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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?

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What is a "natural frequency?" Are you talking about the resonant frequency of a system? –  Chris White Dec 15 '12 at 22:16
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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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Thanks. Would you mind explaining how the equation is derived and what $j$ is? It's not $\sqrt(-1)$, is it? –  Brownish Monster Dec 16 '12 at 15:31
    
@BrownishMonster: Yes, $j=\sqrt{-1}$ (engineering habit, sorry). The result is derived by looking for steady-state solutions of the form $B e^{j \omega t}$ to the second-order linear differential equation with constant coefficients $y + \dot{y}/(Q \omega_n) + \ddot{y}/\omega_n^2 = y_{in}$, with $\omega=2 \pi f$ the driving frequency in radians per second and $\omega_n = 2 \pi f_n$ the natural frequency, also in radians per second. One takes the real part of the complex result to get the magnitude and phase. –  Art Brown Dec 16 '12 at 18:13
    
It's fine, I'm an (mechanical) engineering student myself and just needed to be sure. –  Brownish Monster Dec 17 '12 at 16:31
    
Why does the output fall off? Wouldn't $y_{out}$ be higher when in phase with $f_n$ –  Brownish Monster Dec 17 '12 at 17:51
    
I think I wasn't clear. It's for driving frequencies frequencies well above the natural frequency ($f>f_n$) that the output falls off. At resonance ($f=f_n$), the output is a maximum (for $Q>1$). –  Art Brown Dec 17 '12 at 17:56
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