## Step-by-Step Solution ### **Step 1: Write the Equations of Motion** Given: - \( c_1 = c_2 = c_3 = \) (No damping) - \( F_1(t) = F_{10} \cos \omega t \) - \( F_2(t) = F_{20} \cos \omega t \) Assuming a **two-degree-of-freedom system** (since we have two forces and two masses): Let: - \( m_1, m_2 \) = masses - \( k_1, k_2, k_3 \) = stiffnesses - \( x_1(t), x_2(t) \) = displacements The equations of motion (EOM) are: \[ \begin{align*} m_1 \ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 &= F_{10} \cos \omega t \\ m_2 \ddot{x}_2 + (k_2 + k_3)x_2 - k_2 x_1 &= F_{20} \cos \omega t \end{align*} \] --- ### **Step 2: Find the Steady-State Solution** Assume steady-state solutions of the form: \[ x_1(t) = X_1 \cos(\omega t), \quad x_2(t) = X_2 \cos(\omega t) \] Substitute into the EOM: \[ \begin{align*} -m_1 \omega^2 X_1 \cos\omega t + (k_1 + k_2)X_1 \cos\omega t - k_2 X_2 \cos\omega t &= F_{10} \cos \omega t \\ -m_2 \omega^2 X_2 \cos\omega t + (k_2 + k_3)X_2 \cos\omega t - k_2 X_1 \cos\omega t &= F_{20} \cos \omega t \end{align*} \] Divide both sides by \( \cos \omega t \): \[ \begin{align*} \left[ (k_1 + k_2) - m_1 \omega^2 \right] X_1 - k_2 X_2 &= F_{10} \\ - k_2 X_1 + \left[ (k_2 + k_3) - m_2 \omega^2 \right] X_2 &= F_{20} \end{align*} \] Write in matrix form: \[ \begin{bmatrix} (k_1 + k_2) - m_1 \omega^2 & -k_2 \\ -k_2 & (k_2 + k_3) - m_2 \omega^2 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} F_{10} \\ F_{20} \end{bmatrix} \] **Solve for \( X_1 \) and \( X_2 \):** \[ \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} (k_1 + k_2) - m_1 \omega^2 & -k_2 \\ -k_2 & (k_2 + k_3) - m_2 \omega^2 \end{bmatrix}^{-1} \begin{bmatrix} F_{10} \\ F_{20} \end{bmatrix} \] --- ## **Example Calculation** Let: - \( m_1 = m_2 = 1 \, \text{kg} \) - \( k_1 = k_2 = k_3 = 100 \, \text{N/m} \) - \( F_{10} = 10 \, \text{N} \) - \( F_{20} = 20 \, \text{N} \) - \( \omega = 10 \, \text{rad/s} \) Plug in values: \[ A = (100 + 100) - 1 \cdot 100 = 200 - 100 = 190 \\ B = -100 \\ C = (100 + 100) - 1 \cdot 100 = 200 - 100 = 190 \] So the matrix is: \[ \begin{bmatrix} 190 & -100 \\ -100 & 190 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} 10 \\ 20 \end{bmatrix} \] Solving: \[ \Delta = 190 \cdot 190 - (-100)\cdot(-100) = 361000 - 100000 = 261000 \] \[ X_1 = \frac{1}{\Delta} [190 \cdot 10 + 100 \cdot 20] = \frac{19000 + 20000}{261000} = \frac{39000}{261000} \approx .0149 \] \[ X_2 = \frac{1}{\Delta} [190 \cdot 20 + 100 \cdot 10] = \frac{38000 + 10000}{261000} = \frac{48000}{261000} \approx .0184 \] --- ## **Final Answer** \[ \boxed{ \begin{aligned} x_1(t) &= .0149 \cos(10 t) \\ x_2(t) &= .0184 \cos(10 t) \end{aligned} } \] --- ## **Graph** Below is a sample plot of \(x_1(t)\) and \(x_2(t)\): ```python import numpy as np import matplotlib.pyplot as plt t = np.linspace(, 2, 500) x1 = .0149 * np.cos(10 * t) x2 = .0184 * np.cos(10 * t) plt.plot(t, x1, label='x1(t)') plt.plot(t, x2, label='x2(t)') plt.xlabel('Time (s)') plt.ylabel('Displacement (m)') plt.title('Steady-State Vibration Response') plt.legend() plt.grid(True) plt.show() ``` **Sample Plot:**  *Steady-state vibration response for \( x_1(t) \) and \( x_2(t) \).* --- ### **Summary** - **Step 1:** Write and solve the equations of motion for steady-state. - **Step 2:** Substitute values and compute amplitudes. - **Final Answer:** The steady-state displacements are \( x_1(t) = .0149 \cos(10 t) \), \( x_2(t) = .0184 \cos(10 t) \), as shown in the graph.