Question

(a) Show that the phase of the forced response of Eq. ( 1) satisfies tan \(\delta=\gamma \omega / m\left(\omega_{0}^{2}-\omega^{2}\right)\) (b) Plot the phase \(\delta\) as a function of the forcing frequency \(\omega\) for the forced response of \(u^{\prime \prime}+0.125 u^{\prime}+u=3 \cos \omega t\)

Short answer

Provide a short answer summarizing step-by-step solution to find the phase of the forced response of a second-order differential equation and then plot it as a function of the forcing frequency. To find the phase of the forced response of a second-order differential equation, follow these steps: 1. Write the general solution of the equation as the sum of the complementary function and the particular solution for the forced response. 2. Find the forced response with a trial solution of the form \(A \cos(\omega t) + B \sin(\omega t)\) and calculate its derivatives. 3. Substitute the trial solution and its derivatives into the given differential equation. 4. Solve for A and B in terms of omega by equating the coefficients of \(\cos(\omega t)\) and \(\sin(\omega t)\) on both sides of the equation. 5. Calculate the phase angle \(\delta\) by taking the tangent inverse of the expression involving A and B, i.e., \(\delta(\omega) = \arctan\left(\frac{\gamma \omega}{m\left(\omega_{0}^{2}-\omega^{2}\right)}\right)\). 6. Plot the phase function as a function of the forcing frequency \(\omega\) using a graphing tool or a programming language.

Step-by-step solution

Write the general solution for the forced response

The general solution of a non-homogeneous linear second-order differential equation can be written as the sum of the complementary function (homogeneous equation) and the particular solution (forced response). Mathematically, it can be written as: \(u(t) = u_c(t) + u_p(t)\) Here, \(u_c(t)\) is the complementary function and \(u_p(t)\) is the particular solution for the forced response.

Find the forced response with a trial solution

Given that the forcing function is \(3\cos(\omega t)\), we can take a trial solution of the form: \(u_p(t) = A \cos(\omega t) + B \sin(\omega t)\) Now, we find the first and second derivatives of \(u_p(t)\) for substitution into the given differential equation: \(u_p'(t) = -A \omega \sin(\omega t) + B \omega \cos(\omega t)\) \(u_p''(t) = -A \omega^2 \cos(\omega t) - B \omega^2 \sin(\omega t)\)

Substitute the trial solution and its derivatives into the given differential equation

Now we substitute \(u_p(t)\), \(u_p'(t)\), and \(u_p''(t)\) into the given equation: \(u''+0.125u'+u=3\cos(\omega t)\) which results in: \((-A \omega^2 \cos(\omega t) - B \omega^2 \sin(\omega t)) + 0.125(-A \omega \sin(\omega t) + B \omega \cos(\omega t)) + (A \cos(\omega t) + B \sin(\omega t)) = 3\cos(\omega t)\)

Solve for A and B in terms of omega

Now, we solve for A and B in terms of omega by equating the coefficients of \(\cos(\omega t)\) and \(\sin(\omega t)\) on both sides of the above equation. This results in the following system of equations: \((1 - \omega^2)A + 0.125\omega B = 3\) \(0.125\omega A + (1 - \omega^2)B = 0\) Solving this system of equations, we get: \(A(\omega) = \frac{3\left(1-\omega^2\right)}{\left(1-\omega^2\right)^2+\left(0.125\omega\right)^2}\) \(B(\omega) = -\frac{3\left(0.125\omega\right)}{\left(1-\omega^2\right)^2+\left(0.125\omega\right)^2}\)

Calculate the phase delta

Now that we have the expressions for A and B in terms of omega, we can find the tangent of the phase angle \(\delta\) by taking the ratio of B and A: \(\tan \delta=\gamma \omega/m\left(\omega_{0}^{2}-\omega^{2}\right)=\frac{-B(\omega)}{A(\omega)}\) Substituting the expressions for A and B from Step 4, we get: \(\tan \delta =\frac{-\left(-\frac{3\left(0.125\omega\right)}{\left(1-\omega^2\right)^2+\left(0.125\omega\right)^2}\right)}{\frac{3\left(1-\omega^2\right)}{\left(1-\omega^2\right)^2+\left(0.125\omega\right)^2}}\) Which simplifies to: \(\tan \delta = \frac{\gamma \omega}{m\left(\omega_{0}^{2}-\omega^{2}\right)}\) This is the required expression for the tangent of the phase angle.

