Let's solve the problem step by step as requested. --- ## Step 1: **Write the Differential Equation and Find General Solution** ### **System Setup** - Mass \( M = 1\, \text{kg} \) - Each spring constant \( K = 2\, \text{N/m} \) (two in parallel, so total \( K_{total} = 4\, \text{N/m} \)) - Each damper \( C = 1\, \text{kg/s} \) (two in parallel, so total \( C_{total} = 2\, \text{kg/s} \)) ### **Equation of Motion** \[ M\ddot{x} + C_{total}\dot} + K_{total}x = \[ 1\ddot{x} + 2\{x} + 4x = ] **Characteristic Equation** \[ r^2 + 2r + 4 = \] [ r = \frac{-2 \pm \sqrt^2 - 4 \cdot 1 \cdot 4}}{2} = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3} \] **General Solution:** \[ x(t) = e^{-t} (A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t)) \] **Explanation:** We set up the equation of motion for the mass-spring-damper system, combined the parallel components, and solved the characteristic equation for the roots to write the general solution. --- ## Step 2: **Apply Initial Conditions and Solve for Constants** ### **Initial Conditions:** - \( x() = -.1 \) - \( \dot{x}() = 1 \) #### At \( t = \): \[ x() = e^{}(A \cdot 1 + B \cdot ) = A = -.1 \] #### First Derivative: \[ \dot{x}(t) = \frac{d}{dt} \left[e^{-t} (A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t))\right] \] \[ = -e^{-t}(A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t)) + e^{-t}(-A\sqrt{3}\sin(\sqrt{3}t) + B\sqrt{3}\cos(\sqrt{3}t)) \] \[ = e^{-t} \left[ -A\cos(\sqrt{3}t) - B\sin(\sqrt{3}t) - A\sqrt{3}\sin(\sqrt{3}t) + B\sqrt{3}\cos(\sqrt{3}t) \right] \] \[ = e^{-t} \left[ (-A + B\sqrt{3})\cos(\sqrt{3}t) + (-B - A\sqrt{3})\sin(\sqrt{3}t) \right] \] At \( t = \): \[ \dot{x}() = (-A + B\sqrt{3}) = 1 \] Recall \( A = -.1 \): \[ -(-.1) + B\sqrt{3} = 1 \] \[ .1 + B\sqrt{3} = 1 \] \[ B\sqrt{3} = .9 \implies B = \frac{.9}{\sqrt{3}} = .5196 \] **Explanation:** We applied the initial conditions to solve for \( A \) and \( B \). --- ## **Final Answers** ### **a) Find \( x(t) \):** \[ \boxed{ x(t) = e^{-t} \left[ -.1\cos(\sqrt{3}t) + .52\sin(\sqrt{3}t) \right] } \] ### **b) Frequency in Hertz** The damped natural frequency is: \[ \omega_d = \sqrt{3} \approx 1.732\, \text{rad/s} \] \[ f = \frac{\omega_d}{2\pi} \approx \frac{1.732}{6.283} \approx .276\, \text{Hz} \] \[ \boxed{.28\ \text{Hz}} \] --- **Summary:** - **a)** \( x(t) = e^{-t} \left[ -.1\cos(\sqrt{3}t) + .52\sin(\sqrt{3}t) \right] \) - **b)** Frequency is \( .28\, \text{Hz} \)

## Step 1: **Write the Differential Equation and Find General Solution** ### **System Setup** - **Mass** \( M = 1\, \text{kg} \) - **Spring Constants**: Each spring \( K = 2\, \text{N/m} \) (two springs in parallel yield \( K_{total} = 4\, \text{N/m} \)) - **Damping Coefficient**: Each damper \( C = 1\, \text{kg/s} \) (two dampers in parallel yield \( C_{total} = 2\, \text{kg/s} \)) ### **Equation of Motion** The equation of motion for the mass-spring-damper system is: \[ M\ddot{x} + C_{total}\dot{x} + K_{total}x = 0 \] Substituting the values: \[ 1\ddot{x} + 2\dot{x} + 4x = 0 \] ### **Characteristic Equation** The characteristic equation is: \[ r^2 + 2r + 4 = 0 \] Using the quadratic formula: \[ r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-2 \pm \sqrt{-12}}{2} \] This simplifies to: \[ r = -1 \pm i\sqrt{3} \] ### **General Solution** The general solution of the system is: \[ x(t) = e^{-t} \left( A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t) \right) \] **Explanation:** We derived the equation of motion and solved for the characteristic equation to find the general solution, identifying the damping and oscillatory components. --- ## Step 2: **Apply Initial Conditions and Solve for Constants** ### **Initial Conditions:** - \( x(0) = -0.1 \) - \( \dot{x}(0) = 1 \) #### At \( t = 0 \): Substituting \( t = 0 \) into \( x(t) \): \[ x(0) = e^{0} (A \cdot 1 + B \cdot 0) = A = -0.1 \] #### First Derivative Calculation: The first derivative is given by: \[ \dot{x}(t) = \frac{d}{dt} \left[e^{-t} (A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t))\right] \] Using the product rule: \[ \dot{x}(t) = e^{-t} \left[-(A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t)) + (-A\sqrt{3})\sin(\sqrt{3}t) + B\sqrt{3}\cos(\sqrt{3}t)\right] \] This