Physical reason behind having greater amplitude when driving frequency$ < $ natural frequency than that when driving frequency $>$ natural frequency

Question

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 1

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.

Answer 2

Let's say

$$m\dfrac{d^2x}{dt^2} + kx = \text{force}$$

The transfer function will provide insight about how the amplitude changes, depending on the frequency of force. Now what's the transfer function $\dfrac{X}{\text{Force}}$?

$$ms^2X + kX = \text{Force}$$ $$(ms^2 + k)X = \text{Force}$$ $$\frac{1}{ms^2 + k} = \frac{X}{\text{Force}}$$ with $s = j\omega = j2\pi f$ ($f$ being the frequency, not the $force$ as you can tell by the missing "orce") $$\frac{1}{-m(2\pi f)^2 +k} = \frac{X}{\text{Force}}$$

The transfer function depends on the frequency. For small frequencies $f=0$ it is $$\frac{1}{k} \approx \frac{X}{\text{Force}}$$ For large frequencies ($-m(2\pi f)^2 >> k$) it is $$\frac{1}{-m(2\pi f)^2} \approx \frac{X}{\text{Force}}$$

Answer 3

1. In the underdamped oscillator, it is physically intuitive that if the driven frequency aka. the forced frequency $\omega$ deviates significantly from the damped natural frequency aka. the sinusoid frequency aka. the ringing frequency $\omega_{\rm ring}$, then the external force spends half time supporting and half time counteracting the oscillation, leading to a small amplitude.

2. Interestingly, when $\omega$ is close to the $\omega_{\rm ring}$, the effect is lopsided, cf. OP's title question. If $\omega $ is slightly bigger than $\omega_{\rm ring}$, the system cannot react fast enough; while if $\omega$ is slightly smaller than $\omega_{\rm ring}$, the system adapts.

3. Mathematically, the characteristic frequencies are $$ \omega_{\pm}~=~\pm\overbrace{\underbrace{\omega_{\rm ring}}_{\text{oscillatory}}}^{\text{real}}-\overbrace{\underbrace{ib}_{\text{exp. decay}}}^{\text{imaginary}},\tag{1}$$ where $b$ is the friction coefficient.

4. The characteristic polynomial $$P(\omega)~=~(\omega-\omega_+)(\omega-\omega_-) ~\stackrel{(1)}{=}~(\omega+ib-\omega_{\rm ring})(\omega+ib+\omega_{\rm ring})\tag{2}$$ is the reciprocal transfer function.

5. It follows that the absolute square $$|P(\omega)|^2~\stackrel{(2)}{=}~(\omega^2+b^2-\omega_{\rm ring}^2)^2+4b^2\omega_{\rm ring}^2\tag{3}$$ is a parabola in the variable $\omega^2$.

6. The resonance frequency $$\omega_{\rm peak}~\stackrel{(3)}{=}~\sqrt{(\omega_{\rm ring}^2-b^2)_+}~:=~\sqrt{\max(\omega_{\rm ring}^2-b^2,0)}\tag{4}$$ is given by the minimum of $|P(\omega)|$.

7. In particular, we see that $$\omega_{\rm peak}~\stackrel{(4)}{<}~ \omega_{\rm ring},\tag{5}$$ which is essentially the mathematical content of OP's title question.

8. For more details, see e.g. my Phys.SE answer here.