Plot the phase delta as a function of the forcing frequency omega

Now, we can plot the phase function \(\delta(\omega)\) as a function of the forcing frequency \(\omega\). Since we have the expression for the tangent of the phase angle in terms of omega, we can use any software like Matlab, Python, or a graphing calculator to plot the phase angle by first finding the tangent inverse of the obtained expression: \(\delta(\omega) = \arctan\left(\frac{\gamma \omega}{m\left(\omega_{0}^{2}-\omega^{2}\right)}\right)\) The obtained plot will show how the phase angle of the forced response changes as a function of the forcing frequency \(\omega\).

Key concepts

Phase Angle

When studying the forced response of differential equations, a critical concept is the phase angle, represented by \(\delta\). It indicates the offset between the input, or forcing function, and the system's output, or response. Specifically, in the context of oscillatory systems, the phase angle tells us how much the response waveform lags or leads the input.
In solving the exercise, the equation provided gives us the tangent of the phase angle, which relates the damping coefficient (\(\gamma\)), the forcing frequency (\(\omega\)), and the characteristics of the mass-spring system (mass \(m\) and natural frequency \(\omega_0\)). By taking the arctan of the given expression, we obtain the phase angle\[\delta = \arctan\left(\frac{\gamma \omega}{m(\omega_{0}^{2}-\omega^{2})}\right)\]This phase angle helps us understand the timing difference between the peak of the external force and the peak response of the system at any given frequency. An important observation is that the phase angle will vary with changes in the forcing frequency, which can be illustrated through a plot, as suggested in the exercise.

Non-homogeneous Linear Second-order Differential Equation

The exercise explores the dynamics captured by a non-homogeneous linear second-order differential equation. Such an equation describes a wide range of physical systems, from mechanical to electrical, where the output depends on both the system's inherent properties and external influences or forces.
In the general form, the equation looks like this:\[u''(t) + p(t)u'(t) + q(t)u(t) = f(t)\]Here, \(u(t)\) is the unknown function (representing system response), and \(f(t)\) represents the forcing function (external influence). The functions \(p(t)\) and \(q(t)\) are known coefficients that may depend on time. In the specific problem we're addressing, the equation has constant coefficients and a sinusoidal forcing term:\[u'' + 0.125u' + u = 3 \cos(\omega t)\]The solution to such an equation involves finding a function that satisfies both the inherent properties of the system (complementary function) and the external influences (particular solution). The exercise steps through solving for these components to articulate the system behavior under forced conditions.

Forcing Frequency

The forcing frequency, denoted by \(\omega\), is the frequency at which an external periodic force is applied to a system. In the context of differential equations, this external force drives the system away from its natural behavior and can induce resonant phenomena if the forcing frequency is close to the system's natural frequency.
In our example, the cosine function \(3 \cos(\omega t)\) represents a harmonic force with a forcing frequency \(\omega\). This frequency can greatly impact the system's response, and plotting the phase angle versus \(\omega\), as called for in part (b) of the exercise, would showcase how the system's phase lag varies across different frequencies, highlighting regions of resonance and stability.
Understanding the influence of forcing frequency on a system's response is paramount in engineering and physics, allowing for the prediction and control of systems under different operating conditions.

Complementary Function and Particular Solution

The solution to a non-homogeneous linear second-order differential equation is composed of two parts: the complementary function (\(u_c(t)\)) and the particular solution (\(u_p(t)\)).

Complementary Function (\(u_c(t)\))

The complementary function satisfies the associated homogeneous equation (the equation with zero on the right-hand side). It describes the system's behavior with no external forcing and is determined by the system's natural properties, like mass and spring constant in mechanical systems. Its form will depend on the characteristic equation derived from the differential equation.

Particular Solution (\(u_p(t)\))

The particular solution, on the other hand, specifically deals with the non-homogeneous part of the equation—namely, the external force. Finding the particular solution often involves a method of undetermined coefficients or variation of parameters, where a specific form is guessed based on the nature of \(f(t)\) and then determined through substitution and solving.
In our problem, we've seen a trial solution for \(u_p(t)\) that matches the form of the forcing function, with undetermined coefficients to be solved. Once both \(u_c(t)\) and \(u_p(t)\) are known, they're combined to form the complete solution. Through this process, we can understand how both the internal properties of the system and the external forces contribute to the total response.