Robin04 said:
No, absolutely not. Sorry for my late reaction.
I tried it with a specific ##\Phi_1 = Asin{\omega t}## and I got ##\Phi_i=sin{\omega t}[A(1-\frac{ml^2\omega^2}{k})^{i-1}]## and it surprises me as I got no phase difference. Do you think this is correct in the g=0 case?
You can do something more general: let ##\omega^2 = \frac{k}{ml^2}##. Then, setting ##D = d/dt##, your DE reads as
$$(D^2 - \omega^2) \Phi_i + \omega^2 \Phi_{i+1} = 0.$$ If you assume
$$ \Phi_0 = A \cos(\mu t) + B \sin (\mu t)\hspace{4ex} (1)$$ we have
$$-(\mu^2+\omega^2) \Phi_0 + \omega^2 \Phi_1 = 0,$$ hence
$$\Phi_1 = \beta \Phi_0, \; \text{where} \; \beta = 1 + \frac{\mu^2}{\omega^2}.$$Similarly, ##\Phi_2 = \beta \Phi_1 = \beta^2 \Phi_0,## etc.
In general, ##\Phi_n = \beta^n \Phi_0,## for the specific form of ##\Phi_0## in eq. (1).
More generally, if you have a Fourier series expansion for ##\Phi_0## of the form
$$\Phi_0 = \frac{1}{2} A_0 + \sum_{i=1}^\infty \left( A_i \cos(i\, \mu t) + B_i \sin(i \, \mu t) \right), $$ then you have
$$\Phi_n = \frac{1}{2} A_0 + \sum_{i=1}^\infty \beta_i^n \left( A_i \cos(i\, \mu t) + B_i \sin(i\, \mu t) \right),$$
where
$$\beta_i = 1 + \frac{i^2 \,\mu^2}{\omega^2}.$$
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