Question

Use Laplace transforms to solve, for \(t \geq 0\), the differential equations $$ \begin{aligned} \ddot{x}+2 x+y &=\cos t \\ \ddot{y}+2 x+3 y &=2 \cos t \end{aligned} $$ which describe a coupled system that starts from rest at the equilibrium position. Show that the subsequent motion takes place along a straight line in the \(x y\)-plane. Verify that the frequency at which the system is driven is equal to one of the resonance frequencies of the system; explain why there is no resonant behaviour in the solution you have obtained

Short answer

Use Laplace transforms to find X(s) and Y(s), then inverse transform to find x(t) and y(t).

Step-by-step solution

- Take the Laplace transform of both equations

Recall that the Laplace transform of \(\frac{d^2 x}{dt^2}\) is \s^2 X(s) - sx(0) - \dot{x}(0)\. Similarly for y(t). Given that both x(0) and y(0) along with their first derivatives are zero, we transform the system: \[ s^2 X(s) + 2X(s) + Y(s) = \frac{s}{s^2 + 1} \] \[ s^2 Y(s) + 2X(s) + 3Y(s) = \frac{2s}{s^2 + 1} \]

- Simplify the equations

We need to isolate X(s) and Y(s): \[ (s^2 + 2)X(s) + Y(s) = \frac{s}{s^2 + 1} \] \[ 2X(s) + (s^2 + 3)Y(s) = \frac{2s}{s^2 + 1} \] Solve the first equation for Y(s): \[ Y(s) = \frac{s}{s^2 + 1} - (s^2 + 2)X(s) \]

- Substitute Y(s) into second equation

Replace Y(s) in the second equation: \[ 2X(s) + (s^2 + 3)\left(\frac{s}{s^2 + 1} - (s^2 + 2)X(s)\right) = \frac{2s}{s^2 + 1} \] Simplify and solve for X(s): \[ 2X(s) + \frac{(s^3 + 3s - s^3 - 2s^2 - s}{s^2 + 1} - (s^2 + 3)(s^2 + 2)X(s) = \frac{2s}{s^2 + 1} \] \[ 2X(s) + \frac{-2s^2 - s}{s^2 + 1} - (s^4 + 5s^2 + 6)X(s) = \frac{2s}{s^2 + 1} \] Combine and simplify terms to solve for X(s).

- Solve for Y(s)

Once X(s) is found, substitute it back into \[ Y(s) = \frac{s}{s^2 + 1} - (s^2 + 2)X(s) \] to find Y(s).

- Use inverse Laplace transform

Take the inverse Laplace transform of X(s) and Y(s) to get x(t) and y(t).

Key concepts

Coupled Differential Equations

Coupled differential equations are a set of equations where multiple dependent variables (like x(t) and y(t) in this problem) depend on each other and their derivatives. These systems often describe physical phenomena where two or more properties are interconnected, such as electrical circuits, mechanical systems, or chemical reactions.
In our problem, we have:
\ \( \ddot{x} + 2x + y = \cos(t) \ \) \ \( \ddot{y} + 2x + 3y = 2 \cos(t) \ \)
Here, \( x \) and \( y \) are functions of time \( t \). The second derivatives of \( x \) and \( y \) describe their accelerations. We start solving these by transforming them into a simpler form using the Laplace transform.
This transformation converts the differential equations into algebraic equations that are easier to manipulate and solve.

Resonance Frequencies

Resonance frequencies are the natural frequencies at which a system tends to oscillate in the absence of any driving force. When an external force matches a system's resonance frequency, the system can exhibit large oscillations.
In this problem, our equations are driven by terms involving \( \cos(t) \), which have a natural driving frequency of 1. When we solve the differential equations, we find the motion frequencies and compare them to the natural (resonant) frequency of the system.
Even though the driving frequency equals one of the system’s natural frequencies, resonance might not occur. That’s because the initial conditions, or other specific factors, can dampen these large oscillations, preventing resonant behavior.
In this case, the system starts from rest, which means the initial position and velocity are zero, potentially avoiding resonance effects.

Inverse Laplace Transform

The inverse Laplace transform is used to convert a function from the Laplace domain back to the time domain. It’s the reverse process of the Laplace transform.
After transforming our coupled differential equations into algebraic equations, we solve for functions \( X(s) \) and \( Y(s) \). The solutions in the Laplace domain don't directly tell us the time-behavior, so we revert them using the inverse Laplace transform.
To find \( x(t) \) and \( y(t) \) from \( X(s) \) and \( Y(s) \), we use inverse Laplace transform tables or techniques, turning Laplace solutions like \( \frac{s}{(s^2+1)} \) back into time functions like \( \cos(t) \).

Equilibrium Position

The equilibrium position of a system is a state where it remains at rest if not disturbed. In this problem, the system starts from rest at the equilibrium position, meaning initial velocities and displacements are zero.
We transform both equations considering these initial conditions. This simplifies calculations since the initial terms in the Laplace transform equations become zero.
Once we find \( x(t) \) and \( y(t) \), these solutions describe how the system moves over time from its equilibrium. We can then verify that this motion is along a straight line in the \(xy\)-plane, meaning the displacement stays in a simple, predictable path.