6.9: Discontinuous Forcing
- Page ID
- 388
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In the last section we looked at the Heaviside function its Laplace transform. Now we will use this tool to solve differential equations.
A 1 kg bar is attached to a spring with spring constant 5 and damping constant 2. It is pulled down 3 inches from it equilibrium position and the released. After one second a constant force of 10 Newtons is exerted on the bar. The force remains turned on indefinitely. Determine the equation of motion of the bar.
Solution
We have the differential equation
\[ y'' + 2y' + 5y = 10u_1(t)\]
with
\[ y(0) = 3 \]
\[ y'(0) = 0 .\]
We solve this by the method of Laplace transforms. We have
\[ \begin{align} L\{y''\} + 2L\{y'\} + 5L\{y\} &= 10L\{u_1(t)\} \\ \implies s^2L\{y\} - 3s - 0 + 2(sL\{y\} - 3) + 5L\{y\} &= 10e^{-s} /s \\ \implies (s^2 + 2s + 5)L\{y\} - 3s - 6 &= 10e^{-s} /s \\ \implies L\{y\} &= \dfrac{3s + 6 }{s^2 + 2s + 5} + \dfrac{10 }{s(s^2 + 2s + 5)} . \end{align} \]
We use partial fractions on the second term
\[ \dfrac{10}{s(s^2 + 2s + 4)} = \dfrac{A}{s} + \dfrac{Bs+C}{s^2+2s+5} \]
\[ A(s^2 + 2s + 5) + (Bs + C)s = 10 \]
\[ (A + B)s^2 + (2A + C)s + 5A = 10 \]
The constant term gives
\( A =2 \)
Thus
\( B = -2 \) \(C = -4 \)
We can complete the square
\[ s^2 + 2s + 5 = (s+1)^2 +4 \]
Putting all the algebra together we get
\[ \begin{align*} L\{y\} &= \dfrac{3s + 6}{(s + 1)^2 + 4} + e^{-s}\dfrac{2}{s} + e^{-s}\dfrac{-2s - 4}{ (s + 1)^2 + 4 } \\ &= 3\dfrac{s + 1 + 1}{(s + 1)^2 + 4} + e^{-s}\dfrac{2}{s} - 2e^{-s}\dfrac{(s + 1) + 1 }{(s + 1)^2 + 4 } \\ &= 3\dfrac{s + 1}{(s + 1)^2 + 4} + 3/2 \dfrac{2}{(s + 1)^2 + 4} + e^{-s}\dfrac{2}{s} - 2e^{-s}\dfrac{(s + 1) }{(s + 1)^2 + 4 } - e^{-s}\dfrac{2 }{(s + 1)^2 + 4 } \end{align*}\]
Now use the table to take the inverse Laplace transform to get
\[ y = 3e^{-t}\cos(2t) + 3/2 e^{-t -1}\sin(2t) + 2u_1 - 2u_1(t)e^{-t +1}\cos(2t - 2) - u_1(t)e^{-t +1} \sin(2t - 2). \]
Below is the graph
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.