When an oscillating object undergoing SHM is damped the peaks of the curves becomes flatter and they shift towards the lower frequencies. I understand that the damping causes a loss in energy and this is reflected in reduced amplitude but why do the peaks tend towards the lower frequencies (left on the graph)?
[Physics] Resonance graphs
resonance
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Your understanding of resonance seems about right on a qualitative level. If one were to ignore losses like friction, drag, or the like, "driving" a system at its resonance frequency would indeed result in feeding it more and more energy which is stored in the form of a large amplitude of the oscillation. For a completely lossless system, the amplitude would grow to infinity. In reality, every system is lossy to a certain degree but for "high-Q"$^1$ systems (meaning the losses are small), the amplitude can grow so big that they rip the system apart. This is referred to as a resonance catastrophy and is what happened to the Tacoma Narrows Bridge.
Generally, the losses will be proportionally to the amplitude of the oscillation and also to their frequency. I.e. at resonance where the amplitudes are big, so are the losses. Another way to think about it is in terms of energy transfer. At resonance, it is easy for the object to pick up energy from the driver. The energy stored in the object grows at first, but it will eventually be balanced by the increasing losses. Instead, if the system is driven off resonance, it has trouble picking up energy from the driver, so the stored energy decreases (and so do the losses) until the losses balance the little energy intake that is still provided by the driver.
In summary, at equilibrium (meaning in the steady-state), the object will always dissipate as much energy as it picks up from the driver, otherwise its energy content (and thus the amplitude) would change. At resonance, it is easy for the object to do that, so there also the energy dissipation is greatest.
$^1$ Quality factor or $\mathit{Q}-\text{factor}$
Keep in mind that the bold text isn't a derivation, its a way to qualitatively understand, so it's going to be very very unconvincing. Instead of imaging that you are trying to compute a number imagine you are trying to estimate it by a factor of 10 or 100 or even a factor of 1000 or more. That's how unconvincing it will be.
So. At equilibrium amplitudes, power supplied equals power lost to friction. How much energy is lost to friction? Well, when no power is supplied it basically dissipates almost all of the available energy $E$ in time $\tau$. So friction saps away something like $E/\tau$ as a power lost to friction. So the power supplied is something like $E/\tau$ too.
The conclusion is that we expect the power supplied and the power lost to be somewhat equal to the energy stored divided by $\tau.$
But this is unconvincing because the power lost to friction changed and wasn't constant, if we waited longer then closer to 100% of the energy would be dissipated but the average power would be much smaller because we included more time of lower power loss.
The whole thing is incredibly wild estimates, not much different than dimensional analysis.
Most of the energy $E$ is lost by $\tau$ if the driving force is absent. Okay. But how is it related to the stored energy? Even if the driving force is present, it will have to supply $E/\tau$ so as to compensate the loss. How can it be stored, then?? it is just nullifying the dissipation; it is not stored?
The energy stored is defined to be $E$. Because the letter $E$ is the symbol I made up for the total stored energy. And $\tau$ is an almost arbitrary time related to how long it takes for the friction force to dissipate almost all of the energy $E$. Why is it that almost all of the energy is what is dissipated? Because we choose the zero of energy to be located at that place the friction approaches. Its not the absolute zero. There is more energy there, there is, force instance rest energy associated with every particle. There is chemical binding energy associated with each electron being in some atom/molecule. There is some energy associated with the thermal motion of the system. You could extract that energy with antimatter, chemical reactions with more reactive elements, and thermal contact with colder objects. But there isn't any more macroscopic kinetic energy or potential energy available besides $E$ that's all that is available. And that's the state the friction drives the system to. And it never gets perfectly there but it gets close in time $\tau$ and so in time $\tau$ almost $E$ is dissipated. Because of the definition of the two symbols.
Edit At resonance the amplitude can grow. As it grows the friction increases. When the amplitude grows it grows and grows and grows until the amplitude is so large that the friction dissipated is now exactly equal to the driving power. And that friction power is somewhat approximately equal to $E/\tau$ where $E$ is the total available energy stored.
At another frequency the amplitude doesn't grow so it doesn't get up to a nonzero steady state. In a complicated and real system there might be many modes that can be resonant and there might be shifting of modes over time.
Best Answer
The damped resonance frequency must tend towards zero as the damping increases because, at and above critical damping, the undriven system will not oscillate for any set of initial conditions.
This is most easily seen in the Laplace domain. The poles of an undamped SHO system have zero real part, i.e., the poles are on the imaginary axis.
As the damping increases, the (complex conjugate) poles move into the left half plane with increasingly negative real part while the imaginary part moves towards zero.
At critical damping, the poles are real and equal, i.e., there is no oscillatory component.
Above critical camping, the poles are real and distinct and move apart along the real line as the damping increases.
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In the time domain, one can look at the step response. For the undamped SHO, the step response is a semi-infinite sinusoid.
For an underdamped SHO, the step response is an damped sinusoid with lower frequency than the undamped case.
For a critically or higher damped SHO, the step response is monotonic, i.e., no oscillation.
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