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Magazine Chapter 39: Problem 20
The equation of motion of a body performing damped forced vibrations is \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+5 \frac{\mathrm{d} x}{\mathrm{~d} t}+6 x=\cos t .\) Solve this equation, given that \(x=0 \cdot 1\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}=0\) when \(t=0\). Write the steady-state solution in the form \(K \sin (t+a)[\) see 13\(]\).
Short Answer
Expert verified
The steady-state solution is expressed as \( x_{ss}(t) = \frac{1}{29} \sqrt{26} \sin(t + \tan^{-1}(\frac{1}{5})) \).
Step by step solution
01
Understand the Equation
The given differential equation is \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} + 5 \frac{\mathrm{d} x}{\mathrm{~d} t} + 6x = \cos t \). This is a second-order linear non-homogeneous differential equation with constant coefficients and a forcing function on the right side.
02
Find the Complementary Solution
To find the complementary solution, solve the associated homogeneous equation: \( \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} + 5 \frac{\mathrm{d} x}{\mathrm{~d} t} + 6x = 0 \). The characteristic equation is \( r^2 + 5r + 6 = 0 \). Solve this quadratic equation to find \( r = -2 \) and \( r = -3 \). Thus, the complementary solution is \( x_c(t) = C_1 e^{-2t} + C_2 e^{-3t} \).
03
Determine the Particular Solution
Since the non-homogeneous term is \( \cos t \), try a particular solution of the form \( x_p(t) = A \cos t + B \sin t \). Substitute into the differential equation and equate coefficients to find \( A \) and \( B \). After substituting, solve for the coefficients: \([A - 5B] (-1)\sin t + [6A - B + 6B] \cos t = \cos t \), giving \( A - 5B = 0 \) and \( 6A - 4B = 1 \). Solving these equations yields \( A = \frac{1}{29} \) and \( B = \frac{5}{29} \). The particular solution is therefore \( x_p(t) = \frac{1}{29} \cos t + \frac{5}{29} \sin t \).
04
Write the General Solution
The general solution is the sum of the complementary and particular solutions: \( x(t) = x_c(t) + x_p(t) = C_1 e^{-2t} + C_2 e^{-3t} + \frac{1}{29} \cos t + \frac{5}{29} \sin t \).
05
Apply Initial Conditions
Use the initial conditions \( x(0) = 0.1 \) and \( \frac{\mathrm{d} x}{\mathrm{~d} t}(0) = 0 \) to find \( C_1 \) and \( C_2 \). Substituting \( x(0) = 0.1 \) yields \( C_1 + C_2 + \frac{1}{29} = 0.1 \). The derivative \( \frac{\mathrm{d} x}{\mathrm{~d} t} = -2C_1 e^{-2t} -3C_2 e^{-3t} - \frac{5}{29} \cos t + \frac{1}{29} \sin t \). Substituting \( \frac{\mathrm{d} x}{\mathrm{~d} t}(0) = 0 \) yields \(-2C_1 - 3C_2 = 0\). Solve these equations to find \( C_1 = 0.211 \) and \( C_2 = -0.140 \).
06
Identify the Steady-State Solution
The steady-state solution corresponds to the particular solution as \( t \to \infty \). Therefore, the steady-state solution is \( x_{ss}(t) = \frac{1}{29} \cos t + \frac{5}{29} \sin t \).
07
Write the Steady-State Solution in Given Form
Use the trigonometric identity to express \( x_{ss}(t) = K \sin(t + a) \). Calculate \( K = \sqrt{\left(\frac{1}{29}\right)^2 + \left(\frac{5}{29}\right)^2} = \frac{1}{29} \sqrt{26} \), and \( \tan(a) = \frac{1/29}{5/29} = \frac{1}{5} \). Thus, \( a = \tan^{-1}\left(\frac{1}{5}\right) \). The steady-state solution is \( x_{ss}(t) = \frac{1}{29} \sqrt{26} \sin(t + \tan^{-1}(\frac{1}{5})) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Linear Differential Equation
A second-order linear differential equation is a type of equation that involves derivatives up to the second degree of a function. In mathematical terms, it takes the form: \\[ a \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} + b \frac{\mathrm{d} x}{\mathrm{~d} t} + cx = g(t) \\] \
This type of equation often models physical systems like mechanical vibrations where forces, positions, and velocities interact. The solution usually requires finding the complementary (homogeneous) and particular (non-homogeneous) solutions.
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- Order: The highest derivative is the second derivative \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}\), making it a second-order equation. \
- Linearity: Each term involves either \(x\) or its derivatives to the first power, meaning no products or powers of \(x\) and its derivatives. \
- Homogeneous vs Non-homogeneous: When \(g(t) = 0\), the equation is homogeneous; when \(g(t) eq 0\), it is non-homogeneous. \
This type of equation often models physical systems like mechanical vibrations where forces, positions, and velocities interact. The solution usually requires finding the complementary (homogeneous) and particular (non-homogeneous) solutions.
Steady-State Solution
The steady-state solution is a concept applied to systems that undergo periodic forces, like the vibrations modeled by our differential equation. It's the component of the solution that describes the long-term behavior of the system once transient effects have faded away. \
The steady-state solution is essential in predicting a system's response to persistent external influences, coming into play when assessing systems like electrical circuits or mechanical structures experiencing harmonic forces.
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- Relevance: It signifies how the system behaves under perpetual conditions after initial effects dissipate. \
- Derivation: In the equation \(\frac{1}{29} \cos t + \frac{5}{29} \sin t\), this steady state is defined by the particular solution due to the external force represented by \(\cos t\). \
- Form: The steady-state can be transformed using a trigonometric identity into the form \(K \sin(t + a)\), where \(K\) represents the amplitude and \(a\) the phase shift relative to the external force. \
The steady-state solution is essential in predicting a system's response to persistent external influences, coming into play when assessing systems like electrical circuits or mechanical structures experiencing harmonic forces.
Initial Conditions
Initial conditions are the specific values provided at the start of a problem, allowing one to find a unique solution to a differential equation. For the given equation, the initial conditions are \(x(0) = 0.1\) and \(\frac{\mathrm{d} x}{\mathrm{~d} t}(0) = 0\), specified at time \(t = 0\). \
Initial conditions are a critical aspect of solving differential equations, providing a foundation upon which the specific solution is built and tailored to fit the problem's context.
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- Importance: They provide the necessary constraints to solve for integration constants in the general solution, ensuring the solution accurately reflects the physical scenario. \
- Application: By applying \(x(0) = 0.1\), both constants \(C_1\) and \(C_2\) are adjusted within the complementary solution to meet these initial values. \
- Derivatives: Initial conditions often include both the function's value and its derivatives at specific points to fully characterize the motion or behavior of the system. \
Initial conditions are a critical aspect of solving differential equations, providing a foundation upon which the specific solution is built and tailored to fit the problem's context.
