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Here is a different way to couple two oscillators. The two
Chapter 11, Problem 11.12(choose chapter or problem)
Here is a different way to couple two oscillators. The two carts in Figure 11.16 have equal masses m (though different shapes). They are joined by identical but separate springs (force constant k) to separate walls. Cart 2 rides in cart 1, as shown, and cart 1 is filled with molasses, whose viscous drag supplies the coupling between the carts.
(a) Assuming that the drag force has magnitude \(\beta m v\) where v is the relative velocity of the two carts, write down the equations of motion of the two carts using as coordinates \(x_{1}\) and \(x_{2}\), the displacements of the carts from their equilibrium positions. Show that they can be written in matrix form as \(\ddot{\mathbf{x}}+\beta \mathbf{D} \dot{\mathbf{x}}+\omega_{\mathrm{o}}^{2} x=0\), where x is the \(2 \times 1\) column made up of \(x_{1}\) and \(x_{2}\), \(\omega_{\mathrm{o}}=\sqrt{k / m}\), and D is a certain \(2 \times 2\) square matrix.
(b) There is nothing to stop you from seeking a solution of the form x(t) = Re z(t), with \(\mathbf{z}(t)=\mathbf{a} e^{r t}\). Show that you do indeed get two solutions of this form with \(r=i \omega_{\mathrm{o}}\) or \(r=-\beta+i \omega_{1}\) where \(\omega_{1}=\sqrt{\omega_{0}^{2}-\beta^{2}}\). (Assume that the viscous force is weak, so that ,\(\beta<\omega_{0}\).)
(c) Describe the corresponding motions. Explain why one of these modes is damped but the other is not.
Questions & Answers
QUESTION:
Here is a different way to couple two oscillators. The two carts in Figure 11.16 have equal masses m (though different shapes). They are joined by identical but separate springs (force constant k) to separate walls. Cart 2 rides in cart 1, as shown, and cart 1 is filled with molasses, whose viscous drag supplies the coupling between the carts.
(a) Assuming that the drag force has magnitude \(\beta m v\) where v is the relative velocity of the two carts, write down the equations of motion of the two carts using as coordinates \(x_{1}\) and \(x_{2}\), the displacements of the carts from their equilibrium positions. Show that they can be written in matrix form as \(\ddot{\mathbf{x}}+\beta \mathbf{D} \dot{\mathbf{x}}+\omega_{\mathrm{o}}^{2} x=0\), where x is the \(2 \times 1\) column made up of \(x_{1}\) and \(x_{2}\), \(\omega_{\mathrm{o}}=\sqrt{k / m}\), and D is a certain \(2 \times 2\) square matrix.
(b) There is nothing to stop you from seeking a solution of the form x(t) = Re z(t), with \(\mathbf{z}(t)=\mathbf{a} e^{r t}\). Show that you do indeed get two solutions of this form with \(r=i \omega_{\mathrm{o}}\) or \(r=-\beta+i \omega_{1}\) where \(\omega_{1}=\sqrt{\omega_{0}^{2}-\beta^{2}}\). (Assume that the viscous force is weak, so that ,\(\beta<\omega_{0}\).)
(c) Describe the corresponding motions. Explain why one of these modes is damped but the other is not.
ANSWER:Step 1 of 6
(a)
Let's assume the position of carts 1 and 2 are \(x_{1}\) & \(x_{2}\) and their velocities are \(\dot{x}_{1}\) & \(\dot{x}_{2}\).
Then the relative velocity between carts 1 and 2, \(v=\dot{x}_2-\dot{x}_1\)
The damping force is \(-\beta m v\).
the net force on the cart 1 is given by,
\(\begin{array}{l} F_{1}=-k_{1} x+\beta m v \\ m \ddot{x}_{1}=-k_{1} x+\beta m v \\ m \ddot{x}_{1}=-k_{1} x+\beta m\left(\dot{x}_{2}-\dot{x}_{1}\right) \\ \ddot{x_{1}}-\beta \dot{x}_{1}+\beta \dot{x}_{2}+\frac{k}{m} x_{1}=0 \end{array}\)
Similarly for the cart 2,
\(\ddot{x}_{2}-\beta \dot{x}_{1}+\beta \dot{x}_{2}+\frac{k}{m} x_{2}=0\)