This is quoted from A.P.French's Vibrations & Waves:
If the driving force is of low frequency relative to the natural frequency, we would expect the particle to move essentially with the driving force. This is equivalent in saying that $m\dfrac{d^2x}{dt^2}$ in $$m\dfrac{d^2x}{dt^2} + kx = F_0\cos\omega t$$ plays small role compared to the term $kx$. The amplitude is controlled by spring constant.
On the other hand, at frequencies very large compared to the natural frequency, the opposite situation holds. The $kx$ becomes small compared to $m\dfrac{d^2x}{dt^2}$. In this case, we expect a relatively smaller amplitude of oscillation than the above case.
$\bullet$ Why does $kx$ become prominent in the first case & not in the second case?
$\bullet$ What is the physical reason behind that the first case has greater amplitude & the second one smaller?
Answer
This is just a footnote to Name's answer (which you should accept because it's correct) to give a slightly more intuition based argument.
If the driving force changes slowly compared to the natural frequency of the system then the system can move fast enough to stay in phase with the driving force. So most of the time the system is already moving in the direction the driving force is pushing it, and the force will accelerate the motion so the resulting amplitude of the oscillation will be big.
If the frequency of the driving force is a lot higher than the natural frequency of the system then the system cannot move fast enough to stay in phase with the driving force. This means some of the time the driving force is acting opposite to direction the system is moving, so it's slowing the motion not accelerating it. This means the amplitude of the motion will be reduced.
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