3.5.31. When the forcing's frequency is small as compared to the oscillator's natural frequency, its displacement can be considered equal to $F(t)/k$, where $F(t)$ is the oscillating force, $k$ is the rigidity of the oscillating system. At high frequencies of the forcing force, the acceleration of the oscillator can be considered equal to $F(t)/m$, where $m$ is the mass of the oscillator. Explain this.
Solution
Assuming that the forced oscillation dominates the free oscillation, when the particular solution $x=A\sin(\omega t)$ is substituted into the equation of motion
$$m\ddot x+kx=F(t)=F_0\sin(\omega t),$$
we see that the relative magnitude of the two terms on the left are $m\omega^2:k=\omega^2:\omega_0^2$. If $\omega\ll\omega_0$, then $kx\approx F(t)$, so $x\approx F(t)/k$. On the other hand, if $\omega\gg\omega_0$, then $m\ddot x\approx F(t)$, so $a=\ddot x\approx F(t)/m$.