Let's solve the problem step-by-step as requested. --- ## (a) **Derive the 2nd Order Differential Equation** ### **Step 1: Find the Spring Constant \( k \) Using Hooke’s Law** Hooke's Law: \[ F = k \Delta x \] Given: - \( F = 25.6 \) N - Natural length = 0.5 m - Stretched length = 0.7 m - \( \Delta x = 0.7 - 0.5 = 0.2 \) m Plug into Hooke’s Law: \[ k = \frac{F}{\Delta x} = \frac{25.6}{0.2} = 128 \, \text{N/m} \] --- ### **Step 2: Write the Equation of Motion** For a damped spring-mass system, by Newton’s second law: \[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 \] Given: - \( m = 2 \) kg, - \( c = 40 \) Ns/m, - \( k = 128 \) N/m, Plug in the values: \[ 2\frac{d^2x}{dt^2} + 40\frac{dx}{dt} + 128x = 0 \] Divide both sides by 2: \[ \frac{d^2x}{dt^2} + 20\frac{dx}{dt} + 64x = 0 \] --- ### **Final Differential Equation:** \[ \boxed{ \frac{d^2x}{dt^2} + 20\frac{dx}{dt} + 64x = 0 } \] --- ## (b) **Solve for \( x(t) \) Given Initial Conditions** ### **Step 1: Write the Characteristic Equation** \[ r^2 + 20r + 64 = 0 \] Solve for \( r \): \[ r = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot 64}}{2} = \frac{-20 \pm \sqrt{400 - 256}}{2} = \frac{-20 \pm \sqrt{144}}{2} = \frac{-20 \pm 12}{2} \] So, - \( r_1 = \frac{-20 + 12}{2} = -4 \) - \( r_2 = \frac{-20 - 12}{2} = -16 \) ### **Step 2: Write the General Solution** \[ x(t) = A e^{-4t} + B e^{-16t} \] --- ### **Step 3: Apply Initial Conditions** #### **Initial Position:** The spring is stretched to 0.7 m, equilibrium is at 0.5 m, so initial displacement from equilibrium: \[ x(0) = 0.7 - 0.5 = 0.2 \text{ m} \] So, \[ x(0) = A + B = 0.2 \] #### **Initial Velocity:** Given: initial velocity is **0.6 m/s** (released with velocity). \[ \frac{dx}{dt} = -4A e^{-4t} - 16B e^{-16t} \] At \( t = 0 \): \[ \frac{dx}{dt}\bigg|_{t=0} = -4A - 16B = 0.6 \] --- ### **Step 4: Solve for \(A\) and \(B\)** We have: \[ A + B = 0.2 \tag{1} \] \[ -4A - 16B = 0.6 \tag{2} \] From (1): \( A = 0.2 - B \) Plug into (2): \[ -4(0.2 - B) - 16B = 0.6 \] \[ -0.8 + 4B - 16B = 0.6 \] \[ -0.8 - 12B = 0.6 \] \[ -12B = 0.6 + 0.8 = 1.4 \] \[ B = -\frac{1.4}{12} = -0.1167 \] Now, \( A = 0.2 - (-0.1167) = 0.3167 \) --- ### **Step 5: Write the Final Solution** \[ \boxed{ x(t) = 0.3167\, e^{-4t} - 0.1167\, e^{-16t} } \] --- ## **Summary Table** | Step | Result | |------|--------| | Differential Equation | \(\frac{d^2x}{dt^2} + 20\frac{dx}{dt} + 64x = 0\) | | General Solution | \(x(t) = A e^{-4t} + B e^{-16t}\) | | Initial Displacement | \(x(0) = 0.2\) m | | Initial Velocity | \(x'(0) = 0.6\) m/s | | Constants | \(A = 0.3167, B = -0.1167\) | | **Final Solution** | \(\boxed{x(t) = 0.3167\, e^{-4t} - 0.1167\, e^{-16t}}\) | **This function gives the position of the mass at any time \( t \).** Let me know if you need the plot or further analysis!

