Chapter 12, Problem 12.1P

To determine

The characteristic frequencies of the two coupled oscillators with all spring constants different, and to compare it with the natural frequencies of the two oscillators in the absence of coupling.

Answer to Problem 12.1P

The characteristic frequencies of the two coupled oscillators with all spring constants different are ${\omega }_{+}={\left(\frac{k_1+k_2+2k_{12}}{2M}+\frac{1}{2M}\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right)}^{1/2}$ and ${\omega }_{-}={\left(\frac{k_1+k_2+2k_{12}}{2M}-\frac{1}{2M}\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right)}^{1/2}$. In the absence of coupling ${\omega }_{+}>{\omega }_{01}$ and ${\omega }_{-}<{\omega }_{02}$ where ${\omega }_{01}$ and ${\omega }_{02}$ are the frequencies when mass $m_2$ and $m_1$ are held fixed respectively.

Explanation of Solution

The system of the two coupled oscillators with all spring constants different are shown in Figure 1.

In the figure, the spring constants of the springs are marked as $k_1$, $k_2$, and $k_{12}$.

Write the equations of motions for the system.

  $\begin{array}{l}M{\ddot{x}}_1+\left(k_1+k_{12}\right)x_1-k_{12}{x}_2=0\\ M{\ddot{x}}_2+\left(k_2+k_{12}\right)x_2-k_{12}{x}_1=0\end{array}\}$        (I)

Let us attempt the solutions of the form;

  $\begin{array}{l}{x}_1\left(t\right)=B_1{e}^{i\omega t}\\ x_2\left(t\right)=B_2{e}^{i\omega t}\end{array}\}$        (II)

Here, $B_1$ and $B_2$ are constants.

Use the solutions in (II) in (I).

  $\begin{array}{c}\left(k_1+k_{12}-M{\omega }^2\right)B_1-k_{12}{B}_2=0\\ -k_{12}{B}_1+\left(k_2+k_{12}-M{\omega }^2\right)B_2=0\end{array}\}$        (III)

The condition for obtaining a non-trivial solution is that, the determinant of the coefficients of $B_1$ and $B_2$ must be zero. Thus, it yields;

  $\left[k_1+\left(k_{12}-M{\omega }^2\right)\right]\left[k_2+\left(k_{12}-M{\omega }^2\right)\right]=k_{12}^2$        (IV)

Solve equation (IV) to get an expression for ${\omega }^2$.

  ${\omega }^2=\frac{k_1+k_2+2k_{12}}{2M}\pm \frac{1}{2M}\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}$        (V)

Thus, the characteristic frequencies will be;

  ${\omega }_{+}={\left(\frac{k_1+k_2+2k_{12}}{2M}+\frac{1}{2M}\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right)}^{1/2}$

  ${\omega }_{-}={\left(\frac{k_1+k_2+2k_{12}}{2M}-\frac{1}{2M}\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right)}^{1/2}$

Consider the case; $k_1=k_2=k$, then equation (V) reduces to;

  ${\omega }^2=\frac{\left(k+k_{12}\pm k_{12}\right)}{M}$

If the mass $m_2$ is not moving, then the frequency of oscillation of $m_1$ will be;

  ${\omega }_{01}^2=\frac{1}{M}\left(k_1+k_{12}\right)$        (VI)

If the mass $m_1$ is not moving, then the frequency of oscillation of $m_2$ will be;

  ${\omega }_{02}^2=\frac{1}{M}\left(k_2+k_{12}\right)$        (VII)

Comparing equation (VI) and (VII) with the two frequencies given by equation (V), it can be observed that;

  $\begin{array}{c}{\omega }_{+}^2=\frac{1}{2M}\left[k_1+k_2+2k_{12}+\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right]\\ {\omega }_{+}^2>\frac{1}{2M}\left[k_1+k_2+2k_{12}+\left(k_1-k_2\right)\right]\\ {\omega }_{+}^2>\frac{1}{2M}\left[2k_1+2k_{12}\right]\\ {\omega }_{+}^2>\frac{1}{M}\left[k_1+k_{12}\right]\\ {\omega }_{+}^2>{\omega }_{01}^2\end{array}$        (VIII)

Similarly;

  $\begin{array}{c}{\omega }_{-}^2=\frac{1}{2M}\left[k_1+k_2+2k_{12}-\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right]\\ {\omega }_{+}^2<\frac{1}{2M}\left[k_1+k_2+2k_{12}-\left(k_1-k_2\right)\right]\\ {\omega }_{+}^2<\frac{1}{2M}\left[2k_2+2k_{12}\right]\\ {\omega }_{+}^2<\frac{1}{M}\left[k_2+k_{12}\right]\\ {\omega }_{+}^2<{\omega }_{02}^2\end{array}$        (IX)

From equation (IX) and (X) it is clear that;

  $\begin{array}{l}{\omega }_{+}>{\omega }_{01}\\ {\omega }_{-}<{\omega }_{02}\end{array}$

If $k_1>k_2$, then the ordering of the frequencies will be as;

  ${\omega }_{+}>{\omega }_{01}>{\omega }_{02}>{\omega }_{-}$

Conclusion:

Therefore, the characteristic frequencies of the two coupled oscillators with all spring constants different are ${\omega }_{+}={\left(\frac{k_1+k_2+2k_{12}}{2M}+\frac{1}{2M}\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right)}^{1/2}$ and ${\omega }_{-}={\left(\frac{k_1+k_2+2k_{12}}{2M}-\frac{1}{2M}\sqrt{{\left(k_1-k_2\right)}^2+4k_{12}^2}\right)}^{1/2}$. In the absence of coupling ${\omega }_{+}>{\omega }_{01}$ and ${\omega }_{-}<{\omega }_{02}$ where ${\omega }_{01}$ and ${\omega }_{02}$ are the frequencies when mass $m_2$ and $m_1$ are held fixed respectively.