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While studying about the Resonance and Forced Oscillations, I came across a graph in my textbook that is given below:-enter image description here

Now, the author writes

As the amount of damping increases, the peak shifts to lower frequencies.

Why does this happen? And does this imply that at higher damping levels one cannot achieve a higher amplitude by setting the period of the forced oscillation to be equal to the natural frequency? This seems strange because, the highest amplitude is achieved when the natural frequency is equal to the driving frequency. But, I guess that the rules are somehow different for the damping case.

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Mathematical demonstration

It's straightforward to see why this happens if you use a bit of linear response theory. Consider a generic damped harmonic oscillator. There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$. Newton's law says $F(t) = m \ddot{x}(t)$ which gives $$-k x(t) - \mu \dot{x}(t) + F_\text{drive}(t) = m \ddot{x}(t) \, .$$ Dividing through by $m$ and defining $\phi(t) \equiv x(t)/m$, $\omega_0^2 \equiv k/m$, $2 \beta \equiv \mu/m$, and $J(t) \equiv F_\text{drive}(t)/m$, we get $$ \ddot{\phi}(t) + 2\beta \dot{\phi}(t) + \omega_0^2 \phi(t) = J(t) \, .$$ This is a nice general form of the damped driven harmonic oscillator.

Writing $\phi(t)$ as a Fourier transform $$\phi(t) = \int_{-\infty}^\infty \frac{d\omega}{2\pi} \, \tilde{\phi}(\omega) e^{i \omega t}$$ and plugging into the equation of motion, we find $$\left( - \omega^2 + i 2 \beta \omega + \omega_0^2 \right) = \tilde{J}(\omega)$$ which can be rewritten as $$\tilde{\phi}(\omega) = - \frac{\tilde{J}(\omega)}{\left( \omega^2 - i 2 \beta \omega - \omega_0^2 \right)} \, .$$

Let's take the case where the drive is a cosine, i.e. $J(t) = A \cos(\Omega t)$. In this case $\tilde{J}(\omega) = (1/2)\left(\delta(\omega - \Omega) + \delta(\omega + \Omega) \right)$ so if you work it all out you find $$\phi(t) = \Re \left[ - \frac{A e^{i \Omega t}}{\Omega^2 - i 2 \beta \Omega - \omega_0^2} \right] \, .$$ It's easy to check that $\phi(t)$ has the largest amplitude when $\Omega = \omega_0 \sqrt{1 - 2 (\beta / \omega_0)^2}$, which decreases as $\beta$ increases. Remember that $\beta$ is just proportional to the coefficient of friction $\mu$ so we've shown that more friction makes the peak move to lower frequency.

Resonance

We've shown that the amplitude of the oscillator depends on the damping coefficient. However, this does not mean that the resonance moves to lower frequency. Resonance is a condition defined by unidirectional flow of energy from the drive to the system. It turns out (easy to show with the math we already did) that this happens when $\Omega = \omega_0$, i.e. the drive is at the same frequency as the undamped oscillation frequency. There's already a nice post on this issue which I recommend reading.

Original questions

Why does this happen?

Well, we showed why mathematically. Intuitively it's because the friction takes away kinetic energy so the oscillator doesn't make it as far from equilibrium on each cycle.

And does this imply that at higher damping levels one cannot achieve a higher amplitude by setting the period of the forced oscillation to be equal to the natural frequency?

Assuming constant amplitude of the drive, yes.

This seems strange because, the highest amplitude is achieved when the natural frequency is equal to the driving frequency. But, I guess that the rules are somehow different for the damping case.

Indeed, damping changes things a bit.