Certainly! Here are the final answers summarized: --- ## **(a) Differential Equation of the System** Given data: - Mass, \( m = 2\, \text{kg} \) - Natural length of spring = 0.5 m - Force to stretch to 0.7 m = 25.6 N - Damping constant, \( c = 40\, \text{Ns/m} \) **Step 1:** Find spring constant \( k \): \[ \Delta x = 0.7 - 0.5 = 0.2\, \text{m} \] \[ k = \frac{F}{\Delta x} = \frac{25.6}{0.2} = 128\, \text{N/m} \] **Step 2:** Write the equation of motion: \[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \] \[ 2 \frac{d^2x}{dt^2} + 40 \frac{dx}{dt} + 128 x = 0 \] Divide through by 2: \[ \boxed{ \frac{d^2x}{dt^2} + 20 \frac{dx}{dt} + 64 x = 0 } \] --- ## **(b) Solution for \( x(t) \) with initial conditions** **Initial conditions:** - The spring is stretched from equilibrium (0.5 m) to 0.7 m, so initial displacement: \[ x(0) = 0.7 - 0.5 = \boxed{0.2\, \text{m}} \] - Initial velocity is 0.6 m/s (upon release): \[ x'(0) = 0.6\, \text{m/s} \] --- ### **Step 1:** Find characteristic roots: \[ r^2 + 20 r + 64 = 0 \] \[ r = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 64}}{2} = \frac{-20 \pm \sqrt{400 - 256}}{2} = \frac{-20 \pm \sqrt{144}}{2} \] \[ r = \frac{-20 \pm 12}{2} \] Roots: \[ r_1 = -4,\quad r_2 = -16 \] ### **Step 2:** Write the general solution: \[ x(t) = A e^{-4t} + B e^{-16t} \] ### **Step 3:** Apply initial conditions: - At \( t=0 \): \[ x(0) = A + B = 0.2 \] - Derivative: \[ x'(t) = -4A e^{-4t} - 16 B e^{-16t} \] At \( t=0 \): \[ x'(0) = -4A - 16 B = 0.6 \] ### **Step 4:** Solve for \(A\) and \(B\): From \(A + B = 0.2\): \[ A = 0.2 - B \] Substitute into the velocity equation: \[ -4(0.2 - B) - 16 B = 0.6 \] \[ -0.8 + 4B - 16 B = 0.6 \] \[ -0.8 - 12 B = 0.6 \] \[ -12 B = 1.4 \] \[ B = -\frac{1.4}{12} = -0.1167 \] Then, \[ A = 0.2 - (-0.1167) = 0.3167 \] --- ## **Final Position Function:** \[ \boxed{ x(t) = 0.3167\, e^{-4t} - 0.1167\, e^{-16t} } \] **This describes the displacement from equilibrium at any time \( t \).** --- **Summary:** | Part | Final Answer | |--------|--------------| | Differential Equation | \(\displaystyle \frac{d^2x}{dt^2} + 20 \frac{dx}{dt} + 64 x = 0\) | | General Solution | \(\displaystyle x(t) = A e^{-4t} + B e^{-16t}\) | | Constants | \(\displaystyle A \approx 0.3167,\quad B \approx -0.1167\) | | Initial Conditions | \(x(0) = 0.2\, \text{m}\), \(x'(0) = 0.6\, \text{m/s}\) | | Final Position Function | \(\boxed{x(t) = 0.3167\, e^{-4t} - 0.1167\, e^{-16t}}\) | Let me know if you'd like further assistance!

**Final Answers:** **(a) Differential Equation:** \[ \boxed{ \frac{d^2x}{dt^2} + 20\frac{dx}{dt} + 64x = 0 } \] **(b) Solution for \(x(t)\):** \[ \boxed{ x(t) = 0.3167\, e^{-4t} - 0.1167\, e^{-16t} } \]