}_j - \sum_{k=0}^{j-1} \ddot{x}_k \;\; \;(j \ge 1) \hspace{2cm} \ddot{d}_0 \; = \; \ddot{x}_0$ If the tension in the $j$ th string is $T_j$ then, since the $j$ th pulley is weightless, $T_{j-1} = 2T_j$ , and so we can write $T_j \; =\; M\tau 2^{-j} \hspace{2cm} j \ge 0$ The equation of motion of the $j$ particle is $\begin{aligned} M\ddot{d}_j & = \; Mg - T_j \\ \ddot{x}_j - \sum_{k=0}^{j-1}\ddot{x}_k & = \; g - \tau2^{-j} \end{aligned}$ Thus $\ddot{x}_{j+1} - 2\ddot{x}_j \; = \; \left(\ddot{x}_{j+1} - \sum_{k=0}^j \ddot{x}_k\right) - \left(\ddot{x}_j - \sum_{k=0}^{j-1}\ddot{x}_k\right) \; = \; \tau2^{-j-1}$ and we can solve this recurrence relation, together with $\ddot{x}_0 = g - \tau$ , to deduce that $\ddot{x}_j \; = \; 2^j\Big[g - \tfrac23\tau - \tfrac13\tau2^{-2j}\Big] \hspace{2cm} j \ge 0$ Since these accelerations need to remain bounded, we deduce that $\tau = \tfrac32g$ , so that $\ddot{d}_0 = \ddot{x}_0 = -\tfrac12g$ , making the answer $\boxed{\tfrac12}$ .

Were the first mass not moving at all, that would apply that the first pulley (not the one hanged to the ceiling) doesn't accelerate, and the tension of the first string equals mg. Hence, the tension of the second string, which equals half the first string, is $\frac{m}{2g}$

Solving the motion equation for the second mass, we get that its acceleration is $\frac{g}{2}$

But what we've got here is actually the same problem as was presented!

That's true because the first pulley doesn't move, that is, it behaves as a ceiling. And there are infinite masses connected to the second mass from the right side, that is, the same situation as for the first mass.

Therefore we deduce that the acceleration of the first mass is $\frac{g}{2}$

Looking the top mass you have a force Mg pulling down and the force from the series of pullys pulling up.

The net force can be written as F=Mg-(Mg-(Mg-...

Divide by factor M to get, a=g-(g-(g-...

This can be written at a=g-a.

Solve a=(1/2)g.

lambda = 1/2

Let us inspect the pulley hung directly from the ceiling. We can imagine that the pulley is supporting two masses, one on either side.

The mass on the left has a mass of $M$ and the mass on the right $nM$ , where $n$ represents the total number of masses on that side - since they are all part of the system, their individual masses will add up.

We are now ready to construct equations from Newton's law of motion. Let's start from the left-hand side first.

(We are going to set the direction in which the right-hand side masses are moving as positive.)

$T - Mg = (n+1)Ma$

and the right-hand side:

$nMg - T = (n+1)Ma$

Add both equations together.

$(n-1)Mg = 2(n+1)Ma$

Divide both sides by $M$

$g(n-1) = 2a(n+1)$

Rearrange so that $a$ became the subject.

$a = \frac{g(n-1)}{2(n+1)}$

When $n \rightarrow \infty$ :

$\frac{g(n-1)}{2(n+1)} \rightarrow \frac{g}{2}$

Thus, $\lambda = \frac{1}{2}$

Isolate the second mass -- examine its motion relative to the second pulley. It's easy to see that it also has acceleration $\lambda g$ (with acceleration toward the ground taken in the positive sense) with respect to the pulley, because the pulley setup is infinite and it mirrors the system as a whole. The first mass pulls the pulley upward with force $M \lambda g$ by Newton's second law, and gravity pulls down with force $M g$ . By Newton's third law, the forces $M \lambda g$ and $M g - M \lambda g$ are equal. Setting these two quantities equal and dividing by $M g$ shows us that $\lambda = 0.5$