Doesn't the subsequent change in effective nuclear charge and atomic radii disrupt the standing wave in which electron moves? Electrons move in standing waves around the nucleus. However, as the Atomic No. along a period increases, the effective nuclear charge on the valence electrons increases and the atomic radii decrease. 


*

*As the nucleus pulls the electrons closer, doesn't it disturb the wave by changing its length, considering the fact that standing waves can not be created at any random frequency or wavelength?

 A: In general for a specific nucleus, if you add, lets say one more proton and one  neutron such that the resulting nucleus will stay in the same period as well as add one more electron to the atom, then it will have more effective charge than earlier and will pull the outermost electron more towards itself. However, you need to keep in mind that the angular momentum will also be quantized in terms of 
$$ L = \dfrac{nh}{2\pi}$$
along with the above mentioned =  fact that the outermost electron will now face a different Coulumb potential. If you now approximate the atom as a Bohr atom, then you will have two equations
$$ mvr = \dfrac{nh}{2\pi}$$ 
$$ \dfrac{mv^2}{r} = \dfrac{kZ_{screened}e^2}{r^2}$$
and it is easy to see that these two equations always have a solution and that with increasing $Z_{screened}$  the radius becomes smaller. So the solution to your puzzle in terms of the Bohr atom is that a new stable orbit is formed at a lesser radius than earlier.
One cannot generalize this for a many atom system because of two facts, as the Bohr atom approximation doesn't hold exactly and the Schrodinger equation for a multi electron system isn't solvable analytically. But the intuition can be gained from this basic crude model that stable orbits will be formed at lesser radius than was there earlier and the same conclusion can be applied to standing waves.
