Let's address each part step by step: --- ## **a) Formulate the Differential Equation** ### **Using Newton's Law of Motion:** Newton’s second law: \[ F = m a = m \frac{d^2 x}{dt^2} \] For the spring-mass system: - **Spring force:** \( F_s = -kx \) (restoring force) - **External force:** \( F(t) = 400 \cos(10t) \) Sum of forces: \[ m \frac{d^2 x}{dt^2} = -kx + F(t) \] Substitute the given values: - \( m = 10\, \text{kg} \) - \( k = 400\, \text{N/m} \) - \( F(t) = 400 \cos(10t) \) \[ 10 \frac{d^2 x}{dt^2} = -400x + 400 \cos(10t) \] --- ## **b) Write in Standard Form and State Properties** ### **Standard Form:** \[ \frac{d^2 x}{dt^2} + 400x = 40 \cos(10t) \] (Divide both sides by 10) ### **Properties:** - **Order:** 2 (second-order) - **Degree:** 1 (highest power of derivative is 1) - **Linearity:** Linear (no products or powers of \( x \) or its derivatives) - **Homogeneity:** Nonhomogeneous (right side is not zero) --- ## **c) Find the General Solution Using Undetermined Coefficients** ### **Homogeneous Solution:** Solve: \[ \frac{d^2 x}{dt^2} + 400x = \] The characteristic equation: \[ r^2 + 400 = \implies r^2 = -400 \implies r = \pm 20i \] So, the complementary (homogeneous) solution: \[ x_h(t) = C_1 \cos(20t) + C_2 \sin(20t) \] ### **Particular Solution:** Guess a particular solution of the form: \[ x_p(t) = A \cos(10t) + B \sin(10t) \] Find its derivatives: \[ x_p'(t) = -10A \sin(10t) + 10B \cos(10t) \] \[ x_p''(t) = -100A \cos(10t) - 100B \sin(10t) \] Substitute into the ODE: \[ x_p'' + 400 x_p = 40 \cos(10t) \] \[ (-100A \cos(10t) - 100B \sin(10t)) + 400(A \cos(10t) + B \sin(10t)) = 40 \cos(10t) \] \[ (-100A + 400A)\cos(10t) + (-100B + 400B)\sin(10t) = 40\cos(10t) \] \[ (300A)\cos(10t) + (300B)\sin(10t) = 40\cos(10t) \] Matching coefficients: - \( 300A = 40 \implies A = \frac{2}{15} \) - \( 300B = \implies B = \) So, \[ x_p(t) = \frac{2}{15} \cos(10t) \] ### **General Solution:** \[ x(t) = C_1 \cos(20t) + C_2 \sin(20t) + \frac{2}{15} \cos(10t) \] --- ## **d) Apply Initial Conditions** Given: - \( x() = .1 \) - \( x'() = \) First, calculate \( x'(t) \): \[ x'(t) = -20C_1 \sin(20t) + 20C_2 \cos(20t) - \frac{2}{15} \cdot 10 \sin(10t) \] \[ x'(t) = -20C_1 \sin(20t) + 20C_2 \cos(20t) - \frac{4}{3} \sin(10t) \] At \( t = \): - \( \cos() = 1, \sin() = \) So, \[ x() = C_1 \cdot 1 + C_2 \cdot + \frac{2}{15} \cdot 1 = C_1 + \frac{2}{15} \] \[ x'() = -20C_1 \cdot + 20C_2 \cdot 1 - \frac{4}{3} \cdot = 20C_2 \] Set equal to initial conditions: \[ x() = .1 \implies C_1 + \frac{2}{15} = .1 \implies C_1 = .1 - \frac{2}{15} = .1 - .1333 = -.0333 \] \[ x'() = \implies 20C_2 = \implies C_2 = \] --- ## **Final Solution** \[ \boxed{ x(t) = -.0333 \cos(20t) + \frac{2}{15} \cos(10t) } \] where \( \frac{2}{15} \approx .1333 \). --- ### **Summary of Steps** 1. **Formulate ODE:** \( 10x'' + 400x = 400\cos(10t) \) 2. **Standard Form:** \( x'' + 400x = 40\cos(10t) \); 2nd order, 1st degree, linear, nonhomogeneous. 3. **General Solution:** \( x(t) = C_1 \cos(20t) + C_2 \sin(20t) + \frac{2}{15}\cos(10t) \) 4. **With Initial Conditions:** \( x(t) = -.0333 \cos(20t) + .1333 \cos(10t) \)