[Physics] How would natural (resonant) frequencies affect amplitudes

acousticsfrequencyresonancewaves

I read $y=A\sin(2\pi ft)$, where $A$=Amplitude, $f$=Frequency, $t$=Time and $y$=$Y$ position of the wave.

Since natural frequencies only take the most effect when they are close to the frequency.
How would one natural frequency and several natural frequencies affect the equation?

Would I be correct in thinking it's something to effect of: y=Y_Position*NaturalFrequency
where Y_Postion is the first equation and NaturalFrequency is similar to the first equation but with a low amplitude?

Best Answer

If you are driving a resonant linear system, which is characterized by a natural frequency $f_n$ and quality factor $Q$, with your specified sinusoidal input $y_{in}$ of amplitude $A$ and frequency $f$, the steady-state output $y_{out}$ will be:

$$ y_{out} = \frac{A}{1+j \frac{1}{Q} \frac{f}{f_n} - \left(\frac{f}{f_n} \right)^2} $$

This equation gives a complex phasor quantity, which describes the amplitude and phase (with respect to the input) of the output.

The higher the $Q$ of the system, the higher the output when the driving frequency is near resonance.

At low frequencies (compared with the natural frequency) the output just tracks the input, while at high frequencies, the output falls off like $1/f^2$ and lags the input by a half-cycle (180 degrees).

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