Does Increasing Mass Reduce the Amplitude of a Forced SDOF Oscillator?

[anum]

I'm working on a forced single-degree-of-freedom (SDOF) oscillator. I'm trying to find how an increase in mass will affect the response amplitude? I'm looking for mathematical proof explaining this reduction in amplitude. How to derive \delta x

 

[topsquark]

[math]X = \dfrac{F}{k} \dfrac{1}{\sqrt{ \dfrac{ 4 \zeta ^2 \omega ^2}{ \omega _n ^2 } + \left ( 1 - \dfrac{ \omega ^2 }{ \omega _n ^2 } \right ) ^2}}[/math]

So given a small variation in m, dm, we get a corresponding change in X by dX:
[math]dX = \dfrac{d}{dm} \left \{ \dfrac{F}{k} \dfrac{1}{\sqrt{ \dfrac{ 4 \zeta ^2 \omega ^2}{ \omega _n ^2 } + \left ( 1 - \dfrac{ \omega ^2 }{ \omega _n ^2 } \right ) ^2}} \right \} ~ dm[/math]

It looks pretty bad but the only variable that contains the mass is [math]\zeta[/math]. Do it step by step.

-Dan

 

[anum]

$\omega_n=\sqrt(k/m)$ is also mass-dependent. Thank you for your reply.

 

[anum]

i try on Maple get this.

 

[topsquark]

I'll presume that Maple got it right. Though taking that derivative by hand is good practice!

-Dan

 

[anum]

According to an article, $\Delta X$ on taking mass $m+\Delta m$ is :

 

[anum]

its different from what i got

 

[topsquark]

Yes, it's different. I'm not sure what method your paper is using.

-Dan