Other reading

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  • $\begingroup$ nice. No integration boundaries needed for the FT, or are they implicit? $\endgroup$ – Gert Jan 9 '16 at 2:15
  • $\begingroup$ These are basically the same solutions as for the RLC-circuits, IIRW. $\endgroup$ – Gert Jan 9 '16 at 2:22
  • $\begingroup$ @Gert Yes of course. An RLC (series or parallel, they both work) is a damped harmonic oscillator. $\endgroup$ – DanielSank Jan 9 '16 at 2:32
  • $\begingroup$ Interesting page on FT and DEs, here: thefouriertransform.com/applications/differentialequations.php (if I can grasp it, any id-t can! ;-) ) $\endgroup$ – Gert Jan 9 '16 at 2:59
  • $\begingroup$ Very nice - I like the link to your earlier question about resonance, and the reply it got, as well. $\endgroup$ – Floris Jan 9 '16 at 6:10
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I'll try to answer your question on a conceptual basis, because the math involved would tend to obscure the concept. For a mass that is bouncing up and down on a spring in air, the system is almost totally undamped, and the spring will oscillate at a frequency that depends on the spring constant and the total mass that is oscillating. If I take this same system and place the mass in a liquid such as water, there will be substantial drag forces on the mass, and the frequency of the oscillation will decrease. In addition, the amplitude of the oscillations will exponentially decay because of the amount of damping involved.

For resonance to occur, I would have to drive the undamped or damped system at its natural frequency. In doing so, the maximum amplitude of an undamped system would tend to increase without bound (it would break), while the maximum amplitude of the damped system would be limited by the damping involved, with larger damping resulting in a smaller maximum amplitude, because the damping agent is acting to absorb energy put into the system. This means that the height of the maximum amplitude under resonance conditions will progressively drop as the amount of damping goes up, and the increasing drag forces will also cause the resonant frequency to decrease, as shown in your drawing.

I realize that a spring-mass system is a super simple example, but rest assured, the math that is used to describe this system would be very similar to the math used to describe many other types of oscillating systems. Does this explanation provide enough of a conceptual understanding to explain your question?

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    $\begingroup$ In that crucial sentence: "This means that the height of the maximum amplitude under resonance conditions will progressively drop as the amount of damping goes up, and the increasing drag forces will also cause the resonant frequency to decrease, as shown in your drawing.", can you explain why response peak moves down in frequency? Also note that the resonance does not move down in frequency. As you say, the resonance is always at the undamped frequency. See here. $\endgroup$ – DanielSank Jan 9 '16 at 17:22
  • $\begingroup$ The response peak moves down in frequency due to the drag forces introduced by the damping medium. In addition, the damping medium is not independent of the system, so those same drag forces do affect the resonant frequency. $\endgroup$ – David White Jan 9 '16 at 17:44
  • $\begingroup$ No reply intended - moderators, please delete. $\endgroup$ – David White Jan 9 '16 at 17:48
  • $\begingroup$ The peak in the amplitude response and the resonance do not occur at the same frequeny. I posted a link to another Physics.SE post explaining this at the end of my previous comment. Please take a look. $\endgroup$ – DanielSank Jan 9 '16 at 17:50
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  1. The math of the underdamped oscillator becomes particularly pretty if we use a complex-valued position vector $x$. [I.e. the real part ${\rm Re}(x)$ represents the physical position.] Just use Fourier transformation $$\tag{1}\ddot{x}+2b\dot{x}+\omega_0^2 x~=~f\qquad\stackrel{\text{Fourier transf.}}{\Leftrightarrow}\qquad -\overbrace{\underbrace{(\omega^2+2ib\omega -\omega_0^2)}_{P(\omega)=(\omega-\omega_+)(\omega-\omega_-)}}^{\text{characteristic polynomial}}\tilde{x}~=~\tilde{f}.$$

  2. Let us assume in this answer that the oscillator is underdamped, i.e. that $\omega_0^2>b^2$. The characteristic frequencies $$\tag{2}\omega_{\pm}~=~\pm\overbrace{\underbrace{\omega_{\rm ring}}_{\text{oscillatory}}}^{\text{real}}-\overbrace{\underbrace{ib}_{\text{exp. decay}}}^{\text{imaginary}}$$ are the complex frequencies that the system would choose if there were no external force $f$. Here $$\tag{3} \omega_{\rm ring}~:=~\sqrt{\omega_0^2-b^2} $$ is known as the ringing frequency, sinusoid frequency, or damped natural frequency.

  3. The resonance frequency $$\tag{4} \omega_{\rm peak}~=~\sqrt{(\omega_0^2-2b^2)_+}~:=~\sqrt{\max(\omega_0^2-2b^2,0)}$$ is the minimum point for the absolute value $$\tag{5} |P(\omega)|~=~\sqrt{(\omega^2-\omega^2_0)^2+4b^2\omega^2}, \qquad \omega \geq 0,$$ of the characteristic polynomial. This corresponds to maximum gain (or peak transmissibility) of the forced oscillator.

