Question

The figure shows a system of two equal masses \( m \) and three massless horizontal springs with spring constants \( k_1 \), \( k_2 \), and \( k_1 \). Ignore gravity. The masses can move only in the horizontal direction, and there is no dissipation. If \( m = 1 \), \( k_1 = 2 \), and \( k_2 = 3 \) (all in appropriate units), the frequencies of the normal modes of the system in the same system of units are:

 

• A. \( \sqrt{2}, \sqrt{8} \)
• B. \( \sqrt{2}, \sqrt{6} \)
• C. \( \sqrt{3}, \sqrt{10} \)
• D. \( \sqrt{3}, \sqrt{8} \)

Hint:

In coupled oscillation problems, solving the system of equations of motion and calculating the determinant of the coefficient matrix helps in finding the normal mode frequencies.

Answer 1

The Correct Option is A

We will calculate the frequencies of the normal modes for a system of two equal masses connected by three springs with spring constants \( k_1 \), \( k_2 \), and \( k_1 \).

Step 1: Writing the Equations of Motion

Let the displacements of the two masses be denoted by \( x_1 \) and \( x_2 \). The forces acting on the masses using Hooke's law are:

For mass 1:

\( m \ddot{x}_1 = -k_1 x_1 - k_2(x_1 - x_2) \)

For mass 2:

\( m \ddot{x}_2 = -k_1 x_2 - k_2(x_2 - x_1) \)

Step 2: Normal Mode Solutions

Assume solutions of the form \( x_1 = A_1 e^{i\omega t} \), \( x_2 = A_2 e^{i\omega t} \). Substituting, we get:

\[ m\omega^2 \begin{pmatrix} A_1 \\ A_2 \end{pmatrix} = \begin{pmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_1 + k_2 \end{pmatrix} \begin{pmatrix} A_1 \\ A_2 \end{pmatrix} \]

Step 3: Solving the Eigenvalue Problem

We now solve the determinant equation:

\[ \text{det} \begin{pmatrix} k_1 + k_2 - m\omega^2 & -k_2 \\ -k_2 & k_1 + k_2 - m\omega^2 \end{pmatrix} = 0 \]

Using determinant properties, this simplifies to:

\[ (k_1 + k_2 - m\omega^2)^2 - k_2^2 = 0 \]

Solving this quadratic gives:

\[ \omega^2 = \frac{(k_1 + k_2 \pm k_2)}{m} \Rightarrow \omega_1^2 = \frac{k_1}{m}, \quad \omega_2^2 = \frac{k_1 + 2k_2}{m} \]

Step 4: Substituting Values

Given \( k_1 = 2 \), \( k_2 = 3 \), and \( m = 1 \):

\[ \omega_1 = \sqrt{\frac{2}{1}} = \sqrt{2}, \quad \omega_2 = \sqrt{\frac{2 + 2 \times 3}{1}} = \sqrt{8} \]

Final Answer: The correct frequencies are \( \omega_1 = \sqrt{2}, \omega_2 = \sqrt{8} \).

Therefore, the correct option is (A).