Coupled Oscillator: Jonathan Schreven: Difference between revisions

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Using F=ma we can then find our equations of equilibrium.
Using F=ma we can then find our equations of equilibrium.

:<math>F=ma</math>
:<math>F=m\ddot{x}</math>
:<math>\begin{alignat}{3}
F & = ma \\
:<math>-k_{1}x_{1}-k-{2}(x_1x_2)=m\ddot{x}</math>
F & = m\ddot{x} \\
-k_{1}x_{1}-k_{2}(x_1x_2) & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = m_1\ddot{x_1} \\
-{k_1x_1 \over {m_1}}-{k_2(x_1-x_2) \over {m_1}} & = \ddot{x_1} \\
-{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \ddot{x_1} \\
\end{alignat}</math>

Revision as of 17:27, 9 December 2009

Coupled Oscillator System

In this problem I would like to explore the solution of a double spring/mass system under the assumption that the blocks are resting on a smooth surface. Our system might look something like this.


Using F=ma we can then find our equations of equilibrium.