What will be the reading of spring balance in the following situation? Many of us have seen this question many times ( a classic question of almost every intermediate level mechanics course ) where two blocks of the same mass are join two spring balance as follows :

where $m_{1}=m_{2}$
Now I know that the reading of the spring balance will be $=m_{1}=m_{2}$
But the question is what if $m_{1} \neq m_{2} $
$\text{say} \quad\left(m_{2}>m_{1}\right)$
Let me put my question precisely as follows
Question 1
There is a spring block system as shown in the figure where $m_{1} < m_{2}$. Initially the whole system is at rest as we apply the necessary force at the bottom of $m_{2}$.
What will happen when we release $m_{2}$ ? What will be the reading of the spring balance throughout this motion ? Will this reading be constant or variable ?
Question 2 (related to previous question )
Take the previous operators again this time $m_{1}=m_{2}$. Initially the system is at rest.
Now we start adding small masses at $m_{2}$. What will be the reading of the spring balance throughout this motion ? Will this reading be constant or variable ?
Note : Everything except the blocks are mass-less. There is no friction. You may take spring constant of the spring balance as $k$.
My confusion
I am very confused regarding what will happen to the spring in the first question. Let me explain.
Since it's an ideal mass-less spring show the forces on both of its end must be equal where is in the first question the moment we release $m_{2}$.
According to my wisdom it should look like this   ( I hope the direction of forces are quite clear by their names ) where  $m_{2} \vec{g}>m_{1} \vec{g}$
It is quite clear that this spring cannot be stretched anymore as there's not enough force on the other side to stretch the spring.
Show the net force on the spring balance is equals to
$$
\vec{F}_{\text {net }}=\left(m_{2}-m_{1}\right) g
$$
But remember the spring balance is mass less so it means infinite acceleration which does not make any sense.
But if the the acceleration is taken as finite then the spring will stretch in order to move block $m_{1}$ upwards, and eventually the tension in the string connecting spring balance and block $m_{1}$ will become constant when whole the system will start moving with the same acceleration ( during the elongation of spring the acceleration of block $m_{2}$ decrease ... I will complete this later as my theory is getting confusing to myself.
 A: We will start with much more general case of your question. Where there is a spring block system as shown in the figure $m_{2} > m_{1}$, initially both the blocks are at rest and the spring is not stretched (at its natural length). Now we will try to figure out the motion of both the blocks when they are released.
System in initial state
The blue block is heavier.

NOTE
The following arrangements shown in the diagram below are equivalent ( I have drawn block around Springs just for the sake of my convenience in making diagrams so please ignore the blocks at the ends of the spring ).

Let us now analyze the given situation and try to figure out the motions of the blocks.
When released the blocks will try to move downwards due to gravity but when they start moving the string becomes taut and tension force comes into the play. Since in the string is inextensible so when the block tries to moves downward the string will also try to move but since it is attached to the spring so it will start to stretch the spring. Let us analyze the situation when a small instant of time ( say $dt$ ) is passed.
The spring will stretch more at that end where it is connected to the heavier block as shown in figure below, note that the spring have stretched by half square more on the side where it is connected to the heavier block.
Let's say that the spring is stretched by a small length $dx$ so,
$$
\vec{F}(x)=\vec{T}=k d x
$$
where $k$  is the spring constant. Note that tension in string $\vec{T} \neq m_{1} \vec{g}\neq m_{2} \vec{g}$ because if  $\vec{T} = m_{1} \vec{g}$ then block $m_{1}$ would not move at all which is obviously incorrect.
At this moment both the blocks are moving downward because the forces acting on them are unbalanced. As both the block move down further the elongation in the spring will increase. It will stretch more at end attached to the heavier block and will stretch less at the end attached to the lighter block. Over time following changes will occur in the system:

*

*Tension force will increase due to further elongation of spring. $$
\vec{F}(x)=\vec{T}=k x
$$

*Acceleration of both blocks will decrease.

