Revised Answer (for earlier version click on "edited ... ago")
Reading your question as well as your comments it is still not quite clear what you mean.
I presume that the string of length $L$ with fixed ends A and B is set vibrating, as in the 1st figure below. The tension and mass per unit length determine the speed $v$ of travelling waves. The fundamental frequency is $f=\frac{2v}{L}$ (which comes from speed = wavelength x frequency).
P is a point on the string close to end B. It vibrates with frequency $f$. Note that if we force P to vibrate with frequency $f$ then it does not matter whether or not the section PB of the string exists : we could remove it without destroying the standing wave.

With the string at rest end B is then unfixed and forced to vibrate with the same motion that point P had in the 1st diagram. We now have the situation in the 2nd diagram.
If the moving end B has the same frequency $f$ as P, then a standing wave will not be formed on the string. The forcing frequency corresponds to a string of shorter length $L$, but the effective length $L'$ is now somewhat longer than $L$. (How much longer is difficult to say; see the discussion below.) The forcing frequency $f'$ should be correspondingly lower. The tension in the string and its mass per unit length are the same; the speed $v$ of waves on the string is the same. So :
$v = fL/2 = f'L'/2$.
The amplitude of B is not important. Any amplitude will work, but the frequency is crucial.
How much lower than $f$ does $f'$ need to be? I think this cannot be answered exactly. If $L'$ is only slightly greater than $L$, the section BB' will be very steep. This will require the midpoint of the string to have a very large amplitude. A real string may not have enough elasticity to extend so much. To maintain the same amplitude the increase should be approximately proportional :
$f'/f=L/L'=AP/L$.
If P were the midpoint of AB, then $L'=2L$ and $f'=\frac12 f$.