Question 705047

1. Kinetic and Potential Energy Let $x_1$ and $x_2$ be the coordinates of the two atoms. The kinetic energy ($T$) of the system is: $T = \frac{1}{2}m\dot{x}_1^2 + \frac{1}{2}m\dot{x}_2^2$

The total potential energy ($V$) is the sum of the potential energy of the connecting spring and the external potential given in the question: $V = \frac{1}{2}k(x_1 - x_2)^2 + \frac{1}{2}m\omega_0^2(x_1^2 + x_2^2)$ Expanding the spring potential term: $V = \frac{1}{2}k(x_1^2 + x_2^2 - 2x_1x_2) + \frac{1}{2}m\omega_0^2x_1^2 + \frac{1}{2}m\omega_0^2x_2^2$ Grouping the terms for $x_1^2$, $x_2^2$, and $x_1x_2$: $V = \frac{1}{2}(k + m\omega_0^2)x_1^2 + \frac{1}{2}(k + m\omega_0^2)x_2^2 - kx_1x_2$

2. Matrices for the System From the expressions for $T$ and $V$, we can construct the mass matrix $M$ and the potential energy (stiffness) matrix $K$: $M = \begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}$ $K = \begin{pmatrix} k + m\omega_0^2 & -k \\ -k & k + m\omega_0^2 \end{pmatrix}$

3. Secular Equation To find the normal frequencies ($\omega$), we solve the secular equation $|K - \omega^2 M| = 0$: $\begin{vmatrix} (k + m\omega_0^2) - m\omega^2 & -k \\ -k & (k + m\omega_0^2) - m\omega^2 \end{vmatrix} = 0$

Let $\lambda = k + m\omega_0^2 - m\omega^2$. The determinant simplifies to: $\begin{vmatrix} \lambda & -k \\ -k & \lambda \end{vmatrix} = 0$ $\lambda^2 - (-k)^2 = 0$ $\lambda^2 - k^2 = 0 \implies \lambda = \pm k$

4. Solving for the Frequencies Now, substitute the value of $\lambda$ back into our assumption: $k + m\omega_0^2 - m\omega^2 = \pm k$

• Case 1 (taking $+k$): $k + m\omega_0^2 - m\omega^2 = k$ $m\omega_0^2 - m\omega^2 = 0 \implies \omega^2 = \omega_0^2 \implies \omega = \omega_0$ (This perfectly matches the "one frequency is $\omega_0$" condition given in the question).

• Case 2 (taking $-k$): $k + m\omega_0^2 - m\omega^2 = -k$ $2k + m\omega_0^2 = m\omega^2$ $\omega^2 = \frac{m\omega_0^2 + 2k}{m}$ $\omega^2 = \omega_0^2 + \frac{2k}{m}$ $\omega = \sqrt{\omega_0^2 + \frac{2k}{m}}$

Final Answer: The other frequency of molecular vibration is $\sqrt{\omega_0^2 + \frac{2k}{m}}$. This corresponds to Option (C).