Math Problem Statement

A forcing function Fdrive(t)=-40sin(t) N is applied to a mass of 1 kg in a mass on a spring problem in which the spring constant is 5 N/m and there is damping proportional to velocity with coefficient 4 N*s/m. If the initial mass position is -5 m and the initial velocity is 2 m/s, find the unique solution and identify the steady state and transient terms.  

Solution

To solve this problem, let's model it with a second-order linear differential equation with forcing, damping, and spring constants involved. The equation of motion is:

mx¨+cx˙+kx=Fdrive(t)m \ddot{x} + c \dot{x} + k x = F_{\text{drive}}(t)

where:

  • m=1kgm = 1 \, \text{kg} (mass),
  • c=4Ns/mc = 4 \, \text{Ns/m} (damping coefficient),
  • k=5N/mk = 5 \, \text{N/m} (spring constant),
  • Fdrive(t)=40sin(t)F_{\text{drive}}(t) = -40 \sin(t) (forcing function).

Given initial conditions:

  • x(0)=5mx(0) = -5 \, \text{m},
  • x˙(0)=2m/s\dot{x}(0) = 2 \, \text{m/s}.

Step 1: Set Up the Differential Equation

The equation of motion becomes: x¨+4x˙+5x=40sin(t)\ddot{x} + 4 \dot{x} + 5 x = -40 \sin(t)

Step 2: Solve for the Homogeneous Solution (Transient Term)

The homogeneous equation is: x¨+4x˙+5x=0\ddot{x} + 4 \dot{x} + 5 x = 0

Assume a solution of the form xh(t)=eλtx_h(t) = e^{\lambda t}. Plugging into the homogeneous equation gives: λ2+4λ+5=0\lambda^2 + 4 \lambda + 5 = 0

Solving this quadratic equation: λ=4±16202=2±i\lambda = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i

Thus, the homogeneous solution is: xh(t)=e2t(Acost+Bsint)x_h(t) = e^{-2t} (A \cos t + B \sin t)

This represents the transient solution, which will decay over time due to the e2te^{-2t} factor.

Step 3: Solve for the Particular Solution (Steady-State Term)

Since the driving force is 40sin(t)-40 \sin(t), we look for a particular solution of the form: xp(t)=Csin(t)+Dcos(t)x_p(t) = C \sin(t) + D \cos(t)

Differentiating: x˙p(t)=Ccos(t)Dsin(t)\dot{x}_p(t) = C \cos(t) - D \sin(t) x¨p(t)=Csin(t)Dcos(t)\ddot{x}_p(t) = -C \sin(t) - D \cos(t)

Substitute xp(t)x_p(t), x˙p(t)\dot{x}_p(t), and x¨p(t)\ddot{x}_p(t) into the original equation: (Csin(t)Dcos(t))+4(Ccos(t)Dsin(t))+5(Csin(t)+Dcos(t))=40sin(t)(-C \sin(t) - D \cos(t)) + 4 (C \cos(t) - D \sin(t)) + 5 (C \sin(t) + D \cos(t)) = -40 \sin(t)

Grouping terms involving sin(t)\sin(t) and cos(t)\cos(t): (C+5C+4D)sin(t)+(D+5D4C)cos(t)=40sin(t)(-C + 5C + 4D) \sin(t) + (-D + 5D - 4C) \cos(t) = -40 \sin(t)

This gives the system:

  1. 4C+4D=404C + 4D = -40
  2. 4C+4D=0-4C + 4D = 0

Solving these equations, we find:

  • C=5C = -5
  • D=5D = -5

Thus, the particular solution is: xp(t)=5sin(t)5cos(t)x_p(t) = -5 \sin(t) - 5 \cos(t)

This represents the steady-state solution, which oscillates with the same frequency as the driving force.

Step 4: Write the General Solution

The general solution is the sum of the homogeneous and particular solutions: x(t)=xh(t)+xp(t)x(t) = x_h(t) + x_p(t) x(t)=e2t(Acost+Bsint)5sin(t)5cos(t)x(t) = e^{-2t} (A \cos t + B \sin t) - 5 \sin(t) - 5 \cos(t)

Step 5: Apply Initial Conditions

Using x(0)=5x(0) = -5 and x˙(0)=2\dot{x}(0) = 2:

  1. Initial Position:
    x(0)=A5=5A=0x(0) = A - 5 = -5 \Rightarrow A = 0

  2. Initial Velocity:
    x˙(t)=2e2t(Acost+Bsint)+e2t(Asint+Bcost)5cos(t)+5sin(t)\dot{x}(t) = -2 e^{-2t} (A \cos t + B \sin t) + e^{-2t} (-A \sin t + B \cos t) - 5 \cos(t) + 5 \sin(t)

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Forced Oscillations
Damping
Harmonic Motion

Formulas

m * d²x/dt² + c * dx/dt + k * x = F_drive(t)
Characteristic equation: λ² + 4λ + 5 = 0
Particular solution form for sinusoidal forcing: x_p(t) = C * sin(t) + D * cos(t)

Theorems

Second-Order Linear Differential Equation Solution
Method of Undetermined Coefficients

Suitable Grade Level

Undergraduate - Physics/Engineering