simplifies to: \[ \dot{x}(t) = e^{-t} \left[(-A + B\sqrt{3})\cos(\sqrt{3}t) + (-B - A\sqrt{3})\sin(\sqrt{3}t)\right] \] ### Evaluate at \( t = 0 \): Substituting \( t = 0 \): \[ \dot{x}(0) = (-A + B\sqrt{3}) = 1 \] Substituting \( A = -0.1 \): \[ -(-0.1) + B\sqrt{3} = 1 \] \[ 0.1 + B\sqrt{3} = 1 \] Solving for \( B \): \[ B\sqrt{3} = 0.9 \implies B = \frac{0.9}{\sqrt{3}} \approx 0.5196 \] **Explanation:** We applied the initial conditions to determine the constants \( A \) and \( B \) in the general solution, allowing us to fully describe the motion of the system. --- ## **Final Answers** ### **a) Expression for \( x(t) \):** \[ \boxed{x(t) = e^{-t} \left[ -0.1\cos(\sqrt{3}t) + 0.5196\sin(\sqrt{3}t) \right]} \] ### **b) Frequency in Hertz** The damped natural frequency is: \[ \omega_d = \sqrt{3} \approx 1.732\, \text{rad/s} \] Converting to Hertz: \[ f = \frac{\omega_d}{2\pi} \approx \frac{1.732}{6.283} \approx 0.276\, \text{Hz} \] Final frequency rounded: \[ \boxed{0.28\, \text{Hz}} \] --- **Summary:** - **a)** The position function is \( x(t) = e^{-t} \left[ -0.1\cos(\sqrt{3}t) + 0.5196\sin(\sqrt{3}t) \right] \) - **b)** The frequency is approximately \( 0.28\, \text{Hz} \)

## Step 1: **Write the Differential Equation and Find General Solution** ### **System Setup** - **Mass** \( M = 1\, \text{kg} \) - **Spring Constants**: Each spring \( K = 2\, \text{N/m} \), total \( K_{total} = 4\, \text{N/m} \) (two springs in parallel) - **Damping Coefficient**: Each damper \( C = 1\, \text{kg/s} \), total \( C_{total} = 2\, \text{kg/s} \) (two dampers in parallel) ### **Equation of Motion** The equation of motion for the mass-spring-damper system is: \[ M\ddot{x} + C_{total}\dot{x} + K_{total}x = 0 \] Substituting the values: \[ 1\ddot{x} + 2\dot{x} + 4x = 0 \] ### **Characteristic Equation** The characteristic equation is: \[ r^2 + 2r + 4 = 0 \] Using the quadratic formula: \[ r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-2 \pm \sqrt{-12}}{2} \] This simplifies to: \[ r = -1 \pm i\sqrt{3} \] ### **General Solution** The general solution of the system is: \[ x(t) = e^{-t} \left( A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t) \right) \] **Explanation:** We derived the equation of motion and solved for the characteristic equation to find the general solution, identifying the damping and oscillatory components. --- ## Step 2: **Apply Initial Conditions and Solve for Constants** ### **Initial Conditions** - \( x(0) = -0.1 \) - \( \dot{x}(0) = 1 \) #### At \( t = 0 \): Substituting \( t = 0 \) into \( x(t) \): \[ x(0) = e^{0} (A \cdot 1 + B \cdot 0) = A = -0.1 \] #### First Derivative Calculation The first derivative is given by: \[ \dot{x}(t) = \frac{d}{dt} \left[e^{-t} (A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t))\right] \] Using the product rule: \[ \dot{x}(t) = e^{-t} \left[-(A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t)) + (-A\sqrt{3})\sin(\sqrt{3}t) + B\sqrt{3}\cos(\sqrt{3}t)\right] \] This simplifies to: \[ \dot{x}(t) = e^{-t} \left[(-A + B\sqrt{3})\cos(\sqrt{3}t) + (-B - A\sqrt{3})\sin(\sqrt{3}t)\right] \] ### Evaluate at \( t = 0 \) Substituting \( t = 0 \): \[ \dot{x}(0) = (-A + B\sqrt{3}) = 1 \] Substituting \( A = -0.1 \): \[ -(-0.1) + B\sqrt{3} = 1 \] \[ 0.1 + B\sqrt{3} = 1 \] Solving for \( B \): \[ B\sqrt{3} = 0.9 \implies B = \frac{0.9}{\sqrt{3}} \approx 0.5196 \] **Explanation:** We applied the initial conditions to determine the constants \( A \) and \( B \) in the general solution, allowing us to fully describe the motion of the system. --- ## **Final Answers** ### **a) Expression for \( x(t) \)** \[ \boxed{x(t) = e^{-t} \left[ -0.1\cos(\sqrt{3}t) + 0.5196\sin(\sqrt{3}t) \right]} \] ### **b) Frequency in Hertz** The damped natural frequency is: \[ \omega_d = \sqrt{3} \approx 1.732\, \text{rad/s} \] Converting to Hertz: \[ f = \frac{\omega_d}{2\pi} \approx \frac{1.732}{6.283} \approx 0.276\, \text{Hz} \] Final frequency rounded: \[ \boxed{0.28\, \text{Hz}} \] --- **Summary:** - **a)** The position function is \( x(t) = e^{-t} \left[ -0.1\cos(\sqrt{3}t) + 0.5196\sin(\sqrt{3}t) \right] \) - **b)** The frequency is approximately \( 0.28\, \text{Hz} \)