amplitude and phase shift

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle u'' + 2u' + 6u = 15\cos 3t\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle u'' + 2u' + 6u = 15\cos 3t\)

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle u'' + 2u' + 6u = 15\cos 3t\)

We start by solving this homogeneous differential equation:

\(\displaystyle u'' + 2u' + 6u = 0\)

The complementary solution is:

\(\displaystyle u(t) = c_1e^{-t}\cos \sqrt{5}t + c_2e^{-t}\sin \sqrt{5}t\)

 

[logistic_guy]

The guess for the particular solution is:

\(\displaystyle u_p = A\cos 3t + B\sin 3t\)

Substituting this carefully in the original differential equation will give you:

\(\displaystyle A = -1\)
\(\displaystyle B = 2\)

Then, the general solution to the original differential equation is:

\(\displaystyle u(t) = c_1e^{-t}\cos \sqrt{5}t + c_2e^{-t}\sin \sqrt{5}t - \cos 3t + 2\sin 3t\)

 

[logistic_guy]

\(\displaystyle u(t) = c_1e^{-t}\cos \sqrt{5}t + c_2e^{-t}\sin \sqrt{5}t - \cos 3t + 2\sin 3t\)

I wanna analyze this solution further.

I am interested in the steady state solution. In other words, as \(\displaystyle t\) goes to \(\displaystyle \infty\), the first and the second terms of the solution vanish.

Then, the steady state solution is:

\(\displaystyle u(t) = -\cos 3t + 2\sin 3t\)

 

[logistic_guy]

To find the amplitude and phase shift, we use the following identity.

\(\displaystyle A\cos t + B\sin t = R\sin(t + \alpha)\)

where \(\displaystyle R = \sqrt{A^2 + B^2}\) and \(\displaystyle \alpha = \tan^{-1}\frac{A}{B}\)

Then,

\(\displaystyle u(t) = -\cos 3t + 2\sin 3t = \sqrt{(-1)^2 + (2)^2}\sin\left(3t + \tan^{-1}\frac{-1}{2}\right)\)


\(\displaystyle u(t) = \sqrt{5}\sin\left(3t - 26.57^{\circ}\right)\)

 

[logistic_guy]

\(\displaystyle u(t) = \sqrt{5}\sin\left(3t - 26.57^{\circ}\right)\)

Then,

The \(\displaystyle \bold{amplitude}\) is:

\(\displaystyle \textcolor{blue}{\sqrt{5}}\)

And

The \(\displaystyle \bold{phase \ shift}\) is:

\(\displaystyle \textcolor{red}{-26.57^{\circ}}\)