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Solve the initial value problem and use the result of exercise 21 to find the amplitude and phase shift of the solution. $$y^{\prime \prime}+4 y=0, y(0)=1, y^{\prime}(0)=-2$$

Short Answer

Expert verified
The amplitude and phase shift of the solution are \(sqrt(2)\) and \(-\pi/4\), respectively

Step by step solution

01

Solve the general solution

The general solution of the second order differential equation \(y^{\prime \prime}+4 y=0\) is \(y(x) = C1*cos(2x) + C2*sin(2x)\)
02

Determine the constants C1 and C2

Using the initial condition \(y(0)=1\), we get \(C1 = 1\). From the initial condition \(y^{\prime}(0)=-2\), we find the derivative \(y^{\prime}(x)=-2*C1*sin(2x) + 2*C2*cos(2x)\). At \(x=0\), this becomes \(y^{\prime}(0) =2*C2 = -2\), so \(C2 = -1\)
03

Identify amplitude and phase shift

The amplitude A and the phase shift Φ are given by \(A = sqrt(C1^2 + C2^2) = sqrt(1 + 1) = sqrt(2)\) and \(\Ph=\arctan(\frac{-1}{1})= -\frac{\pi}{4}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Second Order Differential Equations
Second order differential equations are a type of differential equation containing derivatives up to the second order, but no higher. These equations are fundamental in physics and engineering, often representing wave motion, oscillations, and other dynamic systems. The general form for a linear second order differential equation is \(y'' + p(x)y' + q(x)y = g(x)\), where \(p(x)\), \(q(x)\), and \(g(x)\) are functions of \(x\).

When \(g(x) = 0\), the equation is homogeneous and can typically be solved by finding characteristic roots or applying known formulas, especially when coefficients are constant. The solution involves a combination of functions — often sines and cosines for equations with constant coefficients, representing periodic behavior. The crucial part of solving such an equation includes finding the general solution first, followed by applying initial conditions to determine specific constants and thus obtaining the particular solution that satisfies the initial value problem.

In the example provided, \(y''+4y=0\) is a homogeneous second order differential equation with constant coefficients. The general solution derived from the associated characteristic equation reveals a combination of sine and cosine functions. This is typical for equations representing harmonic motion, where solutions describe waves or oscillating systems.
Amplitude and Phase Shift in Oscillations
In the context of oscillating systems described by second order differential equations, amplitude and phase shift are critical parameters that define the nature of the oscillation. The amplitude is a measure of the maximum displacement from the equilibrium position, and the phase shift indicates how far the wave is offset from a reference position.

The amplitude can be found directly from the solution to the differential equation once the constants have been calculated. It is the distance from the centerline to a peak (or trough), and mathematically, it is the square root of the sum of the squares of the coefficients of the sine and cosine terms in the solution (denoted as \(A\) in the solution).

Phase shift, on the other hand, is a little more abstract. It is determined by the argument of the trigonometric functions in the solution and represents a horizontal shift of the wave along the \(x\)-axis. In the solution given, the phase shift \(\Phi\) is found using the arctangent of the ratio of the coefficients, providing the amount by which the sine and cosine functions in the equation are horizontally shifted. Understanding amplitude and phase shift is essential when analyzing the behavior of physical systems such as springs or electrical circuits.
Solving Differential Equations Step by Step
Solving differential equations often involves a step-by-step approach, starting with identifying the type of equation and then proceeding with finding its general solution. Once the general solution is acquired, if it's an initial value problem, the next step is to apply initial conditions to find the particular constants that will satisfy the given conditions.

This systematic process can be seen in the example provided. The initial step was finding the general solution for the homogeneous second order differential equation. This was achieved using the characteristic equation, resulting in a solution that involved a linear combination of cosines and sines. Following that, initial conditions were applied: \(y(0)=1\) and \(y'(0)=-2\). These conditions provided enough information to determine the unique values of the constants (denoted as \(C1\) and \(C2\) in the solution).

Lastly, the solution's amplitude and phase shift were calculated. These values are crucial in understanding the specifics of the oscillatory behavior described by the differential equation. By concluding this step-by-step approach, the distinct oscillatory motion defined by the initial conditions is thoroughly explained.

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Most popular questions from this chapter

In exercise \(10 .\) we sketched the Bode plot of the gain as a function of frequency. The other Bode plot, of phase shift as a function of frequency, is considered here. First, recall that in the general relationship \(a \sin \omega t+b \cos \omega t=B \sin (\omega t+\theta)\) we have \(a=B \cos \theta\) and \(b=B \sin \theta .\) We can "solve" for \(\theta\) as \(\cos ^{-1}\left(\frac{a}{B}\right)\) or \(\sin ^{-1}\left(\frac{b}{B}\right)\) or \(\tan ^{-1}\left(\frac{b}{a}\right) .\) In exercise 10 we have \(a=\frac{\left(5-\omega^{2}\right) A}{\left(5-\omega^{2}\right)^{2}+4 \omega^{2}}\) and \(b=\frac{-2 \omega A}{\left(5-\omega^{2}\right)^{2}+4 \omega^{2}},\) so that \(B=\sqrt{a^{2}+b^{2}}=\frac{A}{\left(5-\omega^{2}\right)^{2}+4 \omega^{2}} .\) For the frequencies \(\omega>0,\) this tells us that \(\sin \theta<0,\) so that \(\theta\) is in quadrant III or IV. Explain why the functions \(\sin ^{-1}\left(\frac{b}{B}\right)\) and \(\tan ^{-1}\left(\frac{b}{a}\right)\) are not convenient for this range of angles. However, \(-\cos ^{-1}\left(\frac{a}{B}\right)\) gives the correct quadrants. Show that \(\theta=-\cos ^{-1}\left(\frac{5-\omega^{2}}{\sqrt{\left(5-\omega^{2}\right)^{2}+(2 \omega)^{2}}}\right)\) and sketch the Bode plot.

Solve the initial value problem. $$y^{\prime \prime}+y^{\prime}-2 y=0, y(0)=3, y^{\prime}(0)=0$$

For the general equation \(a y^{\prime \prime}+b y^{\prime}+c y=0,\) suppose that there is a repeated root \(r_{1}<0\) of the characteristic equation. Show that \(\lim t e^{n t}=0\) and thus, the solution \(y(t) \rightarrow 0\) as \(t \rightarrow \infty\)

Involve frequency response curves and Bode plots. Suppose that the charge in a circuit satisfies the equation \(x^{\prime \prime}(t)+2 x^{\prime}(t)+5 x(t)=A_{1} \sin \omega t\) for constants \(A_{1}\) and \(\omega\) Find the steady-state solution and rewrite it in the form \(A_{2} \sin (\omega t+\delta),\) where \(A_{2}=\frac{A_{1}}{\sqrt{\left(5-\omega^{2}\right)^{2}+4 \omega^{2}}} .\) The ratio \(\frac{A_{2}}{A_{1}}\) is called the gain of the circuit. Notice that it is independent of the actual value of \(A_{1}\).

A spring is stretched 8 inches by a 32 -pound weight. The damping constant equals \(0.4 .\) The weight is released with a downward velocity of \(3 \mathrm{ft} / \mathrm{s}\). Find an equation for the position of the spring at any time \(t\) and graph the position function.

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