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.
 
:'''Equation 1'''
:<math>\begin{alignat}{3}
:<math>\begin{alignat}{3}
                                                         F & = ma \\
                                                         F & = ma \\
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           -{k_1+k_2 \over {m_1}}x_1+{k_2 \over {m_1}}x_2 & = \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>
\end{alignat}</math>
:'''Equation 2'''
:<math>\begin{alignat}{3}
                                                        F & = ma \\
                                                        F & = m\ddot{x} \\
                                            -k_2(x_2-x_1) & = m_2\ddot{x_2} \\
                              {-k_2(x_2-x_1) \over {m_2}} & = \ddot{x_2} \\
              -{k_2 \over {m_2}}x_2+{k_2 \over {m_2}}x_1 & = \ddot{x_2} \\
\end{alignat}</math>
:'''Equation 3'''
:<math>\dot{x_1}=\dot{x_1}</math>
:'''Equation 4'''
:<math>\dot{x_2}=\dot{x_2}</math>

Revision as of 18:41, 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.

Equation 1
F=maF=mx¨k1x1k2(x1x2)=m1x1¨k1x1m1k2(x1x2)m1=m1x1¨k1x1m1k2(x1x2)m1=x1¨k1+k2m1x1+k2m1x2=x1¨
Equation 2
F=maF=mx¨k2(x2x1)=m2x2¨k2(x2x1)m2=x2¨k2m2x2+k2m2x1=x2¨
Equation 3
x1˙=x1˙
Equation 4
x2˙=x2˙