- #1
Does Increasing Mass Reduce the Amplitude of a Forced SDOF Oscillator?
- #2
- 2,020
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\(\displaystyle X = \dfrac{F}{k} \dfrac{1}{\sqrt{ \dfrac{ 4 \zeta ^2 \omega ^2}{ \omega _n ^2 } + \left ( 1 - \dfrac{ \omega ^2 }{ \omega _n ^2 } \right ) ^2}}\)
So given a small variation in m, dm, we get a corresponding change in X by dX:
\(\displaystyle dX = \dfrac{d}{dm} \left \{ \dfrac{F}{k} \dfrac{1}{\sqrt{ \dfrac{ 4 \zeta ^2 \omega ^2}{ \omega _n ^2 } + \left ( 1 - \dfrac{ \omega ^2 }{ \omega _n ^2 } \right ) ^2}} \right \} ~ dm\)
It looks pretty bad but the only variable that contains the mass is \(\displaystyle \zeta\). Do it step by step.
-Dan
So given a small variation in m, dm, we get a corresponding change in X by dX:
\(\displaystyle dX = \dfrac{d}{dm} \left \{ \dfrac{F}{k} \dfrac{1}{\sqrt{ \dfrac{ 4 \zeta ^2 \omega ^2}{ \omega _n ^2 } + \left ( 1 - \dfrac{ \omega ^2 }{ \omega _n ^2 } \right ) ^2}} \right \} ~ dm\)
It looks pretty bad but the only variable that contains the mass is \(\displaystyle \zeta\). Do it step by step.
-Dan
- #3
anum
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- 0
$\omega_n=\sqrt(k/m)$ is also mass-dependent. Thank you for your reply.
- #4
- #5
- 2,020
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I'll presume that Maple got it right. Though taking that derivative by hand is good practice!
-Dan
-Dan
- #6
- #7
anum
- 6
- 0
its different from what i got
- #8
- 2,020
- 833
Yes, it's different. I'm not sure what method your paper is using.
-Dan
-Dan