There is a large difference between the frequencies of the external...

When an external periodic force is applied to a body, it executes forced vibrations. The body vibrates with the frequency of the applied external periodic force, not its natural frequency.

The amplitude of forced vibrations is given by $A = \dfrac{F_0}{m \sqrt{(\omega^2 - \omega_0^2)^2 + (b\omega/m)^2}}$, where $\omega$ is the frequency of the external force and $\omega_0$ is the natural frequency of the body.

Since there is a large difference between the frequency of the external periodic force and the natural frequency, the term $(\omega^2 - \omega_0^2)^2$ is very large.

Consequently, the amplitude $A$ of the vibration becomes very small. Thus, the body will vibrate with a small amplitude at the frequency of the external periodic force.

Answer: vibrate with small amplitude at the frequency of external periodic force.