Question

A two degree of freedom spring–mass system undergoes free vibration with natural frequencies \( \omega_1 = 233.9 \,\text{rad/s} \) and \( \omega_2 = 324.5 \,\text{rad/s} \). The mode shapes are \[ \phi_1 = \begin{bmatrix} 1 \\ -3.16 \end{bmatrix}, \phi_2 = \begin{bmatrix} 1 \\ 3.16 \end{bmatrix}. \] Given zero initial velocities, identify which initial deflections produce pure or mixed mode oscillations. \\

• A. \( x_1(0) = 6.32 \text{ cm},\; x_2(0) = -3.16 \text{ cm} \) gives only the second natural frequency
• B. \( x_1(0) = 2 \text{ cm},\; x_2(0) = -6.32 \text{ cm} \) gives only the first natural frequency
• C. \( x_1(0) = 2 \text{ cm},\; x_2(0) = -2 \text{ cm} \) gives a combination of first and second natural frequencies
• D. \( x_1(0) = 1 \text{ cm},\; x_2(0) = -6.32 \text{ cm} \) gives only the first natural frequency

Hint:

For pure mode vibration, the initial deflection vector must align with one mode shape exactly.

Answer 1

The Correct Option is B, C

The response with zero initial velocities is \[ X(0) = C_1 \phi_1 + C_2 \phi_2. \]

Step 1: Pure mode occurs when \( X(0) \) is a scalar multiple of \( \phi_1 \) or \( \phi_2 \).
Mode shapes: \[ \phi_1 = [1,\;-3.16], \phi_2 = [1,\;3.16]. \] Option (B): \[ (2,\; -6.32) = 2 \cdot (1,\;-3.16) = 2\phi_1. \] Thus only the first natural frequency. True.
Option (D): \[ (1,\; -6.32) \] is not a scalar multiple of \( \phi_1 = (1,-3.16) \). Thus does *not* give pure mode. False.
Option (A): \[ (6.32,\; -3.16) \] is not a scalar multiple of \( \phi_2 = (1,3.16) \). False.
Option (C): \[ (2,\; -2) \] is not a scalar multiple of either mode shape, so it must excite a combination of both modes. True.

Final Answer: (B), (C)