# Solution to Mechanical System: Find \( x_2(t) \) Let's solve for \( x_2(t) \) for the given mechanical system in **2 steps**. --- ## **Step 1: Write the Equations of Motion (EOM)** Let's label the displacements: - \( x_1(t) \): Input displacement (given as \( \bar{X}_1 \sin(\omega t) \)) - \( x_2(t) \): Displacement of mass \( m \) - \( x_3(t) \): Displacement at node between first damper and spring - \( x_4(t) \): Displacement at base spring ### **Free Body Diagram and EOM for \( m \):** By applying Newton’s Second Law to mass \( m \): \[ m \ddot{x}_2 = \text{(Forces to the left)} + \text{(Forces to the right)} \] **Forces from the left:** - Damper: \( c(\dot{x}_3 - \dot{x}_2) \) - Spring: \( k(x_3 - x_2) \) **Forces from the right:** - Damper: \( c(\dot{x}_1 - \dot{x}_2) \) So, \[ m\ddot{x}_2 = c(\dot{x}_3 - \dot{x}_2) + k(x_3 - x_2) + c(\dot{x}_1 - \dot{x}_2) \] Rearrange: \[ m\ddot{x}_2 + 2c\dot{x}_2 + kx_2 = c\dot{x}_1 + c\dot{x}_3 + kx_3 \] ### **Relationship between \( x_3 \), \( x_4 \), and \( x_2 \):** At \( x_3 \), sum forces: - From left: \( k(x_4 - x_3) + c(\dot{x}_4 - \dot{x}_3) \) - From right: \( k(x_2 - x_3) + c(\dot{x}_2 - \dot{x}_3) \) Set sum to zero (no mass at node): \[ k(x_4 - x_3) + c(\dot{x}_4 - \dot{x}_3) + k(x_2 - x_3) + c(\dot{x}_2 - \dot{x}_3) = \] \[ 2k x_3 + 2c \dot{x}_3 = k(x_2 + x_4) + c(\dot{x}_2 + \dot{x}_4) \] \[ 2k x_3 + 2c \dot{x}_3 = k x_2 + k x_4 + c \dot{x}_2 + c \dot{x}_4 \] ### **At \( x_4 \) (attached to wall):** Force at \( x_4 \): \[ k(x_3 - x_4) + c(\dot{x}_3 - \dot{x}_4) = \] \[ k(x_3 - x_4) + c(\dot{x}_3 - \dot{x}_4) = \] --- ## **Step 2: Solve for \( x_2(t) \) using Harmonic Input** Given: \[ x_1(t) = \bar{X}_1 \sin(\omega t) \] Assume steady-state sinusoidal solutions: \[ x_i(t) = X_i \sin(\omega t) \] \[ \dot{x}_i(t) = \omega X_i \cos(\omega t) \] \[ \ddot{x}_i(t) = -\omega^2 X_i \sin(\omega t) \] ### Substitute into the equations Let’s use phasor notation (\( e^{j\omega t} \)), and solve for \( X_2 \): #### Using Laplace Transform (\( s = j\omega \)): Let’s denote \( s = j\omega \). **At \( x_4 \):** \[ k(X_3 - X_4) + cs(X_3 - X_4) = \implies X_3 = X_4 \] **At \( x_3 \):** \[ 2kX_3 + 2csX_3 = kX_2 + kX_4 + csX_2 + csX_4 \] But \( X_3 = X_4 \): \[ 2kX_3 + 2csX_3 = kX_2 + kX_3 + csX_2 + csX_3 \] \[ (2k + 2cs)X_3 - (k + cs)X_3 = (k + cs)X_2 \] \[ (k + cs)X_3 = (k + cs)X_2 \implies X_3 = X_2 \] Therefore, \( X_4 = X_3 = X_2 \). **At \( m \):** \[ m s^2 X_2 + 2cs X_2 + k X_2 = cs X_1 + cs X_3 + k X_3 \] But \( X_3 = X_2 \): \[ m s^2 X_2 + 2cs X_2 + k X_2 = cs X_1 + cs X_2 + k X_2 \] \[ m s^2 X_2 + cs X_2 = cs X_1 \] \[ (m s^2 + cs) X_2 = cs X_1 \] \[ X_2 = \frac{cs}{m s^2 + c s} X_1 \] Substitute \( s = j\omega \): \[ X_2 = \frac{c j\omega}{m (j\omega)^2 + c (j\omega)} X_1 = \frac{c j\omega}{-m \omega^2 + c j\omega} X_1 \] ### **Final Answer** \[ \boxed{ x_2(t) = \left| \frac{c j\omega}{-m \omega^2 + c j\omega} \right| \bar{X}_1 \sin(\omega t + \phi) } \] Where \( \phi \) is the phase angle of the transfer function. --- ## **Example (Numerical Values)** Let \( m = 1\,\mathrm{kg},\ c = 2\,\mathrm{Ns/m},\ \omega = 1\,\mathrm{rad/s},\ \bar{X}_1 = 1 \). \[ X_2 = \frac{2j}{-1 + 2j} \] Magnitude: \[ |X_2| = \frac{2}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{5}} \approx .894 \] Phase: \[ \phi = \arg\left(\frac{2j}{-1 + 2j}\right) = \arctan\left(\frac{2}{-1}\right) = \arctan(-2) + 90^\circ \] So, \[ x_2(t) \approx .894\,\sin(\omega t + \phi) \] --- ## **Graph**  *Illustrative graph: Amplitude of \( x_2 \) vs. frequency \( \omega \) for different \( m, c \) values.* --- ## **Summary** 1. **Write EOMs** for each node and use harmonic input. 2. **Solve for \( x_2(t) \) in frequency domain**, plug in values if needed. **Final formula:** \[ x_2(t) = \left| \frac{c j\omega}{-m \omega^2 + c j\omega} \right| \bar{X}_1 \sin(\omega t + \phi) \] Let me know if you need a plot for specific parameters!