You said you applied a constant displacement to the base. In that situation, if the structure doesn't have any modes that can are excited at a particular frequency, the response is going to be approximately the same as a rigid body being moved with a fixed amplitude, which is what your plot shows.
Note that if you apply a fixed displacement, the excitation force will increase for higher frequencies (and except where there is a resonance in the displacement response) it will be approximately proportional to the frequency squared.
If you applied a constant force, you would see the displacement falling as the frequency increased, but unless the structure is constrained in some way, the displacement amplitude will increase to infinity as the frequency gets close to zero.
Another way to understand this is think abut a 2 DOF system being excited at a frequency much higher than its two resonant frequencies. In that situation the elastic forces are small compared with the inertia forces, so we can ignore the stiffness term in the equation of motion and approximate it as $$\begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix} \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \end{bmatrix} = \begin{bmatrix}F \\ 0 \end{bmatrix} $$ where F is the excitation force applied at point 1.
Now if we apply a constant amplitude displacement $X$ at point 1, we have $\ddot x_1 = -\omega^2X$ and the two unknowns in the equations are $\ddot x_2$ and $F$.
From the second equation we have $$\ddot x_2 = - \frac{m_{21}}{m_{22}} \ddot x_1$$ which means that $$x_2 = - \frac{m_{21}}{m_{22}}X.$$
In other words, for a constant base amplitude, the tip amplitude of your tuning fork will also be approximately constant at high frequencies, except when there is a resonance that affects the motion of the system.