I am going to use ##v## to represent velocities in the lab reference frame and ##u## to represent velocities in the frame of the compression wave ahead of the piston.
Case 1: Impulsively-started piston
Consider first the impulsively started piston. It is initially at rest, and, at some time ##t=0##, it is impulsively accelerated to a velocity of ##v_p## (moving left to right, per the diagram in the original post). No penetration requires that no fluid crosses the face of the piston, so locally at the piston face, the flow is moving with ##v=v_p##, but in the far field it is still moving at ##u = 0##, creating a velocity discontinuity. It is a classical result that such a change in a 1D duct occurs discontinuously, i.e., as a result of a shock. This shock propagates with a speed ##v_s##.
Consider a change in frame of reference such that the coordinate system follows along with the shock, i.e., subtract ##v_s## from everything. Then the velocity upstream of the shock is ##u_1=0-v_s=-v_s## and downstream of the shock is ##u_2 = v_p- v_s##. One of the standard methods for determining the Mach number of a shock is to look at the velocity ratio across it (in its reference frame), which follows the relation
\dfrac{u_2}{u_1} = \dfrac{2 + (\gamma-1)M_1^2}{(\gamma + 1)M_1^2}
or
\dfrac{v_p- v_s}{-v_s} = \dfrac{v_s-v_p}{v_s} = 1 - \dfrac{v_p}{v_s} = \dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2}.
From there,
\dfrac{v_p}{v_s} = 1 -\dfrac{2 + (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{(\gamma + 1)M_s^2 - 2 - (\gamma-1)M_s^2}{(\gamma + 1)M_s^2} = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s^2} \right)
Note that ##M_s = v_s/a_1## where ##a_1## is the speed of sound in the stagnant air in the tube, so you can eliminate ##v_s## above and end up with
\boxed{\dfrac{v_p}{a_1} = M_p = \dfrac{2}{\gamma + 1}\left( \dfrac{M_s^2 - 1}{M_s} \right).}
Here, ##M=u/a## is the Mach number, ##\gamma## is the ratio of specific heats, and ##a## is the local speed of sound. You can uniquely determine the Mach number of the shock that propagates away from the piston based on the velocity of the piston and the conditions in the tube at the beginning. Indeed, the term ##M_p = v_p/a_1## represents a Mach number for the piston relative to the calm air in the tube (i.e., the lab frame).
You can easily solve the boxed equation for ##M_s## for a given piston Mach number.
M_s^2 - \left(\dfrac{\gamma + 1}{2}\right)M_p M_s - 1 = 0.
That's a straightforward quadratic equation. If you pick a very low piston Mach number, say, ##M_p=0.1##, you still get a supersonic shock Mach number of ##M_s=1.06##.
A fundamental truth about shocks is that they increase the temperature, pressure, and density of the air into which the propagate. Thus, the piston face experiences a pressure greater than ambient even if it moves at constant velocity, and a constant force is required to maintain that constant velocity (assuming the tube is sufficiently long).
If it seems odd that the shock is supersonic, that's likely because the shock Mach number is measured based on its speed relative to the medium
into which it is propagating (here, the stagnant air). But the shock is propagating
through the heated air behind it, which has a higher temperature and, thus, a higher speed of sound. It is still subsonic relative to that medium.
Case 2: Gradually-started piston
If the piston instead starts very slowly, it will not necessarily immediately generate a shock. A sufficiently slow ramp-up would produce a weak compression wave that will travel upstream at the speed of sound. Of course, that weak compression wave is also weakly heating the flow, thus increasing the sound speed behind it. As the piston slowly accelerates, each new wave front propagates into a medium which is slightly hotter and therefore propagates slightly faster than the wave in front of it. Thus, those waves all eventually (and fairly quickly) coalesce into a shock anyway, and the Mach number of that shock is driven by the final piston velocity when it stops accelerating. So if it accelerates to the same ##v_p## as before into the same medium, the shock will have the same ##M_s## and the problem is effectively identical once you hit steady piston motion
Of course I am trying to paint this picture as a series of discrete waves leaving the face of the piston at some finite time interval but the real process is continuous. You could certainly solve the 1D transient Euler equation to get an exact solution to the real problem, which would have a continuous generation of increasingly fast waves, but I don't think you'd learn much more than the simple thought experiment above.
What is the ramp-up time? It's going to be the same order of magnitude as the time it takes to ramp up the piston velocity since the waves are propagating faster than the piston.
Back to the question of force...
If you know the shock Mach number, ##M_s##, that uniquely determines the pressure ratio across the shock (for a given gas), ##p_2/p_1 = f(M_s,\gamma)##. If you have pressure and area, then you have force.
Some good sources
[1] Liepman, H. W., Roshko, A.1957. "Elements of gasdynamics." Dover.
[2] Anderson, J. D. 2003. "Modern Compressible Flow with Historical Perspective." McGraw-Hill.
[3]
https://www.grc.nasa.gov/www/k-12/airplane/normal.html
