I apologize for the vagueness. "Hardest to approximate" does require certain definitions first.

The reasoning behind this is simple continued fractions, ie. fractions like a_0 + 1/(a_1 + 1/(a_2 + ...)) = [a_0; a_1, a_2, ...], with a_i\in N. Every irrational number corresponds uniquely to an infinite continued fraction, and the finite "steps" of the SCF are the fractions that best* approximates the irrational numbers.
* a/b is "best" at approximating x, if b|x-a/b|<=d|x-c/d| for all c/d\in Q such that d <= b. The name "best" is to distinguish these from the "good" (without the b&d- weights), and I think its also known as "best approximation of of second kind".

The convergents, the finite "steps" of the SCF, are exactly these "best" approximations[1]. Such convergents are include 355/113 for pi, and are used for many things, like pianos and most of the different systems of leap years. Fascinating stuff, really.
[1] IIRC, http://www.math.hawaii.edu/~pavel/contfrac.pdf contains a full proof.

The size of the a_is determines when there's going to be a jump in denominator size. The 355/113 approx of pi is right before an 292, which is fairly large, and the next convergent is 103993/33102. Phi, being [1;1,1,...], never reaches any such jump in denominator size, and its sequence of convergents (its best approximations, and for phi it's the ratio of fibs) converge slower than any irrational not having a trail of ones at the end. From this, one may consider it the number "least like a rational", or even "the most irrational number".

Its properties are not directly related to the square root of five, as far as I can tell, but it is in this way the uniquely (at least as an infinite tail of a SCF) hardest irrational to approximate.

That being said, my initial comment was intended to point out that phi and the fibonacci numbers is quite special, and its special enough that it "should" occur frequently in nature. I never actually meant to comment too deep on the spiral-parts, because I know fairly little of them. My "reduce periodicity"-argument is only based on the thought that pi with its fourth convergent 355/113 would almost have a period of 113 (off by ~10^{-7}), while phi with its 11th-ish convergent 233/114 would have a not-very-almost-period 114 (off by 0.5). Phi's ~10^{-7}-almost-period would be 1597 from its 16th-ish convergent. While one could just take any number, say 123012/153281=[0; 1, 4, 15, 1, 1, 1, 2, 1, 4, 1, 1, 1, 1, 6] and claim that has a longer period, if I calculated correctly, it has a ~10^{-7}-almost-period of 871 from its seventh convergent 699/871, which is fairly less than phis. Note: one should probably even multiply the error 10^{-7} with the period for a more correct result, but as I hinted to, this is not my strongest subject.

Now, to your example: 1.62 has SCF [1; 1, 1, 1, 1, 1, 2, 2], all quite low, so it should have a fair amount of this "irrationality" that the fibs-ratios have.
It doesn't have to be the fibonacci-ratios exactly (though these would be the best choice), but most* numbers trying to optimize on this property will be close to them.
* I won't say all, because [x,y,1,1,1,1...] could possibly inherit some properties, but the number itself, 1/(x+1/(y+1/phi)), could be far from phi.

EDIT: I only realized now that your plant was hypothetical. Anyway, a quick search yielded this https://www.mathsisfun.com/numbers/nature-golden-ratio-fibon...
which (if you ignore the "for-kids" language) has fairly good display for one of the properties, and actually mentions the continued fraction.
EDIT2: For a more serious article, see
https://plus.maths.org/content/chaos-numberland-secret-life-...
which, all the way down at the bottom, explains that the numbers of SCF ending in [1,1,...] are "noble", and occur frequently as they "are least susceptible to being perturbed into chaotic instability."