*Velocity of blocks will increase since the net force is still downwards. Due to greater acceleration of  $m_{2}$
$$ \vec{V_{2}} >  \vec{V_{1}}$$

Eventually a time will come when  $\vec{T} = kx = m_{1} \vec{g} $
$$\Rightarrow \vec{F_{net}}(m_{1})=m_{1} \vec{g} - \vec{T} = 0$$
$$\Rightarrow \vec{F_{net}}(m_{2})=m_{2} \vec{g} - \vec{T} = (m_{2}-m_{1}) \vec{g} $$
Now, since both the blocks still have their velocity downwards so the spring will stretch further, therefor increasing tension force which will further decrease the acceleration of block $m_{2}$ (still the velocity of $m_{2}$ is increasing) but decelerating block $m_{1}$ (as now the net force on block $m_{1}$ is upwards ).
Eventually a time will come when the velocity of block $m_{1}$ will becomes zero, at this point there are several possibilities for the acceleration of block $m_{2}$ depending on the values of $m_{1}$ & $m_{2}$.
The velocity of block $m_{2}$ will downwards ( for any value of $m_{1}$ & $m_{2}$ since $ \vec{V_{2}} >  \vec{V_{1}}$ and block $m_{1}$ starts to decelerate earlier than $m_{2}$ ) but the net force on the block $m_{2}$ maybe upwards or downwards.
CASE 1 ( when net force on $m_{2}$ is upwards)
In this case, block $m_{2}$ will decelerate and block $m_{1}$ accelerate, the spring will first stretch ( when $ \vec{V_{2}} >  \vec{V_{1}}$) then elongation of spring stop ( when $ \vec{V_{2}} = \vec{V_{1}}$)  and then the spring will shrink ( when $ \vec{V_{2}} <  \vec{V_{1}}$ as the block $m_{1}$ accelerates and the block $m_{2}$ decelerates ) and so on ... the motion will continue. I hope you will be able to figure out the rest of the motion in this case.
Here I have created a simulation for your help. Heavier block is on the right side.

Increasing the length of rope of lighter block.


Note: The spring block system appears to move right, it is because stretching of spring at the right side was more when blocks were moving downwards due to higher acceleration of block $m_{2}$ whereas while coming upwards the compression for spring is more at the left end due to higher velocity and acceleration of block $m_{1}$.

CASE 2 ( when net force on $m_{2}$ is downwards)
In this case, block $m_{2}$ will accelerate downwards and block $m_{1}$ accelerate upwards, the spring will further stretch as $ \vec{V_{2}} >  \vec{V_{1}}$ eventually
$ \vec{V_{2}} = 0  $ then $ \vec{V_{1}}$ will be upwards so the spring start to shrink and so on ... the motion will continue. I hope you will be able to figure out the rest of the motion in this case also.
Here I have created a simulation for your help. Heavier block is on the right side.

This time also the spring block system seem to move right and you can figure out yourselves Why is it so.
After such long discussion you will be able to answer your questions yourselves but I am writing them here in brief.
Question 1
Since in this scenario the extra weight of heavier block is balanced by some external force and whole system is in a state of rest initially so it means that the net force on the lighter block is zero which implies that the tension force in the string is equals to weight of lighter block. When we release the heavier block its speed will increase downward and the speed of lighter block will increase upward since acceleration of heavier block is higher so it will gain speed faster stretching the spring eventually the heavier block will start to decelerate and all this will lead to a motion very similar to the one shown in case one.
Question 2 (I am bit skeptic about this one but here is what I think will happen in this case. Any suggestions would be appreciated.)
In this scenario also since the system is at rest initially it means that the net force on the the blocks is zero which implies that the tension in the string is equals to their weight and the spring is already stretched now as we start adding small weights on one of the blocks this block will try to move down and stretches the spring for the more just a bit more which makes the force on the other block and balance accelerating it up since the spring was stretched only by a very small length so it reaches its natural length very quickly giving both the blocks very small velocities which results in the very small amplitude SHM (simple harmonic motion) of these blocks almost like small vibration and as we keep putting weights this velocity is increased so is the vibrations.
A: If the spring balance has zero mass then the net force on it must always be zero (otherwise it would have infinite acceleration). So the two tensions on either side of the spring balance must always be equal and opposite even when the masses $m_1$ and $m_2$ are accelerating.
Suppose $m_2 > m_1$ and so $m_2$ is accelerating downwards with acceleration $a$ and $m_1$ is accelerating upwards with acceleration $a$ (so this is a steady state solution). If the tension in the right hand string is $T_1$ and the tension in the left hand string is $T_2$ then we have
$T_1 - m_1g = m_1a 
\\ m_2g - T_2 = m_2a
\\\Rightarrow (m_2-m_1)g +(T_1-T_2) = (m_1+m_2)a$
But our reasoning above shows that $T_1=T_2$, so we have
$\displaystyle a = \frac {(m_2-m_1)}{m_1+m_2}g
\\ \displaystyle \Rightarrow T_1=T_2=m_1(g+a)=\frac {2m_1m_2}{m_1+m_2}g$
and this is what the spring balance will read (note that a spring balance actually measures force not mass).
