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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)?

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  • $\begingroup$ Please note for the mechanical system you are asking about displacement resonance not velocity resonance for which the resonance peaks all stay at the same frequency, the natural frequency of undamped oscillations of the system. $\endgroup$ – Farcher Sep 5 '16 at 13:16
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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.

enter image description here

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