Topic: Control Systems
Consider a mechanical system shown in figure. Masses are free to slide over frictionless horizontal surface. The equation of motion of mass $m_1$ is
- A. $m_1\ddot{x}_1+(\lambda_1+\lambda_2)\dot{x}_1+\lambda_2\dot{x}_2-(k_1+k_2)x_1-k_2x_2=F_2$
- B. $m_2\ddot{x}_1+(\lambda_1-\lambda_2)\dot{x}_1+\lambda_2\dot{x}_2-(k_1-k_2)x_1-k_2x_2=F_1$
- C. $m_1\ddot{x}_1+(\lambda_1+\lambda_2)\dot{x}_1-\lambda_2\dot{x}_2+(k_1+k_2)x_1-k_2x_2=F_1$
- D. $F_i=m_i\ddot{x}_i+r_i\dot{x}_i+\lambda_i\dot{x}_i$
Answer: C. $m_1\ddot{x}_1+(\lambda_1+\lambda_2)\dot{x}_1-\lambda_2\dot{x}_2+(k_1+k_2)x_1-k_2x_2=F_1$
Explanation: Applying Newton's second law to mass $m_1$: The sum of forces is $F_1(t) - k_1 x_1 - \lambda_1 \dot{x}_1 - k_2(x_1 - x_2) - \lambda_2(\dot{x}_1 - \dot{x}_2) = m_1 \ddot{x}_1$. Rearranging all displacement and velocity terms to the left side gives: $m_1 \ddot{x}_1 + (\lambda_1 + \lambda_2)\dot{x}_1 - \lambda_2\dot{x}_2 + (k_1 + k_2)x_1 - k_2x_2 = F_1$.