What are the values of $W'$ so that the particles keep passing through the disk with its new height?

So I’m trying to solve this problem (which has some physics elements but it’s mostly a math problem, I think). Here it is:

Consider the image below:

In the picture, we have a device $A$ that realeases particles from rest with a period of $T = 3$ seconds. Just below the device, at a distance $H$ from it, a disk has a hole that allows the passage of all the particles released by the device.
It’s known that between the passage of two particles, the disc fully completes exactly three $360^circ$ revolutions around its axis.
If one raises the disc to a new distance of $displaystylefrac{H}{4}$ from the device, what are all the possible values of $W’$ so that all of the particles keep on passing through its hole?

I would put some of my tries here if I had something concrete; but the thing is, I couldn’t even start. Could someone help?

EDIT:
After the answer gave by Alex Ravsky I was able to kinda find my way, thanks to the direction he gave me. So, just for a brief, here’s how I’m thinking right now.

The question I had for him then was: why does the time between two passages through a plane parallel to the disk independs on either its distance to the device or the projectile’s velocity?

Here’s his answer (its important, for the notations he used I used later on my scribbling):

We have a sequence of particles, moving one-by-one. Each next particle
$p_n$, released at a moment $t+Delta t$ moves synchronous (with a time shift by
$Delta t$) with a previous particle $p_n$ released at a moment $t$. So the
particle $p_n$ crosses any plane parallel to the disc exactly with a
delay $Delta t$ after the particle $p_p$ crossed this plane, and this delay in
independent on the particles speed and the distance from the device to
the plane. Thus the particles pass the disc iff its rotation period $T′$
(and an initial phase) is adjusted to $Delta t$.

After that, I was able to make the following reasoning:

Let $pi_1$ be one of such planes parallel to the disk. Let’s make $h_1$ its distance to the device. Let’s also assume that the device realeases a particle every $T$ seconds. Then, $p_p$ would take $Delta t_1 = k$ to cross plane $pi_1$, making $t_0 = 0$. With that, it’d cross plane $pi_1$ at moment $t_1 = k$ (not that it matters, but on ideal conditions, $k = sqrt{frac{2h_1}{g}}$, just for the record).

• If $T > Delta t_1$, then $p_n$ wouldn’t have been released yet. Then, $Delta t’ = T – k$ would pass before $p_n$ had been released and it’d take more $Delta t_2 = k$ for it to cross plane $pi_1$. Thus, the instant $t_2$ that it’d pass though plane $pi_1$ would be:

$$t_2 = t_1 + Delta t’ + Delta t_2.$$

So, the $Delta t$ he mentioned would be:

$$Delta t = t_2 – t_1 = Delta t’ + Delta t_2 = Delta t’ + k.$$

But as $Delta t’ = T – k$, we have:

$$Delta t = T – k + k = T.$$

• If, however, $T < Delta t_1$, then $p_n$ would have already been released after $p_p$ crosses plane $pi_1$. Then, after $t = t_1$, it’d have already fallen for $Delta t’ = k – T$ and, after that, it’d still take more $Delta t_2 = k – Delta t’$ for $p_n$ to cross the plane. Thus, the instant $t_2$ that it’d pass though plane $pi_1$ would be:

$$t_2 = t_1 + Delta t_2.$$

So, the $Delta t$ he mentioned would be:

$$Delta t = t_2 – t_1 = Delta t_2 = k – Delta t’.$$

But as $Delta t’ = k – T$, we have:

$$Delta t = k – (k – T) = T.$$

• And, of course, if $T = Delta t_1$, then we already have our independence.

With that, we can see that indeed the time between two passages through a plane parallel to the disk depends only on the period of releases, which is intrinsecal to the device.