Coupled Oscillations: no beating effect Consider the famous coupled oscillation problem of 2 spring pendulum:

In a special case the solution can be given as follows:
$x_1 = \displaystyle \frac{C_1}{2}\,\cos(\omega_1\cdot t)+\frac{C_1}{2}\,\cos(\omega_2\cdot t)$
$x_2 = \displaystyle \frac{C_1}{2}\,\cos(\omega_1\cdot t)-\frac{C_1}{2}\,\cos(\omega_2\cdot t)$
where the starting conditions where be chosen like:
$x_1(0) = 2\,C_1 \quad x_2(0) = 0 \quad\dot{x}_1(0) = 0 \quad \dot{x}_2(0) = 0$
Notably the frequencies are expressed by:
$\omega_1 = \displaystyle \sqrt{\frac{D_1}{m}}$ and $\omega_2 = \displaystyle \sqrt{\frac{D_1+2\,D_2}{m}}$

But here is my problem: under these conditions the phenomena is called beating effect.
Assuming $D_1 = D_2 = 1\,\frac{N}{m}$ and plotting the graph
I do not get the typical patterns:

Only if I chose $\omega_1$ fairly close to $\omega_2$, thus $D_2$ to be small, the model emerges:

Classical Question: why is that only occurring under certain constraints?
 A: You can use trig identities to write
$$
x_1 = C_1 \cos [( \omega _1 +\omega _2) t/2] \cos [(\omega_1 -\omega _2)t/2]. 
$$
This function can be interpreted as a “fast” oscillation with frequency $(\omega_1 + \omega_2)/2$ and a “slow” oscillation with frequency $(\omega_1 - \omega_2)/2$.
This much is true for any value of the spring constants. But to really perceive this signal as a “beating” signal, the “fast” oscillation must oscillate many times during each oscillation of the “slow” oscillation.  In your first graph, the ratio of the fast oscillation to the slow oscillation is only about 2.5, and so it is difficult to perceive the slow oscillation as a modulation of the fast oscillation. In the second example, in contrast, the ratio is about 15, the time scales are well separated, and it is straightforward to interpret the signal as a modulated oscillation.
It would be instructive to superimpose the envelope of the modulation,
$$
\pm  \cos [(\omega_1 -\omega _2)t/2], 
$$
on each of your graphs to see how this manifests. (I’m on a phone now or I would create such graphs myself.)
