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Magazine Chapter 4: Problem 6
A system has the equation of motion \(\ddot{x}+5 \dot{x}+6 x=F(t)\) where, at \(t=0, x=0\) and \(\dot{x}=2\). If \(F(t)\) is an impulse of 20 units applied at \(t=4\), determine an expression for \(x\) in terms of \(t\).
Short Answer
Expert verified
The expression for \( x(t) \) is \[ x(t) = -2 e^{-2t} + 2 e^{-3t} + 20 [-e^{-2(t-4)} + e^{-3(t-4)}] u(t-4). \]
Step by step solution
01
- Identify the equation of motion
The given equation of motion is \[ \frac{d^2x}{dt^2} + 5 \frac{dx}{dt} + 6x = F(t) \] with initial conditions, \[ x(0) = 0 \text{ and } \frac{dx}{dt}(0) = 2 \] and an impulse function, \( F(t) = 20 \delta(t-4) \).
02
- Analyze the differential equation
The homogeneous part of the equation is \[ \frac{d^2x}{dt^2} + 5 \frac{dx}{dt} + 6x = 0 \] its characteristic equation is \[ r^2 + 5r + 6 = 0 \].
03
- Solve the characteristic equation
Solving \( r^2 + 5r + 6 = 0 \) gives roots \( r_1 = -2 \) and \( r_2 = -3 \). Thus, the general solution for the homogeneous equation is \[ x_h(t) = C_1 e^{-2t} + C_2 e^{-3t} \].
04
- Determine constants using initial conditions
Using initial conditions: \( x(0) = 0 \) leads to \( C_1 + C_2 = 0 \) and \( \frac{dx}{dt}(0) = 2 \) leads to \( -2C_1 - 3C_2 = 2 \). Solving these we get \( C_1 = -2 \) and \( C_2 = 2 \). Therefore, the homogeneous solution is \[ x_h(t) = -2 e^{-2t} + 2 e^{-3t} \].
05
- Consider the impulse function
For an impulse \(20 \delta(t-4)\), the response is the impulse response function multiplied by the impulse strength: \[ x_i(t) = 20 \times g(t-4) \] where \( g(t) \) is the response of the system to an impulse at \( t=0 \).
06
- Find impulse response
The impulse response \( g(t) \) is found by solving the homogeneous equation with initial conditions \( x(0)=0 \) and \( \frac{dx}{dt}(0)=1 \). This gives: \[ g(t) = -e^{-2t} + e^{-3t} \].
07
- Formulate the total solution
The impulse response shifted by 4 units in time and scaled by 20 units is \[ x_i(t) = 20 [-e^{-2(t-4)} + e^{-3(t-4)}] u(t-4) \], where \( u(t-4) \) is the unit step function ensuring the impulse effect starts at \( t = 4 \). The total solution is the sum of homogeneous and impulse responses.
08
- Combine homogeneous and impulse responses
Combining the two parts, the total solution is \[ x(t) = -2 e^{-2t} + 2 e^{-3t} + 20 [-e^{-2(t-4)} + e^{-3(t-4)}] u(t-4) \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of Motion
In engineering and physics, the equation of motion describes how a system evolves over time. In this problem, we have a second-order differential equation: \[\ddot{x}+5 \dot{x}+6 x=F(t)\] Where:
- \(\ddot{x}\) is the second derivative of \(x(t)\), representing acceleration.
- \(\dot{x}\) is the first derivative of \(x(t)\), representing velocity.
- \(x(t)\) is the displacement.
- \(F(t)\) is the external force applied to the system.
Initial Conditions
Initial conditions specify the state of the system at the beginning of the observation, typically at \(t = 0\). For our problem, we have:
- \(x(0) = 0\): The system starts from rest, with no initial displacement.
- \(\dot{x}(0) = 2\): The initial velocity of the system is 2 units.
Impulse Response
An impulse is a sudden force applied over a short time interval. Here, an impulse of 20 units is applied at \(t = 4\). Impulse response is how the system reacts to such an impulse. The given impulse can be represented by the Dirac delta function: \(F(t) = 20 \delta(t-4)\). Using the impulse response function \(g(t)\), which is the system鈥檚 response to an impulse at \(t=0\), we can compute the overall response as: \[x_{i}(t) = 20 \cdot g(t-4)\].
Homogeneous Solution
The homogeneous solution corresponds to the solution of the homogeneous differential equation: \[\ddot{x} + 5 \dot{x} + 6 x = 0\].Solving for the roots of its characteristic equation \[r^2 + 5r + 6 = 0\],we find \(r_1 = -2\) and \(r_2 = -3\). The general homogeneous solution is then: \[x_h(t) = C_1 e^{-2t} + C_2 e^{-3t}\].Using the initial conditions, we determine the constants \(C_1\) and \(C_2\), forming the complete homogeneous solution.
Characteristic Equation
The characteristic equation is obtained by solving the homogeneous differential equation: \[r^2 + 5r + 6 = 0\].This quadratic equation is derived by assuming a solution of the form \(x(t) = e^{rt}\). Solving for \(r\) provides the roots \(r_1 = -2\) and \(r_2 = -3\). These roots are used to construct the general solution to the homogeneous equation \[x_h(t) = C_1 e^{-2t} + C_2 e^{-3t}\], dependent on the system鈥檚 dynamics.
