Question

The equations of motion for a two degrees of freedom undamped spring-mass system are: \[ m \ddot{x}_1 + 2k x_1 - k x_2 = 0 \] \[ m \ddot{x}_2 - k x_1 + 2k x_2 = 0 \] where $m$ and $k$ represent mass and stiffness, respectively, in SI units. The larger of the two natural frequencies is given by: \[ \omega = \alpha \sqrt{\frac{k}{m}} \, \, \text{rad/s} \] The value of $\alpha$ is .............. (rounded off to 2 decimal places).

Hint:

In 2-DOF vibration problems, always form the characteristic equation $\det([K] - \omega^2 [M])=0$. The eigenvalues give $\omega^2$, and the higher root corresponds to the larger natural frequency.

Answer 1

Step 1: Write equations in matrix form.
The given system can be expressed as:
\[m \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{bmatrix} + k \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0 \]

So, the mass and stiffness matrices are:
\[[M] = m \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad [K] = k \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \]

Step 2: General eigenvalue problem.
We assume a solution of the form:
\[\mathbf{x}(t) = \mathbf{X} e^{i \omega t} \]
where $\mathbf{X} = \begin{bmatrix} X_1 \\ X_2 \end{bmatrix}$ is the mode shape vector and $\omega$ is the natural frequency.
 

This leads to: \[ \big( [K] - \omega^2 [M] \big) \phi = 0 \] The characteristic equation is: \[ \det \Big( [K] - \omega^2 [M] \Big) = 0 \] Step 3: Determinant expansion. \[\det \begin{bmatrix} 2k - m \omega^2 & -k \\ - k & 2k - m \omega^2 \end{bmatrix} = 0 \]
 \[ (2k - m \omega^2)^2 - (-k)(-k) = 0 \] \[ (2k - m \omega^2)^2 - k^2 = 0 \] \[ 2k - m \omega^2 = \pm k \] Step 4: Solve for eigenvalues. Case (i): \[ 2k - m \omega^2 = +k \quad \Rightarrow \quad m \omega^2 = k \quad \Rightarrow \quad \omega^2 = \frac{k}{m} \] Case (ii): \[ 2k - m \omega^2 = -k \quad \Rightarrow \quad m \omega^2 = 3k \quad \Rightarrow \quad \omega^2 = \frac{3k}{m} \] Step 5: Natural frequencies. \[ \omega_1 = \sqrt{\frac{k}{m}}, \quad \omega_2 = \sqrt{\frac{3k}{m}} \] The larger one is: \[ \omega = \sqrt{\frac{3k}{m}} \] Step 6: Express in given form. Given: \[ \omega = \alpha \sqrt{\frac{k}{m}} \] Thus, \[ \alpha = \sqrt{3} = 1.732 \, \approx \, 1.73 \] \[ \boxed{\alpha = 1.73} \]