Forced response of a mass-spring-damper system

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SUMMARY

The discussion focuses on finding the forced response of a mass-spring-damper system using the Laplace Transform. The equation of motion provided is \(\ddot{x} + 2 \zeta \omega_n \dot{x} + {\omega_n}^2 x = \frac{F}{m} \cos(\omega t)\). The solution approach involves transforming the equation into the Laplace domain, resulting in \(X(S) = \frac{F/m}{S^2 + 2 \zeta \omega_n S + {\omega_n}^2} \cdot \frac{S}{S^2 + \omega^2}\). The recommended methods for solving include using partial fractions and the convolution theorem to derive the time response.

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Homework Statement



I need to find the forced response of a mass-spring-damper system by using Laplace Transform.

Homework Equations



Equation of motion:
\ddot{x} + 2 \zeta \omega_n \dot{x} + {\omega_n}^2 x = \frac{F}{m} cos \left( \omega t \right)

The Attempt at a Solution



At the Laplace domain, we have:

X \left( S \right) = \frac{F/m}{S^2+ 2 \zeta \omega_n S + {\omega_n}^2} \frac{S}{S^2+\omega^2}

After this point, I try to make partial fractions, but I must be missing something. It results in a page full of algebra, and it does not look like the correct solution.

Please, could you help me on figuring out what I should do?
 
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You could complete the squares in the denominator of the first term and then use the convolution theorem. Or use partial fractions on the first term and then use the convolution theorem.
 
nickjer said:
You could complete the squares in the denominator of the first term and then use the convolution theorem. Or use partial fractions on the first term and then use the convolution theorem.

You mean doing the convolution between the impulse response and the cosine forcing function?

I'd like to solve the problem in the Laplace domain, then apply the inverse transform to get the time response.
 
You did solve it in the Laplace domain, assuming that equation you gave in the first post is correct. Now you need to take the inverse transform of that equation. I just recommended using the convolution theorem when taking the inverse.
 
If you want to break it all up by partial fractions you can, but it will get very messy. Another trick would be taking the partial fraction of just the first term and then using the convolution theorem.
 

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