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

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Increasing the mass of a forced single-degree-of-freedom (SDOF) oscillator reduces the response amplitude, which can be mathematically derived. The relationship involves the natural frequency, damping ratio, and external force, leading to a complex expression for amplitude change. A small variation in mass results in a corresponding change in amplitude, which is influenced by the damping ratio. The discussion highlights the importance of correctly applying derivatives and understanding the mass dependence of natural frequency. The conversation emphasizes the need for clarity in mathematical derivations related to oscillator behavior.
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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
 

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[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
 
$\omega_n=\sqrt(k/m)$ is also mass-dependent. Thank you for your reply.
 
i try on Maple get this.
 

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

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

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

-Dan
 
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