Each side of the contact point has a 2×2 stiffness matrix defined in local (body) coordinates

$$ \begin{align} {\bf k}_1^{\rm body} & =
\begin{vmatrix} k_1^{\rm long} & 0 \\ 0 & k_1^{\rm lat}
\end{vmatrix} & {\bf k}_2^{\rm body} & =
\begin{vmatrix} k_2^{\rm long} & 0 \\ 0 & k_2^{\rm lat}
\end{vmatrix} \end{align}$$

Now if each body has a 2×2 orientation (rotation) matrix defined by the angles $\psi_1$ and $\psi_2$ then

$$ \begin{align} {\bf R}_1 &= \begin{vmatrix} \cos\psi_1 & -\sin\psi_1 \\ \sin\psi_1 & \cos\psi_1 \end{vmatrix} & {\bf R}_2 &= \begin{vmatrix} \cos\psi_2 & -\sin\psi_2 \\ \sin\psi_2 & \cos\psi_2 \end{vmatrix} \end{align} $$

The effective 2×2 stiffness (symmetric) matrix in world coordinates is calculated with

$$ {\bf k} = {\bf R}_1 {\bf k}_1^{\rm body} {\bf R}_1^\intercal +{\bf R}_2 {\bf k}_2^{\rm body} {\bf R}_2^\intercal = \begin{vmatrix} k_{xx} & k_{xy} \\ k_{xy} & k_{yy} \end{vmatrix}$$ where $\Box^\intercal$ is the matrix transpose.

What you are looking for is the direction of minimum and maximum stiffness (which are 90°) apart where the contact point would try to move towards. This is found by finding the angle $\varphi$ which causes

$$\begin{vmatrix} \cos\varphi & \sin\varphi \\ -\sin\varphi & \cos\varphi \end{vmatrix} \begin{vmatrix} k_{xx} & k_{xy} \\ k_{xy} & k_{yy} \end{vmatrix} \begin{vmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{vmatrix} = \begin{vmatrix} k_{A} & 0 \\ 0 & k_{B} \end{vmatrix} $$

The solution is $$\varphi = \frac{1}{2} \tan^{-1} \left( \frac{2 k_{xy}}{k_{xx}-k_{yy}} \right ) $$

In addition, every integer $i$ multiple offset $\varphi+i \frac{\pi}{2}$ is also a solution.

I think you are misunderstanding what they mean by

If the springs are initially stretched very little from their relaxed length

You take it to mean "large displacement from equilibrium" - but I think it means "both springs are under tension in equilibrium". In that case there is a significant tension $T$ without displacement, and for small displacements $y$ the restoring force is approximately given by

$$F = T \sin\theta \approx T ~\frac{y}{a}$$

Because $T$ will not change much if there was quite a bit of initial tension, the system will be linear.

However, if there is little initial tension, then the restoring force will be mostly due to the additional stretching of the spring. In the case of zero initial tension, the restoring force (for small displacements) is given by

$$\begin{align}\\
F &= k\left(\sqrt{y^2+a^2}-a\right)\sin\theta\\
&=ka\left(\sqrt{1+\frac{y^2}{a^2}}-1\right)\frac{y}{a}\\
&\approx\frac{ky^3}{2a^2}
\end{align}$$

So when the increase in tension due to the lateral displacement is significant, the motion becomes non-linear.

**UPDATE**

We can look at this for both horizontal and vertical displacements. Assume the unstretched length of the springs is $L_0$, and the stretched length (in equilibrium) is $L\gg L_0$. If we displace by a small amount $dx$ horizontally, and a small amount $dy$ vertically (both $\ll LL$), then we can compute the horizontal and vertical forces on the mass.

First we compute the new length of the spring. This will be

$$L_1 = \sqrt{(L+dx)^2+dy^2}$$

for the spring on the left, and

$$L_2 = \sqrt{(L-dx)^2+dy^2}$$

for the spring on the right.

The net force on the mass is given by the horizontal and vertical components of the tension in the spring.

For the horizontal component, we notice that

$$F_x = T_1 \cos\alpha_1 - T_2\cos\alpha_2$$

for small angles this simplifies to

$$F_x = T_1 - T_2\\
= kL\left[\left(\left((1+\frac{dx}{L}\right)^2+\left(\frac{dy}{L}\right)^2\right)^{1/2}-\left(\left((1-\frac{dx}{L}\right)^2+\left(\frac{dy}{L}\right)^2\right)^{1/2}\right]\\
\approx 2kdx$$

For the vertical component of force, we approximate the sum of $T_1$ and $T_2$, and find that it varies only with higher orders of $dx$ and $dy$ - so we consider it ($T_1+T_2$) constant for small displacements. The vertical force $F_y = 2T\sin\alpha\approx 2T\frac{dy}{L}$ will then depend only on $dy$ and not on $dx$. So horizontal and vertical oscillations will be independent, and linear.

## Best Answer

## Summary

We do not know if the hanging weight will have a an instant of going downward or notIt depends on the mass and spring constant of the lower spring and mass of the hanging weight.

## Logic

We cut the rope. There is gravitational force on the weight. The springs begin to contract rapidly and contract until the strings are tight. (John Hunter’s answer assumes the mass has to descend to make the strings taut, but the contracting springs do most of the moving to tighten the strings).

Assume the strings have no mass.

After cutting but before tight strings, the [lower spring / hanging weight] is a free system not affected by the upper spring or weightless un-taut strings. Let’s consider this system during that time:

The system

as a wholeis only acted on by gravity so as mentioned by alephzero it will fall - it’s center of mass will accelerate downward at F/M = g, the acceleration of gravity, where M = m_spr + m_hw (spring and hanging weight).In addition to that, the spring (which has mass) will contract. As alephzero said, if the hanging weight has a much much higher mass than the lower spring, the mass will go down as this system falls (until the strings are tight).

But what if the spring doesn’t have a mass that is negligible compared to the hanging weight? There are cases where the spring constant and/or spring mass are high enough that the mass never (even instantaneously) has a downward acceleration of velocity. The spring contracts with enough inertial force that it’s instantaneous force on the weight is greater than m_hw g.

To be sure there are such cases, take an extreme case where the mass of the hanging weight is negligible compared to the spring’s mass and the spring constant is high. The spring will begin to fall under its own weight while contracting rapidly about its center and its lower end will go up (hence the hanging weight will)