  4. Assume for simplicity that $\omega_0^2>2b^2$, so that the resonance frequency $\omega_{\rm peak}>0$ is non-zero. Then the square $$\tag{6} \omega_{\rm ring}^2~=~\omega_0^2-b^2$$ of the ringing frequency sits precisely between the square $$\tag{7} \omega_{\rm peak}^2~=~\omega_0^2-2b^2$$ of the resonance frequency and the square $\omega_0^2$ of the undamped natural frequency.

    $$ \begin{array}{rccc ccl} &--|--&--&--&--&--|-- \hspace{2ex} --|-- \hspace{2ex} --|--&--> \quad\omega^2\cr &0&&&&\omega_{\rm peak}^2 \hspace{7ex} \omega_{\rm ring}^2 \hspace{7ex} \omega_0^2&\cr &&&&& \underbrace{\hspace{11ex}}_{b^2} \underbrace{\hspace{11ex}}_{b^2} &\end{array} $$ $\uparrow $Fig. 1: The resonance frequency $\omega_{\rm peak}$ and the ringing frequency $\omega_{\rm ring}$ decrease with increasing friction $b$.

  5. Now let us return to OP's question:

    Why does the peak shifts to lower frequencies as the amount of damping increases?

    Answer: OP is essentially asking for intuition behind the negative coefficient in front of $b^2$ in eq. (7). Here is one argument: It is intuitive that the resonance frequency (7) and the ringing frequency (6) would shift in the same direction for growing friction $b$, i.e. it is intuitive that the signs in front of $b^2$ in eqs. (6) and (7) are the same. Moreover, the negative coefficient in front of $b^2$ in eq. (6) has a physical meaning: When we increase friction $b$, at some point the oscillator becomes overdamped with purely imaginary characteristic frequencies (2). This transition only happens if the coefficient in front of $b^2$ is negative, which answers OP's question.

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You already have a good mathematical answer, so I will focus on an answer with almost no equations.

I take it you understand the basic mathematics of the simple harmonic oscillator.

When you add damping, the amount of energy you lose per cycle depends on the velocity: the faster you go, the more energy you lose (at the same amplitude) because the force scales with $k\dot{x}$.

Of course the velocity is proportional to the frequency - so an oscillator being driven at a higher frequency will lose more energy per cycle than an oscillator driven at a lower frequency.

On the other hand, the best coupling of energy into the system happens when the driving force is exactly 90 degrees out of phase with the amplitude (so force is in phase with the velocity) - which happens at the undamped resonant frequency.

As you increase the amount of damping, the "more energy lost per cycle" factor starts to beat the "more energy coupled in per cycle" factor. And that means that the largest response amplitude shifts to lower frequencies.

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  • $\begingroup$ The term "resonance" is unfortunately abused when discussing damped systems (and all systems are damped, so the term gets a lot of abuse). I'd remove the phrase "effective resonance" in favor of the clearer phrase immediately following it in parentheses. $\endgroup$ – DanielSank Jan 9 '16 at 9:04
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But, I guess that the rules are somehow different for the damping case.

In both cases the outcomes are the solutions of the (differential) equations of motion. In the case of the driven, undamped oscillator:

$$m\frac{d^2 x}{dt^2}+kx=F_0\cos(\omega t + \phi_d)$$

$\omega$ is the angular speed of the driving force. Solving it gives the result in your graph.

Source.

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  • $\begingroup$ Just wanted to know why do the peaks shift to the left for higher damping... $\endgroup$ – model_checker Jan 8 '16 at 19:25
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    $\begingroup$ @ShreyAryan: well, I look forward to DS's answer but for me the DEs do the talking. I'm not sure a more intuitive understanding is possible here. $\endgroup$ – Gert Jan 8 '16 at 20:02
  • $\begingroup$ This doesn't really answer the question. Do you think you could provide some qualitative analysis of the differential equation so that we can see why the peak response goes down in frequency as friction increases? The friction is the key, but this answer doesn't even include a friction term in the equation of motion! $\endgroup$ – DanielSank Jan 9 '16 at 9:02

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