# 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?

• Typo: When you give the frequencies you wrote $ω_1$ twice Aug 15, 2021 at 21:51
• Beats will occur for a slowly varying envelope amplitude, which will beat with frequency $\omega_1-\omega_2$, so choose your parameters accordingly. Already you see in your last example that a beat structure becomes apparent. Aug 16, 2021 at 0:52

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 $$(\omega_1 - \omega_2)/2$$.
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.)