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I am trying to derive the general transfer function for a second order dynamic system, shown below:

[tex]\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}[/tex]

In order to do this I am considering a mass-spring-damper system, with an input force of f(t) that satisfies the following second-order differential equation:

[tex]m\frac{d^2y}{\dt^2}+c\frac{dy}{dt}+ky=f(t)[/tex]

Using the following two relationships:

[tex]c=2\zeta\omega_nm[/tex]

[tex]\frac{k}{m}=\omega_n^2[/tex]

I get this:

[tex]\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=\frac{f(t)}{m}[/tex]

[tex]\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=\frac{1}{m}\mathcal{L}\left\{f(t)\right\}[/tex]

[tex]Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=\frac{F(s)}{m}[/tex]

[tex]\frac{Y(s)}{F(s)}=\frac{1}{m(s^2+2\zeta\omega_ns+\omega_n^2)}[/tex]

Wheras my lecturer has the following in his notes:

[tex]\frac{d^2y}{dt^2}+2\zeta\omega_n\frac{dy}{dt}+\omega_n^2y=K\omega_n^2x(t)[/tex]

[tex]\mathcal{L}\left\{\frac{d^2y}{dt^2}\right\}+2\zeta\omega_n\mathcal{L}\left\{\frac{dy}{dt}\right\}+\omega_n\mathcal{L}\left\{y\right\}=K\omega_n^2\mathcal{L}\{x(t)\}[/tex]

[tex]Y(s)\left[s^2+2\zeta\omega_ns+\omega_n^2\right]=K\omega_n^2X(s)[/tex]

[tex]\frac{Y(s)}{X(s)}=\frac{K\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}[/tex]

This obvisously gives the correct transfer function. So, from the two approaches, I have come to the conclusion that:

[tex]\frac{f(t)}{m}=K\omega_n^2x(t)[/tex]

But I do not understand the physical reasoning behind this. Can anyone offer any help with this?

Thanks,

Ryan